If the velocity of upward projection is increased, a stage will be reached when the velocity given to the body is such that, the kinetic energy of the body is sufficient to overcome the gravitational influence of the earth. This velocity is known as escape velocity.

The escape velocity of a body which is at rest on the earth’s surface Is defined as that minimum velocity with which It should be projected from the surface of the earth so that it escapes from the earth’s gravitational influence.

Escape the velocity of a body does not depend on the direction in which the body is thrown from the surface of the planet.

Expression for Escape Velocity of a Satellite:

Consider a body of mass ‘m’ which is at rest on the surface of the earth.  Let M be the mass of the earth and R be the radius of the earth.  Then the binding energy of the body on the surface of the earth is given by

Where G is Universal gravitational constant.

Let V_e be the escape velocity, then the kinetic energy given to the body is given by

It means Satellite should be given this much kinetic energy so that it can go out of earth’s gravitational influence.

K.E. = B.E.

This is an expression for the escape velocity of a satellite on the surface of the earth.

This equation shows that the escape velocity of a satellite is independent of the mass of the satellite (as term ‘m’ is absent).  The escape velocity is the same for all bodies from the given planet.  Escape velocity depends on the mass of the planet and its radius. For earth escape velocity is 11.2 km/s.

Expression in terms of Acceleration Due to Gravity:

We know that   GM = R²g

This is the expression for the escape velocity of a satellite in terms of acceleration due to gravity

Factors Affecting Escape Velocity:

• The escape velocity of a body is directly proportional to the square root of mass (M) of the planet (earth) around which the satellite is orbiting.
• The escape velocity of a body is inversely proportional to the square root of the radius of the Planet
• The equation does not contain the term, ‘m’ which shows that the critical velocity is independent of the mass of the satellite.
• The escape velocity of a body is independent of the direction of projection.

Relation Between Escape Velocity and Critical Velocity of a Satellite Moving Very Close to the Earth’s Surface:

For an orbiting satellite, critical velocity is given by

Where G = Universal gravitational constant

M = the mass of the earth

R = the radius of the earth

h = height of the satellite above the earth’s surface.

For a satellite orbiting very close to the earth, h can be neglected as h < < R. Hence it can be neglected. Therefore the critical velocity of the satellite orbiting very close to the earth’s surface.

The escape velocity of a satellite oh the surface of the earth is given by

Dividing equation (1) by (2)

 Thus the escape velocity of a body from the surface of the planet (earth) is √2  times the critical velocity of the body when it is orbiting close to the planet’s (earth’s) surface.

Expression in Terms of Density of Material of a Planet (earth):

The escape velocity of a satellite on the surface of the planet is given by

Where G = Universal gravitational constant

M = the mass of the Planet

R = the radius of the Planet

Let d    = density of the material of the planet

Now, Mass of Planet   = Volume of Planet x Density

This is an expression for escape velocity in terms of the density of the material of the planet.

Numerical Problems:

Example – 01:

The radius of earth is 6400 km, calculate the velocity with which a body should be projected so as to escape earth’s gravitational influence. Does the escape velocity depend upon the direction in which the body is projected? g =9.8 m/s².

Given: radius of earth = R = 6400 km = 6.4 x 10⁶ m, Acceleration due to gravity = 9.8 m/s².

To Find: v_e =?

Solution:

Ans: Thus the body should be thrown with a speed of 11.2 km/s. We are supplying kinetic energy to the body by throwing it. Hence it is independent of the direction of the throw.

Example – 02:

Calculate the escape velocity on the surface of the earth if the radius of the earth = 6400 km, G = 6.67 x 10^-11 Nm² /kg², and Density of the earth is 5500 kg /m³.

Given: Radius of earth = R = 6400 km = 6.4 x 10⁶ m, G = 6.67 x 10^-11 Nm² /kg². d = 5500 kg /m3, density = d = 5500 kg /m³.

To Find:  v_e =?

Solution:

Ans: Escape velocity is 11.2 km/s

Example – 03:

Calculate the escape velocity on the surface of the planet having radius 1100 km and acceleration due to gravity on the surface of the planet 1.6 m/s².

Given: radius of planet = R = 1100 km = 1.1 x 10⁶ m, Acceleration due to gravity = 1.6 m/s².

To Find: v_e =?

Solution:

Ans: Escape velocity on the planet is 1.876 km/s

Example – 04:

Calculate the escape velocity on the surface of the planet having radius 2000 km and acceleration due to gravity on the surface of the planet 2.5 m/s².

Given: radius of planet = R = 2000 km = 2 x 10⁶ m, Acceleration due to gravity = 2.5 m/s².

To Find: v_e =?

Solution:

Ans: Escape velocity on the planet is 3.162 km/s

Example – 05:

A satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the earth. Find its height above the earth’s surface.

Given: v_c =1/2 v_e

To Find: height of satellite above the surface of earth = h =?

Solution:

Ans: The height above the earth’s surface is ‘R’.

Example – 06:

Taking the mass of moon as 7.35 x 10²² kg and radius as 1750 km and G = 6.67 x 10^-11 Nm² /kg², find g at the surface of the moon and the escape velocity of a body from the surface of the moon.

Given: Mass of moon = M = 7.35 x 10²² kg, Radius of moon = R = 1750 km = 1.750 x 10⁶ m, G = 6.67 x 10^-11 Nm² /kg²,

To find: acceleration due to gravity =? v_e = ?

Solution:

Ans: Acceleration on the surface of moon is 1.6 m/s², Escape velocity on the planet is 2.367 km/s

Example – 07:

A satellite is revolving around the earth in a circular orbit of radius 7000 km. Calculate its period given that the escape velocity from the earth’s surface is 11.2 km/s and g = 9.8 ms/s²

Given: radius of orbit = r = 7000 km = 7 x 10⁶ m, g = 9.8 ms/s², escape velocity = v_e = 11.2 km/s = 11.2 x 10³ m/s,

To find: Period = T =?

Solution:

Ans: Thus the period of the satellite is 5809 s or 1.61 hr

Example – 08:

The mass of the moon is 1/80th that of the earth and the diameter of the moon is 1/4th that of the earth. Given that the escape velocity from the earth’s surface 11.2 km/s, find that from the moon’s surface.

Given: mass of moon = 1/80 mass of earth i.e. M_M = 1/80M_E, Diameter of moon = 1/4 diameter of earth i.e. R_M = 1/4 R_E, V_eE = 11.2 km/s

To find:  V_eM=?

Solution:

Dividing equation (1) by (2) we have

Ans: Escape velocity on the surface of the moon is 2.504 km/s

Example – 09:

A planet A has a mass and radius twice that of planet B, find the ratio of the escape velocities from A & B

Given: M_A = 2 M_B, R_A = 2 R_B,

To find: the ratio of escape velocities =?

Solution:

Dividing equation (1) by (2) we have

Ans: The ratio of escapes velocities on the two planets is 1: 1

Example – 10:

The escape velocity from the earth’s surface is 11.2 km/s. If the mass of Jupiter is 318 times that of earth and its radius is 11.2 times that of earth, find the escape velocity from Jupiter’s surface.

Given: M_J = 318 M_E, R_J = 11.22 R_E, escape velocity on surface earth = v_eE = 11.2 km/s

To find:   v_eJ =?

Solution:

Dividing equation (1) by (2) we have

Ans: The escape velocity on the surface of Jupiter is 59.68 km/s

Previous Topic: Binding Energy of Satellite

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Science > Physics > Gravitation > Escape Velocity of a Body

The post Escape Velocity of a Body appeared first on The Fact Factor.

In this article, we shall study the concept of the binding energy of satellite and its significance.

The binding energy of a satellite can be defined as the minimum amount of energy required to be supplied to it in order to free the satellite from the gravitational influence of the planet (i.e. in order to take satellite from the orbit to a point at infinity).

Binding energy gives us an idea about the energy by which the satellite is bound to the planet. This binding energy will be used to overcome the gravitational force of attraction between the Satellite and the planet.

Expression for Binding Energy of a Satellite Orbiting Around the Earth:

Consider a satellite revolving around the earth in a circular orbit. Necessary centripetal force to keep the satellite orbiting in a stable circular orbit is provided by the force of gravitational attraction between the earth and the satellite.

When the satellite is orbiting around the earth it possesses two types of mechanical energies. The kinetic energy due to its orbital motion and the potential energy due to its position in the gravitational field of the earth.

Let, M  = the mass of the earth

R   = the radius of the earth

h   = the height of the satellite above  the surface of the earth

v_c  =  the critical velocity of the satellite

m  = the mass of the satellite.

r   =  the radius of a circular orbit of the satellite = (R + h)

Kinetic Energy of satellite:

As the gravitational force is providing the necessary centripetal force required for circular motion,

Now, Centripetal force = Gravitational force

Where G is Universal gravitational constant.

The kinetic energy of a satellite orbiting around the earth is given by

Potential Energy:

Now, the satellite is in the gravitational field of the earth. The gravitational potential at a point on the surface of the earth is given by

The potential energy of a body (satellite) Is given by

P.E. = The gravitational potential x the mass of a satellite

Total Energy of Satellite:

The total mechanical energy of the satellite In orbit is given by

T.E.   =  K.E. + P.E.

The negative sign indicates that the satellite is bound to the earth by attractive forces and cannot leave it on its own.  To move the satellite to infinity .we have to supply energy from outside the planet-satellite system.  This energy is known as the binding energy of a satellite.

Binding Energy of Satellite:

B.E.   =     (Energy of a Satellite at Infinity) – (The energy of the satellite in the orbit)

This is an expression for Binding Energy of a satellite orbiting around the earth In stable circular orbit. Numerically, Binding Energy is equal to the total energy of a satellite in the orbit.

Expression for Binding Energy of a Satellite Stationary on the Earth’s Surface:

Consider a satellite of mass ‘m’ which is at rest on the earth’s surface.  As the satellite is at rest, it will not possess any kinetic energy.  i.e. K.E. = 0.

Now, the satellite is in the gravitational field of the earth.  The gravitational potential at a point on the surface of the earth is given by

The potential energy of a body (satellite) Is given by

P.E. = The gravitational potential x the mass of a satellite

The total mechanical energy of the satellite on the surface of the earth is given by

T.E.   =  K.E. + P.E.

The negative sign indicates that the satellite is bound to the earth by attractive forces and cannot leave it on its own.  To move the satellite to infinity we have to supply energy from outside the planet-satellite system.  This energy is known as the binding energy of a satellite.

Binding Energy:

B.E.   =     (Energy of a Satellite at Infinity) – (The energy of the satellite in the orbit)

This is an expression for the binding energy of a satellite stationary on the earth’s surface.

Numerical Problems on Binding Energy:

Example – 01:

Two satellites A and B are moving in circular orbits of radii 3R and 5R respectively around the same planet. If the masses of satellites are in the ratio of 2:1, compute their critical velocities and binding energies and periods of revolution.

Given:  radius orbit of satellite A = r₁ = 3R, radius orbit of satellite B = r₂ = 5R, Ratio of masses of satellite m₁: m₂ = 2 : 1

To find:  ratio of critical velocities = v₁ : v₂ = ?, ratio of binding energies = B.E.₁ : B.E.₂ = ? , ratio of time periods = T₁:T₂ =?

Solution:

Ans: The ratio of critical velocities is √5: √3, The ratio of binding energy is 10 : 3, The ratio of periods is 0.465 : 1

Example – 02:

Calculate the work done in moving a body of mass 1000 kg from a height of 2R to a height 3R above the surface of the earth. Mass of the earth = 6 x 10²⁴ kg; Radius of earth = 6400 km, G = 6.67 x 10^-11 Nm² /kg².

Given: initial height = h₁ = 2R, final height = h₂ = 3R, mass of satellite = m = 1000 kg, Mass of earth = M = 6 x 10²⁴  kg; radius of earth = R = 6400 km = 6.4 x 10⁶ m, G = 6.67 x 10^-11 Nm² /kg².

To Find: work done = W =?

Solution:

r₁ = R + h₁ = R + 2R = 3R,

r₂ = R + h₂ = R + 3R = 4R

Work done = Change in B.E.

Thus, W = B.E.₁ – B.E.₂

Ans: The work done is 2.615 x 10⁹ J.

Example – 03:

What is the binding energy of a satellite of mass 2000 kg moving in a circular orbit around the earth close to its surface and at a height of 600 km? G = 6.67 x 10^-11 S.I. units; Radius of earth = 6400 km; mass of earth = 6 x 10²⁴ kg.

Given: mass of satellite = m = 2000 kg, G = 6.67 x 10^-11 S.I. units; R = 6400 km = 6.4 x 10⁶ m; Mass of earth = M = 6 x 10²⁴ kg

To Find: Binding energies =?

Solution:

For satellite orbiting very close to earth’s surface h₁ = 0

r₁ = R + h₁ = R + 0 = R

For second satellite h₂ = 600 km

r₁   = R + h₁ = 6400 + 600 = 7000 km = =  7 x 10⁶ m;

Ans: Binding energy of satellite orbiting very close to the earth’s surface is 6.25 x 10¹⁰ J and binding energy of satellite orbiting at height 600 km from the surface of the earth is 5.72 x 10¹⁰ J.

Example – 04:

What is the (i) KE (ii) PE (iii) total energy and (iv) binding energy of an artificial satellite of mass 100 kg orbiting at a height of 1600 km above the surface of the earth? Mass of the earth = 6 x 1024 kg; Radius of earth = 6400 km, G = 6.67 x 10-11 Nm2 /kg2.

Given: mass of satellite = m = 100 kg, Mass of earth = M = 6 x 10²⁴ kg; radius of earth = R = 6400 km = 6.4 x 10⁶ m, G = 6.67 x 10^-11 Nm² /kg².

To Find: K.E. =? P.E. = ?, T.E. = ?, B.E. = ?,

Solution:

r = R + h = 6400 + 1600 = 8000 km  = 8 x 10⁶ m;

Binding energy = 2.5 x 10⁹ J

Now total energy = – B.E. = – 2.5 x 10⁹ J

Now potential energy = – 2 x B.E. = – 2 x 2.5 x 10⁹ = – 5 x 10⁹ J

Kinetic energy = B.E. = 2.5 x 10⁹ J

Ans: The kinetic energy of satellite = 2.5 x 10⁹ J, Potential energy of satellite = – 5 x 10⁹ J, Total energy of satellite = – 2.5 x 10⁹ J, Binding energy of satellite =  2.5  x 10⁹ J

Example – 05:

Binding energy of a satellite is 4 x 10⁸ J. Calculate its KE and PE.

Given: B.E = 4 x 10⁸ J

To Find: K.E. =? P.E. = ?,

Solution:

Now Potential Energy = -2 x B.E. = -2 x 4 x 10⁸ = – 8 x 10⁸ J

Kinetic energy = B.E. = 4 x 10⁸ J

Ans: Potential energy of satellite = – 8 x 10⁸ J, the kinetic energy of satellite = 4 x 10⁸ J

Example – 06:

What is the binding energy of a satellite of mass 80 kg moving in a circular orbit around the earth close to its surface and at a height of 1600 km? G = 6.67 x 10^-11 S.I. units; Radius of earth = 6400 km; mass of earth = 6 x 10²⁴ kg.

Given: mass of satellite = m = 80 kg, G = 6.67 x 10^-11 S.I. units; R = 6400 km = 6.4 x 106 m; Mass of earth = M = 6 x 10²⁴ kg; radius of earth = R = 6400 km = 6.4 x 10⁶ m, G = 6.67 x 10^-11 Nm² /kg².

To Find: B.E. =?

Solution:

For satellite orbiting very close to earth’s surface h₁ = 0

r₁ = R + h₁ = R + 0 = R

For second satellite h₂ = 1600 km

r₁   = R + h₁ = 6400 + 1600 = 8000 km = = 8 x 10⁶ m;

Ans: The binding energy of satellite orbiting very close to the earth’s surface is 2.5 x 10⁹ J and binding energy of satellite orbiting at height 600 km from the surface of the earth is 2 x 10⁹ J.

Example – 06:

What is the (1) KE (2) PE (3) total energy and (4) binding energy of an artificial satellite of mass 1000 kg orbiting at a height of 3600 km above the surface of the earth? Mass of the earth = 6 x 10²⁴ kg; Radius of earth = 6400 km, G= 6.67 x 10^-11 Nm² /kg²

Given: mass of satellite = m = 1000 kg, height of satellite above the surface of earth = 3600 km, Mass of earth = M = 6 x 10²⁴ kg; radius of earth = R = 6400 km = 6.4 x 10⁶ m, G = 6.67 x 10^-11 Nm² /kg².

To Find: K.E. =? P.E. = ?, T.E. = ?, B.E. = ?,

Solution:

r = R + h = 6400 + 3600 = 10000 km = 10⁴ m = 10⁷  m;

Ans: Kinetic energy of satellite = 2 x 10⁹ J, Potential energy of satellite = – 4 x 10⁹ J, Total energy of satellite = – 2 x 10⁹ J, Binding energy of satellite = 2 x 10⁹ J

Example – 07:

What is the binding energy of an artificial satellite of mass 1000 kg orbiting a) at a height of 500 km above the surface of the earth and b) close to the earth’s surface? Mass of earth = M = 6 x 10²⁴ kg; radius of earth = R = 6400 km = 6.4 x 10⁶ m, G = 6.67 x 10^-11 Nm² /kg².

Given: mass of satellite = 1000 kg, Mass of earth = M = 6 x 10²⁴ kg; radius of earth = R = 6400 km = 6.4 x 10⁶ m, G = 6.67 x 10^-11 Nm² /kg².

To Find: B.E. =?

Solution:

For first case h₁ = 5600 km

r₁   = R + h₁ = 6400 + 500 = 6900 km = = 6.9 x 10⁶ m;

For satellite orbiting very close to erath’s surface h₂ = 0

r₂   = R + h₂ = 6400 + 0 = 6400 km = = 6.4 x 10⁶ m;

Ans: The binding energy of satellite orbiting at height 500 km from the surface of the earth is 2.9 x 10¹⁰ J and Binding energy of satellite orbiting very close to the earth’s surface is 3.127 x 10¹⁰ J

Example – 08:

A playful astronaut releases a bowling ball of mass 500 g into circular orbit about an altitude of 600 km. What is the mechanical energy of the ball in its orbit. radius of the earth = 6400 km, mass of earth = 6 x 10²⁴ kg.

Given: mass of a ball = 500 g = 0.5 kg, height of satellite above the surface of the earth = 600 km, Radius of the earth = 6400 km, radius of orbit = 6400 + 600 = 7000 km = 7 x 10⁶ m, mass of earth = 6 x 10²⁴ kg.

To Find: Mechanical energy of ball = E_T = ?

Solution:

B.E. = GMm/ 2r = ( 6.67 x 10^-11 x 6 x 10²⁴ x 0.5)/(2 x 7 x 10⁶ )

B.E. = 1.43 x 10⁷ J

Now total energy of satellite = T.E. = – B.E. = – 1.43 x 10⁷ J

Ans: The mechanical energy of the ball in its orbit is – 1.43 x 10⁷ J

Example – 09:

Find the binding energy of a body of mass 50 kg at rest on the surface of the earth. Given: G = 6.67 x 10^-11 N m²/kg², R = 6400 km = 6.4 x 10⁶ m; M = 6 x 10²⁴ kg.

Given: the mass of body = m = 50 kg, Universal gravitational constant = G = 6.67 x 10^-11 N m²/kg², R = 6400 km = 6.4 x 10⁶ m; M = 6 x 10²⁴ kg.

To find: binding energy of satellite = B.E. =?

Solution:

The binding energy of the body at rest on the surface of the earth is given by

B.E. = GMm/R = ( 6.67 x 10^-11 x 6 x 10²⁴ x 50)/(6.4 x 10⁶ )

B.E. = 3.127 x 10⁹ J

Ans: Binding energy of the body is  3.127 x 10⁹ J

Previous Topic: Numerical Problems on Critical Velocity

Next Topic: Escape Velocity

Science > Physics > Gravitation > Binding Energy of Satellite

The post Binding Energy of Satellite appeared first on The Fact Factor.

Example – 18:

A satellite revolves around a planet of the mean density of 10⁴ kg/m³. If the radius of its orbit is only slightly greater than the radius of the planet, find the time of revolution of the satellite. G = 6.67 x 10^-11S. I.  units.

Given: Density of material of planet = ρ = 10⁴ kg/m³, G = 6.67 x 10^-11 N m²/kg² ;

To Find: Period of Satellite = T =?

Solutions:

The time period of a satellite orbiting around the earth is given by

If satellite orbiting very close to the earth  (i.e. h < < R) then h can be neglected. Then R + h = R

Ans: Time of revolution of the satellite is 1.044 hr

Example – 19:

What would be the speed of a satellite revolving in a circular orbit close to the earth’s surface? Given G = 6.67 x 10^-11 S.I. units; density of earth’s matter = 5500 kg/m³ and radius of earth = 6400 km.  Also, find its period.

Given: density of erth’s matter = ρ = 5500 kg/m³, radius of earth = R = 6400 km = 6.4 x 10⁶ m, G = 6.67 x 10^-11 N m²/kg² ;

To Find: orbital velocity = v_c = ?, period = T = ?

Solution:

If satellite orbiting very close to the earth  (i.e. h < < R) then h can be neglected. Then R + h = R

The time period of a satellite orbiting around the earth is given by

T = 2πR/v_c =  2 x 3.142 x 6400 /7.931 = 5071 s

T = 5071/3600 = 1.408 h

Ans: The speed of the satellite is 7.931 km/s and the time of revolution of the satellite is 1.408 h.

Example – 20:

The mean angular velocity of the earth around the sun is 1° per day. The distance from the sun to the earth is 1.5 x 10⁸ km. Determine the mass of the sun. G = 6.67 x 10^-11 S.I. units.

Given: Angular frequency = ω = 1° per day, radius of orbit = r = 1.5 x 10⁸ km = 1.5 x 10¹¹ m, G = 6.67 x 10^-11 S.I. units.

To find: the mass of sun = M =?

Solution:

∴ T = 2 x 24 x 60 x 60 x 180    s

∴ M = 2.062 x 10³⁰ kg

Ans: Mass of the sun is 2.062 x 10³⁰ kg

Example – 21:

Find the mass of the sun given that the radius of earth’s orbit is 1.5 x 108km. G = 6.67 x 10-11 S. I. units and the period of earth’s revolution around the sun is 365 days.

Given: Period of revolution = T= 365 days = 365 x 24 x 60 x 60 s, Radius of orbit = r = 1.5 x 10⁸ km = 1.5 x 10¹¹ m, G = 6.67 x 10^-11 S.I. units.

To find: the mass of sun =M =?

Solution:

∴ M = 2.062 x 10³⁰ kg

Ans: Mass of the sun is 2.062 x 10³⁰ kg

Example – 22:

If the moon revolves around the earth once in 27 days and 7 hours in an orbit which is 60 times the earth’s radius, find g at the earth’s surface. R= 6400 km.

Given: radius of earth = R = 6400 km = 6.4 x 10⁶ m, radius of orbit of moon = r = 60 R , Period of revolution of ,moon = 27 days 7 hours =  = 27 x 24 x 60 x 60+ 7 x 60 x 60 =6556060 s,

To find: acceleration due to gravity = g = ?

Solution:

Ans: g on the Earth’s surface = 9.81 m/s²

Example – 23:

Assuming that the moon describes a circular orbit of radius R about the earth in 27 days and that Titan describes a circular orbit of radius 3.2 R about Saturn in 16 days, compare the mass of Saturn and the earth.

Given: Period of Moon =T_M= 27 days, radius of orbit of moon = r_M = R, time period of titan = T_T =16 days, radius of orbit of titan = r_T = 3.2 R,

To Find:   Ratio of mass of saturn to the earth = M_S /M_E=?

Solution:

Ans: The ratio of the mass of Saturn to the earth is 93.3 :1

Example – 24:

A satellite of mass 1750 kg is moving around the earth in an orbit at a height of 2000 km from the surface of the Earth. Find its angular momentum given G = 6.67 x 10^-11 Nm² kg^-2 and mass of the  earth = 6 x  10²⁴ kg.

Given: h = 2000 km, hence radius of orbit = r = R + h = 6400 + 2000 = 8400 km = 8.4 x 10⁶ m, G = 6.67 x 10^-11 Nm² kg^-2 ; mass of satellite = m = 1750 kg, mass of the earth = M = 6 x  10²⁴ kg;

To find: Angular momentum = L= ?,

Solution:

Ans : The angular momentum of the satellite is 1.015x 10¹⁴ kg m² s^-1.

Example – 25:

The radii of the orbits of two satellites revolving around the earth are in the ratio 3:8. Compare their i) critical speed and ii) periods.

Given: ratio of radii of orbits r₁ :  r₂  =  3:8

To find: v_c1 : v_c2  =? , T₁ : T₂ = ?

Solution:

Ans : The ratio of their critical velocities is 1.633:1, The ratio of their period is 0.2296:1

Example – 26:

The distances of two planets from the sun are 10¹³ m 10¹² m respectively. Find the ratio of their periods and orbital speeds.

Given: r₁ = 10¹³ m ,  r₂ = 10¹² m

To find: T₁ : T₂ = ?, v_c1 : v_c2 = ? ,

Solution:

Ans: The ratio of their period is 31.62:1, The ratio of their critical velocities is 0.3162:1

Example – 27:

Two satellites X and Y are moving in circular orbits of radii r and 2r respectively around the same planet. What is the ratio of their critical velocities?

Given: r_X = r, r_Y = 2r.

To find: v_cX : v_cY  =?

Solution:

Ans: The ratio of their critical velocities is √2 : 1

Example – 28:

A communication satellite is at a height of 36400 km from the surface of the earth. What will be its new period when it is brought down to the height of 20000 km from the surface of the earth? The radius of the earth is 6400 km.

Given: Initial radius of orbit = r₁ = 36400 + 6400 = 42800 km, Final radius of orbit = r₂=   6400 + 20000 = 26400 km, Initial time period of satellite = T₁ =24 hr,

To Find:   New time period of satellite = T₂ =?

Solution:

By Keppler’s law

∴ T₂ = 11.62 h

Ans: The new period of satellite 11.62 hr

Example – 29:

A satellite at a height of 2725 km makes ten revolutions around the earth per day. What is the number of revolutions made per day by a satellite orbiting at a height of 560 km? Radius of earth = 6400 km.

Given: radius of orbit of earth = R = 6400 km, height of satellite above the surface of earth in first case = h₁ = 2725 km, height of satellite above the surface of earth in second case = h₂ = 560 km, number of revolutions of satellite in first case = n₁ = 10 per day

To Find: number of revolutions of satellite in second case = n₂ =?

Solution: 

r₁ = R + h₁ = 6400 + 2725 = 9125 km

r₂ = R + h₂ = 6400 +560 = 6960 km

Ans: The satellite will make 15 revolutions per day.

Example – 30:

India’s most powerful rocket (PSLV) projected a remote sensing 850 kg satellite into an orbit 620 km above the earth’s surface. Assuming that the orbit is circular, find its speed and calculate the number of complete revolutions it makes around the earth in one day. g = 9.8 m/s² and R = 6400 km.

Given: h = 620 km, radius of orbit of satellite = r = R + h = 6400 + 620 = 7020 km = 7.02 x 10⁶ m, g = 9.8 m/s², R = 6400 km  = 6.4 x 10⁶ m

To find: speed of satellite = v_c = ?, N = ?

Solution:

N = 14.81 revolutions per day

Ans: The speed of satellite = 7.561 km/s and number of revolutions of satellite per day = 14.81

Example – 31:

A satellite going in a circular orbit of radius 4 x 10⁴ km around the earth has a certain speed. If this satellite were to move around Mars with the same speed, what would be its orbital radius? The masses of earth and mars are in the ratio of 10:1 and their radii are in the ratio 2:1

Given: speeds are same v_cE = v_cM, radius of satellite around the earth = r_E = 4 x 10⁴ km, M_E:M_M = 10:1 and R_E:R_M = 2:1

To find: radius of the satellite around the mars =r_M =?

Solution:

Ans: Radius of the satellite around Mars is 4000 km.

Example – 32:

Two satellites orbiting around the earth have their initial speeds in the ratio 4 : 5. Compare their orbital radii.

Given: ratio of speeds v_c1 : v_c2  =  4 : 5

To find: r₁ :  r₂  = ?

Solution:

The critical velocity of a satellite orbiting around the earth is given by

Ans : The ratio of their radii is 25 : 16

Example – 33:

Compare the critical speeds of two satellites if the ratio of their periods is 8 : 1.

Given: T₁/T₂ = 8/1

To Find: Ratio of critical velocity = v_c1/v_c2 =?,

Solution:

From equations (1) and (2)

Ans: The ratio of their critical velocities is 1:2

Example – 34:

Calculate the height of communication satellite above the surface of the earth? Given G = 6.67 x 10^-11 N m²/kg²; radius of earth = 6400 km, Mass of the earth = 6 x 10²⁴ Kg

Given: For communication satellite, T = 24 hr = 24 x 60 x 60 s, Universal gravitational Constant = G = 6.67 x 10^-11 N m²/kg²; radius of earth = R = 6400 km = 6.4 x 10⁶ m, Mass of the earth = M = 6 x 10²⁴ Kg

To Find: height of communication satellite above the surface of the earth = h = ?

Solution:

The time period of satellite is given by

Now, r = R + h
Hence, h = r – R = – 6.37 x 106
= 35.95 x 106 m
= 35.95 x 103 x 103 m
= 35950 km
Ans: The height of satellite above the surface of the earth is 35950 km.

Previous Topic: More Problems on Critical Velocity

Next Topic: Binding Energy of Satellite

Science > Physics > Gravitation > Numerical Problems on Critical Velocity and Period of Satellite

The post Numerical Problems on Critical Velocity and Period of Satellite – 02 appeared first on The Fact Factor.

In this article, we shall study to solve problems to calculate time period and orbital speed of satellite.

Example – 01:

Calculate the speed and period of revolution of a satellite orbiting at a height of 700 km above the earth’s surface. Assume the orbit to be circular. Take the radius of the earth as 6400 km and g at the surface of the earth to be 9.8 m/s².

Given: height of satellite above the surface of earth = h = 700 km, Radius of earth = R = 6400 km, G = 6.67 x 10^-11 N m²/kg² ; g = 9.8 m/s².

To Find: speed of satellite = v_c = ?, Period = T = ?

Solution:

r = R + h = 6400 + 700 = 7100 km = 7.1 x 10⁶ m,

The speed of a satellite orbiting around the earth is given by

The time period of a satellite orbiting around the earth is given by

Ans: The speed of the satellite is 7.519 km/s and the period of revolution of the satellite is 5930 s.

Example – 02:

From the following data, calculate the period of revolution of the moon around the earth: Radius of earth = 6400 km; Distance of moon from the earth = 3.84 x 10⁵ km; g = 9.8 m/s².

Given: radius of earth = R = 6400 km = 6.4 x 10⁸ m, radius of orbit of moon = r = 3.84 x 10⁵ km = 3.84 x 10⁸ m, g = 9.8 m/s².

To find: Period of revolution of moon = T =?

Solution:

Ans: Period of the moon around the earth is 27.3 days

Example – 03:

A satellite is revolving around the earth in a circular orbit in the equatorial plane at a height of 35850 km. Find its period of revolution. What is the possible use of such a satellite? In what direction is such a satellite projected and why must it be in the equatorial plane? Given g = 9.81 m/s²; Radius of earth 6.37 x 10⁶ m

Given: Radius of orbit of the earth =R = 6.37 x 10⁶ m, height of satellite above the surface of the earth = h = 35850 km = 35850 x 10³ m = 35.85 x 10⁶ m, G = 6.67  10-11 N m²/kg²,  g = 9.8 m/s².

To Find: T =?

Solution:

r = R + h = 6.37 x 10⁶ m + 35.85 x 10⁶ m = 42.22 x 10⁶ m

Ans: Period of the satellite is 24 hr.

Such a satellite is called geosynchronous satellite and is used for communication, broadcasting and weather forecasting. Such a satellite moves in the same direction as that of the rotation of the earth and its orbit is in the equatorial plane. If this satellite is not in the equatorial plane, it will appear to move up and down of the equatorial plane and thus it will not be stationary w.r.t. an observer on the Earth.

Example – 04:

With what velocity should a satellite be launched from a height of 600 km above the surface of the earth so as to move in a circular path?. Given: G = 6.67 x 10^-11 N m²/kg²; radius of earth = 6.4 x 10⁶ m, Mass of the earth = 5.98 x 10²⁴ Kg.

Given: G = 6.67 x 10^-11 N m²/kg²; radius of earth = R = 6.4 x 10⁶ m = 6400 km, Mass of the earth = M = 5.98 x 10²⁴ Kg, height of satellite above the surface of earth  = 600 km, raius of orbit = 6400 + 600 = 7000 km = .7 x 10⁶ m

To find: critical velocity = v_c = ?

Solution:

The critical velocity of a satellite orbiting around the earth is given by

Ans: The velocity of the satellite for launching is 7.545 km/s.

Example – 05:

A satellite is revolving around the earth in a circular orbit at a distance of 10⁷ m from its centre. Find the speed of the satellite. Given G = 6.67 x 10^-11 N m²/kg², Mass of the earth = M =6 x 10²⁴ Kg,

Given: radius of orbit  = 10⁷ m, G = 6.67 x 10^-11 N m²/kg², Mass of the earth = M = 6 x 10²⁴ Kg,

To Find: speed of satellite = v_c =?

Solution:

The critical velocity of a satellite orbiting around the earth is given by

Ans: The speed of the satellite is 6.326 km/s.

Example – 06:

A body is raised to a height of 1600 km above the surface of the earth and projected horizontally with a velocity of 6 km/s. Will it revolve around the earth as a satellite? Given G = 6.67 x 10^-11 N m²/kg²; radius of earth = 6.4 x 10⁶ m, Mass of the earth = 6 x 10²⁴ Kg.

Given:  radius of earth = 6.4 x 10⁶ m = 6400 km, Mass of the earth = 6 x 10²⁴ Kg, height of satellite above the surface =  h = 1600 km, horizontal velocity given to the satellite = v = 6 km/s, r = R + h = 6400 + 1600 = 8000 km, r = 8 x 10⁶ m .

To find: the condition of orbiting of satellite = ?

Solution:

The critical velocity of a satellite orbiting around the earth is given by

The horizontal velocity given to satellite is 6 km/s which is less than the critical velocity of 7.073 km/s. Hence the satellite will not revolve around the earth in circular orbit but will fall back on the earth in a parabolic path.

Example – 07:

A body is raised to a height equal to the radius of the earth above the surface of the earth and projected horizontally with a velocity 7 km/s. Will it revolve around the earth as a satellite? If yes what is the nature of the orbit? Given G = 6.67 x 10^-11 N m²/kg²; radius of earth = 6.4 x 10⁶ m, Mass of the earth = 5.98 x 10²⁴ Kg.

Given:  radius of earth = 6.4 x 10⁶ m = 6400 km, Mass of the earth = 6 x 10²⁴ Kg, height of satellite above the surface =  h = R, horizontal velocity given to the satellite = v = 7 km/s, r = R + R = 2R = 2 x 6400 = 12800 km, r = 12.8 x 10⁶ m .

To find: the nature of the orbit of satellite =?

Solution:

The critical velocity of a satellite orbiting around the earth is given by

The horizontal velocity given to satellite is 7 km/s which is greater than the critical velocity 5.582 km/s and less than the escape velocity (11.2 km/s). Hence the satellite will revolve around the earth in an elliptical orbit.

Example – 08:

An artificial satellite is revolving around the earth in circular orbit at a height of 1000 km at a speed of 7364 m/s. Find its period of revolution if R = 6400 km

Given: Radius of earth =R = 6400 km = 6.4 x 10⁶ m, height of satellite above the surface of the earth = h = = 1000 km = 1.0 x 10⁶ m, radius of orbit of satellite = r = R + h = 6.4 x 10⁶  +  1.0 x 106 m = 7.4 x 10⁶  m, critical velocity = v_c =7364 m/s..

To Find: T =?

Solution:

Ans: The period of revolution of the satellite is 1.75 hr.

Example – 09:

Moon takes 27 days to complete one revolution around the earth. Calculate its linear velocity. If the distance between the earth and the moon is 3.8 x 10⁵ km.

Given: Radius of orbit of moon = r = 3.8 x 10⁵ km =3.8 x 10⁸ m, Period of moon = T = 27 days = 27 x 24 x 60 x 60 s.

To find: critical velocity = v_c =?

Solution:

Ans:  The linear velocity of the moon is 1.023 km/s

Example – 10:

Find the radius of the moon’s orbit around the earth assuming the orbit to be circular. Period of revolution of the moon around the earth = 27.3 days, g at the earth’s surface = 9.8 m/s². Radius of earth = 6400 km.

Given: Radius of earth = R = 6400 km = 6.4 x 10⁶ m, Time period = T = 27.3 days = 27.3 x 24 x 60 x 60, g = 9.8 m/s⁶.

To Find: radius of the orbit = r =?

Solution:

Ans: The radius of the moon’s orbit is 3.841 x 10⁵ km

Example – 11:

Venus is orbiting around the sun in 225 days. Calculate the orbital radius and speed of the planet.  Mass of sun is   2 x 10³⁰ kg, G = 6.67 x 10^-11 S.I. units.

Given: Mass of sun = M = 2 x 10³⁰ kg, Period of venus = T = 225 days = 225 x 24 x 60 x 60 s, G = 6.67 x 10^-11 S.I. units.

To Find: radius of orbit = r =? Orbital velocity = v_c =?

Solution:

Ans: Orbital radius of Venus = 1.085 x 10¹¹ m and its orbital speed = 3.506 x 10⁴ m/s

Example – 12 

An observer situated at the equator finds that a satellite is always overhead. What must be its distance from the centre of the earth? Given g at earth’s surface = 9.81 m/s²; radius of earth = 6.4 x 10⁶ m. What is the KE of such a satellite w.r.t. an observer on earth? What must be its height above the surface of the earth?

Given: Period of Earth = T = 24 hr = 24 x 60 x 60 s, Radius of the cearth = R = 6.4 x 10⁶ m, g = 9.8 m/s².

To Find: radius of the orbit = r =?

Solution:

Now, r = R + h

Hence, h = r – R = 42350 – 6400 = 35950 km

Ans: Radius of the orbit of the satellite is 42350 km. The height of the satellite above the surface of the earth is 35950 km.

As the satellite is stationary w.r.t. observer, the kinetic energy of satellite w.r.t. the observer is zero.

Example – 13:

Calculate the height of the communication satellite above the surface of the earth? Given G = 6.67 x 10^-11 N m²/kg²; radius of earth  = 6.4 x 10⁶ m, Mass of the earth = 6 x 10²⁴ Kg.

Given: Period of satellite = T = 24 hr = 24 x 60 x 60 s, G = 6.67 x 10^-11 N m²/kg²; radius of earth = R  = 6.4 x 10⁶ m, mass of earth = M = 6 x 10²⁴ Kg

To Find: height of satellite above the surface of the earth = h =?

Solution:

Now, r = R + h

Hence, h = r – R = 42.35 x 10⁶ – 6.4 x 10⁶ = 35.92 x 10⁶ m

h = 35920 x 10⁶  m = 35920 km

Ans: The height of satellite above the surface of the earth is 35920 km.

Example – 14:

A satellite makes ten revolutions per day around the earth. Find its distance from the earth assuming that the radius of the earth is 6400 km and g at the earth’s surface is 9.8 m/s².

Given: radius of earth = R = 6400 km = 6.4 x 10⁶ m, g = 9.8 m/s², Number of revolutions = 10 per day

To Find: radius of orbit of satellite = r =?

Solution:

T = 24/No. of revolutions per day = 24/10 = 2.4 hours

Ans: The radius of the orbit of the satellite is 9125 km

Example – 15:

A satellite is revolving around a planet in a circular orbit with a velocity of 8 km/s at a height where the acceleration due to gravity is 8 m/s². How high is the satellite from the planet’s surface? Radius of planet = 6000 km.

Given: velocity of satellite v_c = 8 km/s, R = 6400 km = 6.4 x 10⁶ m,  acceleration due to gravity at height = g_h = 8 m/s².

To Find: height of the satellite above the surface = h = ?,

The critical velocity of a satellite orbiting around the earth is given by

Now, r = R + h

Thus h = r – R = 8000 – 6400 = 1600 km

Ans: The height of the satellite above the surface of the earth is 1600 km.

Example – 16:

The critical velocity of a satellite is 5 km/s. Find the height of satellite measured from the surface of the earth given G = 6.67 x 10^-11 Nm² kg^-2; R=6400 km and M = 5.98 x 10²⁴ kg

Given: critical velocity = v_c = 5 km/s, 5 x 10³ m, radius of earth = R = 6400 km = 6.4 x 10⁶ m, G = 6.67 x 10^-11 Nm² kg^-2;  mass of earthM = 5.98 x 10²⁴ kg

To find: height of satellite above the surface of the earth = h = ?,

Solution:

The critical velocity of a satellite orbiting around the earth is given by

Thus h = r – R = 15950 – 6400 = 9550 km

Ans: The height of satellite above the surface of the earth is 9550 km.

Example – 17:

A satellite is revolving around a planet in a circular orbit with a velocity of 6.8 km/s. Find the height of the satellite from the planet’s surface and the period of its revolution. g = 9.8 m/s² , R = 6400 km.

Given: velocity of satellite = v_c = 6.8 km/s = 6.8 x 10³ m/s, R = 6400 km = 6.4 x 10⁶ m, g = 9.8 m/s².

To find: height of the satellite above the surface = h =?

Solution:

The critical velocity of a satellite orbiting around the earth is given by

r = 8681 km

r = R + h

Thus h = r – R = 8681 – 6400 = 2281 km

Now,

Ans: The height of the satellite above the surface of the earth is 2281 km and the period of the satellite is 8017 s

Previous Topic: Theory of Critical Velocity and Period of Satellite

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Science > Physics > Gravitation > Numerical Problems on Critical Velocity and Period of Satellite

The post Numerical Problems on Critical Velocity and Period of Satellite – 01 appeared first on The Fact Factor.

In this article, we shall derive expressions for the critical velocity and time period of a satellite in different forms.

Critical Velocity:

The constant horizontal velocity given to the satellite so as to put it into a stable circular orbit around the earth is called critical velocity and is denoted by Vc. It is also known as orbital speed or proper speed.

The Expression for Critical Velocity:

Let us consider a satellite of mass “m” orbiting at height “h” from the surface of earth around the earth with critical velocity V_c as shown in the diagram. Let M and R be the mass and radius of earth respectively. The radius ’r’ of the orbit is    r = R + h

The necessary centripetal force for the circular motion of satellite is provided by the gravitational attraction between the satellite and the earth.

Now, Centripetal force = Gravitational force

This is the expression for the critical velocity of a satellite orbiting around the earth at height h from the surface of the earth. The expression does not contain the term ‘m’. Hence we can conclude that the critical velocity of a satellite is independent of the mass of the satellite.

Expression for the Critical Velocity of Satellite Orbiting Very Close to Earth’s Surface:

If a satellite is orbiting very close to the earth’s surface i.e. h < < R then h may be neglected in comparison of R and Critical velocity may be given as

Expression for the Critical Velocity of Satellite in Terms of Acceleration Due to Gravity:

We have

Where g_h = acceleration due to gravity at height h from the surface of the earth.

This is an expression for the critical velocity of a satellite orbiting at height h in terms of acceleration due to gravity at that height.

If the satellite is orbiting very close to the earth’s surface then h < < R  and thus h can be neglected. Similarly near the surface of the earth g_h =  g i.e. the acceleration due to gravity on the surface of the earth.

This is an expression for the critical velocity of a satellite orbiting very close to the Earth’s surface.

Expression for the Critical Velocity of Satellite Orbiting Very Close to the Earth’s Surface in Terms of Density of the Material of the Earth / Planet:

The critical velocity of a satellite orbiting very close to the earth’s surface is given by

This is the expression for the critical velocity of a satellite orbiting very close to the earth’s surface in terms of density of the material of the earth/planet.

Factors Affecting Critical Velocity of Satellite:

• The critical velocity of the satellite is directly proportional to the square root of mass (M) of the planet (earth) around which the satellite is orbiting.
• It is inversely proportional to the square root of the radius of its orbit.
• The equation does not contain the term, ‘m’ which shows that the critical velocity is independent of the mass of the satellite.

Possibilities of Orbits of Satellites:

Depending upon Magnitude of Horizontal Velocity following four cases can arise for Motion of Satellite

Case – 1 (v_h < v_c): If the horizontal velocity imparted to the satellite is less than critical velocity Vc, then the satellite moves in a long elliptical orbit with the centre of the Earth as the further focus. If the point of projection is apogee and in the orbit, the satellite comes closer to the earth with its perigee point lying at 180o. If enters the earth’s atmosphere while coming towards perigee, it will lose energy and spiral down on the earth. Thus it will not complete the orbit. If it does not enter the atmosphere, it will continue to move in an elliptical orbit.

Case – 2 (v_h = v_c): If the horizontal velocity imparted to the satellite is exactly equal to the critical velocity Vc then satellite moves in a stable circular orbit with the earth as centre as shown in the diagram.

Case – 3 (V_e > V_h > V_c): If the horizontal velocity Vh is greater than critical velocity Vc but less than escape velocity Ve then satellite orbits in
the elliptical path around the earth with the centre of the earth as one of the foci (nearer focus) of the elliptical orbit as shown.

Case – 4 (V_h > V_e): If the horizontal velocity is greater than or equal to escape velocity Ve then satellite overcomes the gravitational
attraction and escapes into infinite space along a hyperbolic trajectory.

Period of a Satellite:

The time taken by the satellite to complete one revolution around the planet is known as the period of revolution of the satellite. It is denoted by ‘T’.  Its S.I. unit is second.

Expression for Period of a Satellite:

Let us consider a satellite of mass “m” orbiting at height “h” from the surface of earth around the earth with critical velocity V_c as shown in the diagram. Let M and R be the mass and radius of earth respectively. The radius ’r’ of the orbit is    r = R + h

The necessary centripetal force for the circular motion of satellite is provided by the gravitational attraction between the satellite and the earth.

Now, Centripetal force = Gravitational force

Now the period of satellite ‘T’ is given by

This is the expression for the time period of a satellite orbiting around the earth at height h from the surface of the earth.

Squaring both sides of the above equation, we get

For the given planet the quantity in the bracket is constant hence we can conclude that

T²  ∝  r³

Thus the square of the period of a satellite is directly proportional to the cube of the radius of Its orbit. (Kepler’s third law of planetary motion)

Expression for the Periodic Time in Terms of Acceleration Due to Gravity:

Let g_h be the acceleration due to gravity at a point on the orbit i.e. at a height h above the earth’s surface.

This is an expression for the period of a satellite orbiting at height h in terms of acceleration due to gravity at that height.

If satellite orbiting very close to the earth (i.e. h < < R) then h can be neglected. Then r = R and    g_h = g

This is an expression for the time period of a satellite orbiting very close to the earth’s surface.

Expression for the Periodic Time of Satellite Orbiting Very Close to the Earth’s Surface in Terms of Density of the Material of the Earth / Planet:

The time period of a satellite orbiting around the earth is given by

If satellite orbiting very close to the earth  (i.e. h < < R) then h can be neglected. Then R + h = R

This is the expression for the period of a satellite orbiting very close to the earth’s surface in terms of density of the material of the earth/planet.

Factors Affecting Period of Satellite:

• The square of the time period of the satellite is directly proportional to the cube of the radius of orbit (r) of the satellite
• The equation does not contain the term, ‘m’ which shows that the critical velocity is independent of the mass of the satellite.

Previous Topic: Artificial Satellites and Their Uses

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Science > Physics > Gravitation > Critical Velocity and Time Period of Satellite

The post Critical Velocity and Time Period of Satellite appeared first on The Fact Factor.

In this article, we shall study about satellites, their types and uses of artificial satellites

Satellites:

Any object that revolves around a given planet in circular orbit under the influence of planet’s gravitational force is called as a satellite.

Types of Satellites:

Satellites are of two types. viz natural satellites and artificial satellites (man-made objects orbiting around the earth/planet). The moon is the natural satellite of the earth. The Earth, the Venus, and Jupiter are the natural satellites of the sun. INSAT-B, INSAT-IC are artificial satellites of the earth.

Projection of Satellite:

To launch a satellite in an orbit around the earth multistage rocket is used.  The launching involves two steps. In the first step, the satellite is taken to the desired height and then in the second step, it is projected horizontally with the calculated speed in a definite direction. If the velocity is proper it starts revolving in a stable circular orbit.

To understand the launching of satellite let us consider a simple case of the use of a two-stage rocket.  The satellite is kept at the tip of the two-stage rocket.

Initially, the first stage of the rocket is ignited on the ground so that the rocket is raised to the desired height, the first stage is detached. Then the rocket is rotated by remote control to point it in the horizontal direction. Then the second stage is ignited so the rocket gets push in the horizontal direction and acquires certain horizontal velocity (Vh).  When the fuel is completely burnt second stage also gets detached and the satellite starts orbiting around the earth.

A Satellite is Placed Outside Earth’s Atmosphere:

The satellite orbiting very close to the earth’s surface has an orbital speed of about 8 km/s. As the height of the satellite from the surface of the earth increases its orbital velocity decreases. If the satellite is placed in the atmosphere, due to the high velocity of satellite and friction between the atmosphere and the satellite, large heat will be produced and the satellite will get burnt.

Types of Artificial Satellites:

Geostationary or Geosynchronous or Communication Satellite:

A communication satellite is an artificial satellite which revolves around the earth in a circular orbit in the equatorial plane such that,

• its direction of motion is the same as the direction of rotation of the ‘earth about its axis.
• its period is the same as the period of rotation of the earth, i.e. 24 hours.

When observed from the earth’s surface, this satellite appears stationary. Therefore, it is called a geostationary satellite. As its motion is synchronous with the rotational motion of the earth, it is called a geosynchronous satellite. The height of communication satellite above the surface of the earth is about 36,000 km.

e.g. INSAT series satellites, APPLE.

Notes:

• The orbit of the geosynchronous satellite is called geosynchronous orbit.
• It lies in an equatorial plane. i.e. the angle between geosynchronous orbit and equatorial plane is 0°.
• The radius of the geosynchronous orbit is about 42400 km.
• The satellite parking strip is an area over the equator is becoming congested with several hundreds of communication, weather, military and transmission satellites.

Uses of Communication Satellites:

• The communication satellites are used for sending microwave and TV signals from one place to another.

• The communication satellites are used for weather forecasting.

• The communication satellites are used for detecting water resource -locations and areas rich in ores.

• The communication satellites are used for spying In enemy countries i.e. It can be used for military purposes

Polar or Sun-synchronous Satellite:

A polar satellite is a low altitude satellite orbit around the earth in north-south orbit passing over the north pole and south pole. The orbit of the polar satellite is called the polar orbit. The polar orbit makes an angle of inclination of 90° with the equatorial plane. Polar satellites cross the equatorial plane at the same time daily.

The height of the polar satellite above the earth is about 500-800 km. Its time period is about 100 minute.

Advantages of Polar Satellite:

Geostationary satellites are fixed at one position w.r.t. the earth at height 36000 km above the Earth. Its long-range helps meteorologist to understand and analyze the weather. But to understand Earth’s atmosphere and changes in the atmosphere, the whole planet must be scanned periodically and most effectively. To this polar satellites are used.

Since its time period is about 100 minutes it crosses any altitude many times a day and its height h above the earth is about 500-800 km, a camera fixed on it can view only small strips of the earth in one orbit. Adjacent strips are viewed in the next orbit so that in effect the whole earth can be viewed strip by strip during the entire day. From the path shown in the figure, we can see that it covers almost all geographical area.

These satellites can view polar and equatorial regions at close distances with good resolution.

Information gathered from such satellites is extremely useful for remote sensing, meteorology as well as for environmental studies of the earth.

Uses of Polar Satellites:

• Information gathered from polar satellites is extremely useful for remote sensing, meteorology as well as for environmental studies of the earth.
• They are used for spying and surveillance.
• They are used for monitoring troop movements i.e. for military purpose.
• They are used to note land and sea temperature variation.
• They are also used to monitor the growth of crops.

Weightlessness in Satellite:

By the definition, the weight of a body is equal to the gravitational force with which the body is attracted towards the centre of the earth. When the astronaut is on the surface of the earth, his weight acts vertically downwards.  At the same time, the earth’s surface exerts an equal and opposite force of reaction on the astronaut.  Due to this force of reaction, the astronaut feels his weight.

When the astronaut is in an orbiting satellite, a gravitational force still acts upon him.  However, in this case, both the astronaut as well as the satellite are now attracted towards the earth and have the same centripetal acceleration due to gravity at that place.

As both astronaut and the surface of the satellite are attracted towards the earth centre with the same acceleration, and hence the astronaut can’t produce any action on the floor of the satellite.  So the floor does not give any reaction on the astronaut.  Hence the astronaut has a feeling of weightlessness. When we are moving down in accelerating lift, we can have the feeling of partial weightlessness.

Actual Weightlessness Condition:

As we go away from the earth’s surface, the acceleration due to gravity decreases. At some point in space, it becomes negligible. At such points in space, weightlessness can be experienced. The acceleration due to gravity is zero at the centre, hence at the centre of the earth, weightlessness can be experienced.

Under free fall of the body, weightlessness can be experienced. Weightlessness can be experienced at some point between the Earth and the Moon where the gravitational field created by them cancel each other.

Effect of Weightlessness Condition:

• One can lift a load of 1000 kg easily.
• One can overturn coffee mug without spilling the coffee in it on the ground.
• It leads to a headache and puffy faces.
• The height can grow by 1 mm to 2 mm.
• The cardiovascular system does less work.
• The muscle can get out of condition, and consequently one may have difficulty in doing routine work after returning back to the earth.

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Science > Physics > Gravitation > Satellites

The post Satellites appeared first on The Fact Factor.

In this article we shall study numerical problems on Keppler’s third law of orbital motion, to calculate period of revolution of a planet around the sun.

Example – 01:

What would be the length of the year if the earth were at half its present distance from the sun?

Given: r₂ = 1/2 r₁, old period T₁ = 365 days

To Find:  New period T₂ =?

Solution:

By Keppler’s law, we have T² ∝ r³

Ans: Thus length of the year would be 129 days

Example – 02:

What would be the duration of the year if the distance of the Earth from the sun gets doubled?

Given: r₂ = 2 r₁, old period T₁ = 365 days

To Find:  New period T₂ =?

Solution:

By Keppler’s law, we have T² ∝ r³

Ans: Thus the length of the year would be 1032 days

Example – 03:

Calculate the period of revolution of the planet Jupiter around the Sun. The ratio of the radius of Jupiter’s orbit to that of earth’s orbit around the Sun is 5.2.

Given: r_J : r_e  = 5.2 , Time period of the Earth  T₁ = 1 year.

To Find:  Period of Jupiter T_J =?

Solution:

By Keppler’s law, we have T² ∝ r³

Ans: The period of revolution of planet Jupiter is 11.86 years.

Example – 04:

A geostationary satellite is orbiting the earth at a height of 6R above its surface. What is the time period of a satellite orbiting at a height of 2.5 R above the earth’s surface? Where R is the radius of the Earth.

Solution:

Given: r₁ = R + 6R = 7R, r₂ = R + 2.5 R = 3.5 R, Time period of geostationary satellite T₁ = 24 hours.

To Find:  Period of second satellite T₂ =?

By Keppler’s law, we have T² ∝ r³

Ans: The period of revolution of a satellite orbiting at a height of 2.5 R above the earth’s surface is 8.458 hr.

Example – 05:

A satellite orbiting around the earth has a period of 8 hrs. If the distance of another satellite from the centre of the earth is four times that of above, what is its period?

Given: Ratio of radii of orbits r₁ = r, r₂ = 4r, Time period of first satellite  T₁ = 8 hours.

To Find:  Period of second satellite T₂ =?

Solution:

By Keppler’s law, we have T² ∝ r³

Ans: The period of the second satellite is 64 hours.

Example – 06:

Calculate the period of revolution of a planet around the sun if the diameter of its orbit is 60 times that of Earth’s orbit around the Sun. Assume both orbits to be circular.

Given: d_P = 60 d_E,  r_P =  60 r_E, Time period of Earth  T_E = 1 year.

To Find:  Period of the planet T_P =?

Solution:

By Keppler’s law, we have T² ∝ r³

Ans: The period of the planet is 464.8 years

Example – 07:

The planet Neptune travels around the sun with a period of 165 years. Show that the radius of its orbit is approximately thirty times that of the Earth.

Given: Period of NeptuneT_N = 165 years, Time period of Earth  T_E = 1 year.

To Show:  Radius of Neptune r_N =30 r_E

Solution:

By Keppler’s law, we have T² ∝ r³

Ans: Thus the radius of its orbit is approximately thirty times that of the Earth.

Example – 08:

The radius of earth’s orbit is 1.5 ×10⁸ km and that of mars is 2.5 × 10¹¹ m. in how many years the mars completes its one revolution.

Given: Radius of the orbit of the Earth r_E = 1.5 ×10⁸ km = 1.5 ×10¹¹ m, Radius of the orbit of the Mars r_M = 2.5 ×10¹¹ m, Time period of Earth T_E = 1 year.

To Find:  Time period of Mars T_M =?

Solution:

By Keppler’s law, we have T² ∝ r³

T_M = 2.15 years

Ans: In 2.15 years Mars completes its one revolution.

Example – 09:

The mean distance of the earth from the Sun is 1.496 ×10⁸ km and its period of revolution around the Sun is 365.3 days. The time periods of revolutions of planets Venus and Mars are 224.7 days and 687.0 days respectively, where day means a terrestrial day. Calculate the distances of Venus and Mars from the Sun.

Given: Radius of the orbit of the Earth r_E = 1.496 ×10⁸ km, Time period of Earth T_E = 365.3 days, Time period of Venus  T_E = 224.7 days, Time period of Mars  T_M = 687.0 days.

To Show:  Radius of the orbit of the Venus r_V =? Radius of the orbit of the Mars r_M = ?,

Solution:

By Keppler’s law, we have T² ∝ r³

Ans: Distance of Venus from the sun is 1.08 ×10⁸ km and distance of Mars from the sun is 2.279 ×10⁸ km

Example – 10

Suppose Earth’s orbital motion around the Sun is suddenly stopped. What time the Earth shall take to fall into the Sun.

Given: Radius of the orbit of the Earth = r, Time period of Earth T = 365 days,

To Show:  Time taken by the Earth to fall into the Sun T₂ = ?,

Solution:

In this problem, we are assuming there is no effect of temperature on the Earth. If the Earth suddenly stops rotating around the sun and falls into the sun and let us assume it comes back to the point on the orbit. Thus it starts orbiting along a highly flattened ellipse with major axis  = r. Thus semimajor axis = a = r₂ = r/2

By Keppler’s law, we have T² ∝ r³

This is the total period of revolution. It should take half the time period to fall into the Sun

Time required = 130/2 = 65 days

Ans: Earth shall take 65 days to fall into the Sun.

Note: Thus, in general, the time taken by a planet to fall into the sun = Period x 0.1768 (This relation can be used for MCQ)

Previous Topic: Keppler’s Laws of Orbital Motion

Next Topic: Satellites and Their Uses

Science > Physics > Gravitation > Numerical Problems on Keppler’s Laws

The post Numerical Problems on Keppler’s Laws appeared first on The Fact Factor.

In this article, we shall study Keppler’s laws of orbital motion and their use to derive Newton’s law of gravitation.

In the second century A.D., Ptolemy proposed a geocentric theory of the universe. According to this theory, the Earth is the centre of the universe and all other planets, sun and other stars revolve around the Earth. Indian astronomer, mathematician and philosopher Aryabhatta in 498 A.D. proposed that not only the Earth revolves around the sun but it rotates around its own axis. Polish astronomer Nicolaus Copernicus (1473 -1543) proposed the heliocentric theory of the universe. According to this theory, the Sun is the centre of the galaxy and all other planets revolve around the Sun. The Danish astronomer Tyco Brahe made the most accurate observations possible before the invention of the telescope. These observations showed discrepancies within Ptolemy’s geocentric theory of the universe.

Johannes Kepler (1571 – 1630) studied data of motion of planets by Danish astronomer Tycho Brahe and put forward his laws of planetary motion. These laws are known as Keppler’s law of orbital motion. He also proposed that the motion of the planet around the sun is not in circular orbit but it is in an elliptical orbit with the Sun at one of the foci of the elliptical orbit. A Netherland’s spectacle maker Hans Lippershey assembled first reflecting telescope. Using this concept Galileo developed his own telescope. Using the telescope he observed skies and substantiated Copernicus’s heliocentric theory of the universe. Further, he observed there are many stars in the universe and the Sun is one of them. He observed that only planets move around the sun. The moon moves around the Earth and Some planets like Jupiter has more than one moon.

There was a curiosity among astronomers about the motion of planets around the sun. For the study of the motion of planets around the sun or satellites around the Earth, we consider gravitational attraction between the sun and the planet or earth and the satellite only. Here we ignore the disturbing effect of the gravitational forces due to other bodies. Another major assumption is that the centre of mass of the sun (earth) and the planet (satellite) system is at their geometrical centre.

Keppler’s Laws:

First Law (Keppler’s Law of Elliptical Orbits – 1609):

Every planet moves around the sun in a closed elliptical orbit with the sun at one of its foci.

The following figure shows the path of the planet around the Earth.

Due to elliptical motion and the Sun at one of the foci, the distance of a planet from the sun changes continuously. The closest point on the orbit of the planet from the sun is B. This closest point on the orbit of the planet to the sun is called perihelion and this minimum distance is called perigee. Thus Perigee = SB = a – ae = a (1 – e). Where ‘a’ is semi-major axis and e is the eccentricity of the ellipse. The farthest point on the orbit of the planet from the sun is A. This farthest point on the orbit of the planet to the sun is called aphelion and this maximum distance is called apogee. Thus Apogee = SA = a + ae = a (1 + e). Where ‘a’ is semi-major axis and e is the eccentricity of the ellipse.

Second Law (Keppler’s Law of Equal Areas – 1609):

The radius vector drawn from the Sun to the planet sweeps out equal areas in equal time. i.e. aerial velocity of the radius vector is constant. Thus the Aerial velocity of the satellite is always constant. i.e.  A/t = Constant.

In this case, the planet covers unequal distances in equal time. Thus the planet has variable speed. When the planet is very close to the Sun, the maximum distance is covered in given time. Thus the planet has maximum velocity when it is at perihelion. Hence it has maximum kinetic energy at perihelion. When the planet is very far to the Sun, the minimum distance is covered in given time. Thus the planet has minimum velocity when it is at aphelion. Hence it has minimum kinetic energy at perihelion.

Keppler’s Third Law (Keppler’s Law of Period – 1618)

The square of the period of a satellite is directly proportional to the cube of the semimajor axis of its elliptical orbit.

If T is the period of satellite and r is the semimajor axis or the radius of the circular orbit of the satellite.

A similar result is obtained for the elliptical orbit and then the radius r of the circular orbit is replaced by semi-major axis ‘a’.

Note: 

Keppler’s laws are empirical laws because they are based on observations and not on theory. Kepler’s laws are applicable whenever an inverse square law is involved. They are applicable to the solar system and artificial satellite systems.

Proof of Keppler’s Law of Equal Areas:

Consider a planet moving in an elliptical orbit from point P₁ to point P₂, Let in a small interval of time Δt, it traces small area ΔA at the focus S. Let the angle traced by radius vector be Δθ at the Focus S.

The area of the sector is given by

Where ω is the angular velocity of the planet.

Now, the instantaneous angular momentum is given by

L = m r ω²

Thus r ω² = L/m

Substituting in equation (1) we have

As net torque acting on the planet is zero, the angular momentum is constant. Thus R.H.S. of equation (2) is constant.

Thus dA / dt = constant

Thus the areal velocity of the planet is constant. This proves Keppler’s second law.

Proof of Newton’s Law of Gravitation Using Keppler’s Law:

Let, m = Mass of planet
M = Mass of sun
R = Radius of the orbit of the planet
ωw= Angular velocity of the planet.
F = Centripetal force exerted on planet by the Sun

We know that

By Keppler’s law, we have T² ∝ r³

T² =  k r³ 

Substituting in equation(1)

The quantities in the bracket are constant.

Thus force exerted by the sun on the planet is a) directly proportional to the mass ‘m’ of the planet. b) Inversely proportional to the square of the distance ‘R’ between the planet and the sun. Now, the planet will exert equal and opposite force on the sun such that

Now, the planet will exert equal and opposite force on the sun such that

Thus force exerted by the planet on the Sun is a) directly proportional to mass ‘M’ of the Sun and b) inversely proportional to the square of the distance ‘R’ between the planet and the sun.

From (2) and (3) we have

This relation is known as Newton’s law of gravitation.

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Science > Physics > Gravitation > Keppler’s Laws of Orbital Motion

The post Keppler’s Laws of Orbital Motion appeared first on The Fact Factor.

The acceleration of gravity is constant at a particular place but it varies from place to place. In this article, we shall study this variation in acceleration due to gravity.

Variation in Acceleration Due to Gravity due to Shape of the Earth:

The acceleration due to gravity on the surface of the earth is given by

We can see that the acceleration due to gravity at a place is inversely proportional to the square of the distance of the point from the centre of the earth. Now, the earth is not perfectly spherical. It is flattened at the poles and elongated on the equatorial region. The radius of the equatorial region is approximately 21 km more than that at the poles. Hence acceleration due to gravity is maximum at the poles and minimum at the equator. As we move from the equator to the poles the distance of the point on the surface of the earth from the centre of the earth decreases. Hence the acceleration due to gravity increases.

Variation in acceleration due to gravity Due to Latitude of the Place:

The latitude of a point is the angle Φ between the equatorial plane and the line joining that point to the centre of the earth. Latitude of the equator is 0° and that of poles is 90°.

Let us consider a body of mass ‘m’ at a point P with latitude ‘Φ’ as shown on the surface of the earth. Let ‘g_Φ’ be the acceleration due to gravity at point P.

OP = Radius of earth = R

O’P = distance of point P from the axis of the earth

Due to rotational motion of the earth about its axis, the body at P experiences a centrifugal force which is given by mrω². Let us resolve this centrifugal force into two rectangular components. Its component along the radius of the earth is mrω² cosΦ.

Now the body is acted upon by two forces its weight mg acting towards the centre of the earth and the component mrω² cosΦ acting radially outward. The difference between the two forces gives the weight of that body at that point.

mg_Φ = mg – mrω²cosΦ ………….. (1)

Now cos Φ = O’P / OP = r/R

∴ r  = R cos Φ

Substituting in equation (1)

mg_Φ = mg – m(R cos Φ)ω²cos Φ

∴   g_Φ = g – R ω²cos² Φ

This is an expression for acceleration due to gravity at a point P on the surface of the earth having latitude Φ.

At the equator Φ = 0°. Hence ‘g’ is minimum on the equator. For the poles Φ = 90°. Hence ‘g’ is maximum on the poles.

Numerical Problems:

Example – 01:

Find the weight of a body of mass 100 kg on the earth at a) equator b) pole c) latitude of 30°. R = 6400 km, g = 9.8 m/s²

Given: m = 100 kg, R = 6400 km = 6.4 × 10⁶ m,

To find: W_E =? W_P = ? W_Φ =?

Solution:

Time period of earth = 24 hours = 24 x 60 x 60 s

The acceleration due to gravity at latitude Φ is given by

g_Φ = g – Rω²cos²Φ

The Weight of body at latitude Φ is given by

W_Φ = mg_Φ = mg – mRω²cos²Φ

At equator Φ = 0°

W_E = 100 x 9.8 – 100 x 6.4 × 10⁶ x (7.273 x 10^-5)²cos²0

W_E = 980 – 100 x 6.4 × 10⁶ x (7.273 x 10^-5)² x (1)²

W_E = 980 – 3.386 = 976.6 N

At poles Φ =90°

W_E = 100 x 9.8 – 100 x 6.4 × 10⁶ x (7.273 x 10^-5)²cos²90

W_E = 980 – 100 x 6.4 × 10⁶ x (7.273 x 10^-5)² x (0)²

W_E = 980 – 0 = 980 N

At latitude Φ = 30°

W_E = 100 x 9.8 – 100 x 6.4 × 10⁶ x (7.273 x 10^-5)²cos²30

W_E = 980 – 100 x 6.4 × 10⁶ x (7.273 x 10^-5)² x (0.866)²

W_E = 980 – 2.539 = 977.5 N

Ans: Weight of the body on the equator, on the pole and on latitude 30° are 976.6 N, 980 N and 977.5 N respectively.

Example – 02:

Find the difference in weight of a body of mass 100 kg on equator and pole. R = 6400 km, g = 9.8 m/s²

Given: m = 100 kg,  R = 6400  km = 6.4 × 10⁶ m,

To find: W_P – W_E =?

Solution:

Time period of earth = 24 hours = 24 x 60 x 60 s

The acceleration due to gravity at latitude Φ is given by

g_Φ = g – Rω²cos²Φ

The Weight of body at latitude Φ is given by

W_Φ = mg_Φ = mg – mRω²cos²Φ

At equator Φ = 0°

W_E = mg – mRω²cos²0° = mg – mRω²(1)² = mg – mRω²    ………… (1)

At poles Φ = 90°

W_P = mg – mRω²cos²90° = mg – mRω²(0)² = mg    ………… (2)

Now, W_P – W_E = mg – (mg – mRω²) = mRω² 

W_P – W_E = mg – (mg – mRω²) = 100 x 6.4 × 10⁶ x (7.273 x 10^-5)²   = 3.386 N

Ans: The required difference in Weight is 3.386 N

Variation in Acceleration Due to Gravity Due to Altitude:

Expanding binomially and neglecting terms of higher power of (h/R) we get

This is an expression for the acceleration due to gravity at small height ‘h’ from the surface of the earth. This expression shows acceleration due to gravity decreases as we move away from the surface of the earth.

Loss in Weight of a Body at Height h:

Numerical Problems:

Example – 03:

A mass of 5 kg is weighed on a balance at the top of a tower 20 m high. The mass is then suspended from the pan of the balance by a fine wire 20 m long and weighed. Find the change in the weight of a body in mgf assuming the radius of the earth as 6330 km.

Given: Mass of body = m = 5 kg, Radius of earth = R = 6330  km = 6.33 x 10⁶ m, height of tower =  h = 20 m.

To find: W – W_h =?

Solution:

Ans: The change in the weight of a body is 31.6 mgf

Example – 04:

At what height will a man’s weight become half his weight on the surface of the earth? Take the radius of the earth as R.

Given: W_h = 1/2W, Radius of earth = R

To find: h =?

Solution:

Now r = R + h

∴   h = r – R = 1.414 R – R = 0.414 R

Ans: At a height of 0.414R, the man’s weight become half his weight on the surface of the earth.

Example – 05:

A meteor is falling. How much gravitational acceleration it will experience when its height from the surface of the earth is equal to three times radius of the earth. Acceleration due to gravity on the surface of the earth is ‘g’.

Given: h = 3R.

To find: g_h =?

Solution:

Ans: The acceleration of the meteor is g/16.

Variation in Acceleration Due to Gravity Due to Depth:

The acceleration due to gravity on the surface of the earth is given by

Let ‘ρ’ be the density of the material of the earth.

Now, mass = volume x density

Substituting in the equation for g we get

Now, let the body be taken to the depth ‘d’ below the surface of the earth. Then acceleration due to gravity g_d at the depth ‘d’ below the surface of the earth is given by

Dividing equation (3) by (2) we get

This is an expression for the acceleration due to gravity at the depth ‘d’ below the surface of the earth. This expression shows acceleration due to gravity decreases as we move down into the earth. At the centre of the earth d = R, hence acceleration due to gravity at the centre of the earth is zero.

Relation between g_d and g_h:

We have

Thus the acceleration due to gravity at a small height ‘h’ from the surface of the earth is the same as the acceleration due to gravity at the depth ‘d = 2h’ below the surface of the earth. It means that the value of acceleration due to gravity at a small height from the surface of the earth decreases faster than the value of the acceleration due to gravity at the depth below the surface of the earth.

Variation of g with Altitude and Depth

Numerical Problems:

Example – 06:

Find percentage decrease in the weight of a body when taken 16 km below the surface of the earth. Take radius of the earth as 6400 km.

Given: depth = d = 16 km, Radius of earth = R = 6400 km, g = 9.8 m/s².

To find: Percentage decrease in weight =?

Solution:

Ans: The percentage decrease in weight is 0.25.

Example – 07:

How much below the surface of the earth does acceleration due to gravity becomes 1 % of the value at the earth’s surface? Assume the radius of the earth as 6380 km.

Given: g_d = 1% g = 0.01 g, Radius of earth = R = 6380 km, g = 9.8 m/s².

To find: depth d =?

Solution:

Ans: At a depth of 6316 km below the surface of the earth the acceleration due to gravity becomes 1 % of the value at the earth’s surface.

Example – 08:

Compare the weight of 5 kg body 10 km above and 10 km below the surface of the earth. Given the radius of the earth = 6400 km.

Given: d = 10 km, d = 10 km, Radius of earth = R = 6400 km

To find:  W_h : W_d = ?

Solution:

Ans: The required ratio of weights is 0.998 : 1

Example – 09:

Compare the weight of body 0.5 km above and 1 km below the surface of the earth. Given the radius of the earth = 6400 km.

Solution:

Given: d = 1 km, d = 0.5 km, Radius of earth = R = 6400 km

To find:  W_h : W_d = ?

Ans: The required ratio of weights is 1:1

Example – 10:

The acceleration due to gravity at a height 1/20 th radius of the earth above the earth’s surface is 9 m/s². Find the value of acceleration due to gravity at an equal distance below the surface of the earth. Given the radius of the earth = 6400 km.

Given: h = 1/20 R, gh = 9 m/s², Radius of earth = R = 6400 km

To find:   g_d =? when d = 1/20 R

Solution:

Ans: The acceleration due to gravity at depth equal to R is 9.5 m/s²

Example – 11:

At what depth below the surface of the earth, a man’s weight becomes half his weight on the surface of the earth. Take the radius of the earth as R = 6400 km.

Given: W_d = 1/2 W, Radius of earth = R = 6400 km

To find:   d =?

Solution:

∴ R = 2R – 2d

∴ d = R i.e. d = R/2 = 6400/2 = 3200 km

Ans: At depth of 3200 km below the surface of the earth a man’s weight becomes half his weight on the surface of the earth.

Example – 12:

Find decrease in value of acceleration due to gravity at a point 1600 km below the earth’s surface. R = 6400 km, g = 9.8 m/s².

Given: d = 1600 km, Radius of earth = R = 6400 km, g = 9.8 m/s².

To find:   decrease in value of acceleration due to gravity = g – g_d =?

Solution:

Ans: The decrease in acceleration due to gravity is 2.45 m/s²

Example – 13:

What is the decrease in the weight of a body of mass 500 kg when it is taken into a mine of depth 1000 m? R = 6400 km, g = 9.8 m/s².

Given: Mass of body = m = 500 kg, depth d = 1000 m = 1 km, Radius of earth = R = 6400 km, g = 9.8 m/s².

To find:   decrease in weight = W – W_d =?

Solution:

Ans: The decrease in the weight of the body is 1 N.

Example – 14:

Find the acceleration due to gravity at a depth of 2000 km from the surface of the earth, assuming earth to be a homogeneous sphere. R = 6400 km, g = 9.8 m/s².

Given: d = 2000 km, Radius of earth = R = 6400 km, g = 9.8 m/s².

To find:   acceleration due to gravity =  g_d =?

Solution:

Ans: The acceleration at depth of 2000 km below the surface of the earth is 6,738 m/s²

Previous Topic: Acceleration Due to Gravity

Next Topic: Keppler’s Laws of Orbital Motion

Science > Physics > Gravitation > Variation in Acceleration Due to Gravity

The post Variation in Acceleration Due to Gravity appeared first on The Fact Factor.

In this article, we shall study the concept of acceleration due to gravity and its characteristics. Also, we shall derive an expression for the same and solve some numerical problems.

Weight of a Body:

Weight of a body is the force with which the body is attracted towards the centre of the earth (planet).

Its unit is newton (N) and dimensions are the same as that of the force [M¹L¹ T^-2]. Mathematically the weight of a body on the surface of the Earth (Planet) is given by

W = F = mg

Where m = mass of the body and

g = acceleration due to gravity on the surface of the earth

Force of Gravitation:

The force of attraction between two material bodies in the universe is known as the force of gravitation. If one of the body is the earth or some other planet or natural satellite then the force of gravitation is called the force of gravity.

Acceleration Due to Gravity:

When a body is released from a height, it gets accelerated towards the earth with constant acceleration, this constant acceleration is called the acceleration due to gravity.

Expression for Acceleration Due to Gravity on the Surface of the Earth (Planet):

Let us consider a body of mass ‘m’ be at rest on the surface of the earth on which the acceleration due to gravity is ‘g’.  If ‘M’ and ‘R’ are mass and radius of the earth (planet) respectively

Now the force of attraction on the body is equal to its weight ‘mg’

This is the expression for acceleration due to gravity on the surface of the earth. Thus acceleration due to gravity on the surface of the earth (planet) is directly proportional to the mass of the earth (planet) and inversely proportional to the square of the radius of the earth (planet).

Characteristics of Acceleration Due to Gravity:

• It is on account of the gravitational force acting on the body.
• Average acceleration due to gravity on the surface is different for different planets.
• At a given place, the value of acceleration due to gravity is the same for all bodies irrespective of their masses.
• It changes from place to place. i.e. it changes with the change in latitude or altitude or depth.
• The acceleration due to gravity at a small height ‘h’ from the surface of the earth is the same as the acceleration due to gravity at the depth, ‘d = 2h’ below the surface of the earth. It means that the value of acceleration due to gravity at a small height from the surface of the earth decreases faster than the value of the acceleration due to gravity at the depth below the surface of the earth.

Notes:

• The value of acceleration due to gravity at the sea level and latitude 45° is taken as the standard.
• The value of acceleration due to gravity at the equator is 9.7804 m/s² and at poles is 9.8322 m/s².
• Unless otherwise stated, the value of ‘g’ is taken as 9.81 m/s² in S.I. system and 981 cm/s².
• The values of ‘g’ for Delhi, Kolkata, and Mumbai are 9.7914 m/s², 9.7876 m/s², 9.7863 m/s² respectively.

Difference Between Universal Gravitation Constant (G) and Acceleration Due to Gravity (g):

• Universal gravitation constant ‘G’ is a scalar quantity while acceleration due to gravity ‘g’ is a vector quantity.
• Universal gravitation constant ‘G’ is universal constant, while acceleration due to gravity ‘g’ changes from place to place and from planet to planet.
• Dimensions of Universal gravitation constant ‘G ‘are [M^-1L³ T^-2], while dimensions of acceleration due to gravity ‘g’ are [M⁰L¹ T^-2].

Expression for acceleration due to gravity At a Height ‘h’ From the Surface of the Earth:

Let us consider a body of mass ‘m’ at rest at height ‘h’ from the surface of the earth. Let the acceleration due to gravity at this height be is ‘g_h’. Let ‘r’ be the distance of the body from the centre of the earth. If ‘M’ and ‘R’ are mass and radius of earth respectively then, r = R +  h

Where r = R + h

This is the expression for the acceleration due to gravity at height h from the surface of the earth.

Relation Between g and g_h:

Numerical Problems on Acceleration Due to Gravity:

Example – 01:

Taking G = 6.67 × 10^-11 N m²/kg², the radius of the earth as 6400 km and mean density of earth as 5500 kg/m³, calculate g at the surface of the earth.

Given: Radius of the Earth = R = 6400 km = 6.4 × 10⁶ m, Density of material of earth = ρ = 5500 kg/m³, G = 6.67 × 10^-11 N m²/kg²

.To find: Acceleration due to gravity = g =?

Solution:

Ans: Acceleration due to gravity = 9.83 m/s².

Example – 02:

At what height will be the acceleration due to the gravity of the earth fall off to one half that at the surface? At what height will the value of g be 8 m/s²? Take radius of earth = 6400 km.

Solution Part – I:

Given: R = 6400 km and g_h = g/2

To find: Height above the surface of the earth = h =?

Now, r = R + h

∴ h = r – R = 9050 – 6400 = 2650 km

Solution Part – II:

Given: R = 6400 km and g_h = 8 m/s²

To find: Height above the surface of the earth = h =?

Now, r = R + h

∴ h = r – R = 7084 – 6400 = 684 km

Ans: At height of 2650 km, the acceleration due to gravity of the earth fall off to one half that at the surface and at a height of 684 km the acceleration due to gravity is 8 m/s².

Example – 03:

How far from the centre of the earth does the acceleration due to gravity reduce by 5 per cent of its value at the surface of the earth? Take the radius of the earth as 6.4 x 10⁶ m.

Given: , R = 6.4 x 10⁶ m  and g_h = g – 5% g = 0.95 g

To find: Distance of point from centre of the earth =r =?

Solution:

Ans: At the distance of 6566 km from the centre of the earth does the acceleration due to gravity reduce by 5 percent of its value at the surface of the earth

Example – 04:

At a certain height above the surface of the earth, the gravitational acceleration is 90 % of its value on the surface of the earth. Determine the height if the radius of the earth is 6400 km.

Solution:

Given: R = 6400 km  and g_h = 90% g = 0.9 g

To find: Height above the surface of the earth =h =?

Now, r = R + h

∴ h = r – R = 6746 – 6400 = 346 km

Ans: At a height of 346 km from the surface of the earth acceleration due to gravity be 90% of the value at the surface of the earth

Example – 05:

At what height above the earth’s surface will the acceleration due to gravity be 4% of the value at the surface of the earth? R= 6400 km.

Given: R = 6400 km  and g_h = 4% g = 0.04 g

To find: Height above the surface of the earth =h =?

Solution:

Now, r = R + h

∴ h = r – R = 32000 – 6400 = 25600 km

Ans: At height of 25600 km from the surface of the earth acceleration due to gravity be 4% of the value at the surface of the earth

Note:

When solving this type of problem, take care of the phrases reduced by and reduced to. For e.g. if the acceleration is reduced by 5 % data is g_h = g – 5% g = 0.95 g and if the acceleration reduces to 5 % data is g_h = 5% g = 0.05 g.

Example – 06:

The mass of the body on the surface of the earth is 100 kg. What will be its mass and weight at an altitude of 1000 km?

Given: Mass of body = 100 kg, R = 6400 km and altitude = h = 1000 km

To find: Mass and weight at altitude of 1000 km =?

Solution:

r = R + h = 6400 + 1000 = 7400 km

Now weight of body at altitude 100 km = W_h = m g_h = 100 x 7.33 = 733 N

The mass is always constant, hence mass at altitude of 100 km = 100 kg

Ans:  At an altitude of 1000 km the mass of the body is 100 kg and its weight is 733 N.

Example – 07:

A body weights 1.8 kg on the surface of the earth. How much will it weigh on the surface of a planet whose mass is 1/9 that of earth and whose radius is half that of earth

Given: Weight of body on earth = W_E = 1.8 kg, mass of planet = 1/9 mass of earth i.e M_P = 1/9 M_E, radius of planet = 1/2 radius of earth i.e.  R_P = 1/2 R_E.

To find: Weight of body on the planet = W_P =?

Solution:

Ans: The weight of the body on the surface of the planet is 0.8 kg or 0.8 kg wt.

Example – 08:

A body weights 4.5 kg on the surface of the earth. How much will it weigh on the surface of a planet whose mass is 1/9 that of earth and whose radius is half that of the earth?

Given: Weight of body on earth = W_E = 4.5 kg, mass of planet = 1/9 mass of earth i.e M_P = 1/9 M_E, radius of planet = 1/2 radius of earth i.e.  R_P = 1/2 R_E.

To find: Weight of body on the planet = W_P =?

Solution:

Ans: The weight of the body on the surface of the planet is 2 kg or 2 kg wt

Example – 09:

A body weights 3.5 kg wt, on the surface of the earth. How much will it weigh on the surface of a planet whose mass is 1/7 that of earth and whose radius is half that of earth

Given: , Weight of body on earth = W_E = 1.8 kg, mass of planet = 1/9 mass of earth i.e M_P = 1/7 M_E, radius of planet = 1/2 radius of earth i.e.  R_P = 1/2 R_E.

To find: Weight of body on the planet = W_P = ?

Solution:

Ans: The weight of the body on the surface of the planet is 2 kg wt.

Example – 10:

The radius of a planet is half that of the earth. The acceleration due to gravity on the planet’s surface is half that on earth’s surface. Find the mass of the planet in terms of mass M of earth

Given: Acceleration due to gravity on planet   = 1/2 Acceleration due to gravity on earth i.e g_P = 1/2 g_E, radius of planet = 1/2 radius of earth i.e.  R_P = 1/2 R_E.

To find: Mass of the planet = M_P =?

Solution:

Ans: The mass of planet in terms of the mass of earth is M/8

Example – 11:

Find the acceleration due to gravity on the surface of the moon. Given that the mass of the moon is 1/80 times that of the earth and the diameter of the moon is 1/4 times that of the earth. g = 9.8 m/s².

Given: Mass of Moon = 1/80 mass of earth i.e M_M = 1/80 M_E, diameter of Moon = 1/4 diameters of earth i.e.  R_M = 1/4 R_E. , acceleration due to gravity on surface of earth = g_E = 9.8 m/s².

To find: acceleration due to gravity on the surface of moon = g_M =?

Solution:

Ans: The acceleration due to gravity on the surface of the moon is 1.96  m/s²,

Example – 12:

A star having a mass 2.5 times that of the sun and collapsed to a size of radius 12 km rotates with a speed of 1.5 rev/s (Extremely compact stars of this kind are called neutron stars. Astronomical objects pulsars belong to this category). Will object placed on its equator remain stuck to its surface due to gravity? Mass of sun is 2 x 10³⁰ kg.

Given: Mass of Star = 2.5 times mass of Sun i.e M_Star = 2.5 M_Sun, Radius of star  = 12 km = 12 x 10³ m, mass of sun = M_Sun =  2 x 10³⁰ kg. Number of revolutions of star = n = 1.5 rev per second.

Solution:

As the gravitational acceleration on the surface of the star is greater than the centripetal acceleration, the weight of the body will be much larger than the centrifugal force acting on the body. Thus body remains stuck to the surface of the star.

Example – 13:

The mass of the Hubble telescope is 11600 kg. What is its weight and mass when it is in an orbit 598 km above the surface of the earth? Mass of earth is 5.98 x 10²⁴ kg, Radius of earth = 6400 km,

Given: Mass of telescope = m = 11600 kg, Mass of earth = M = 5.98 x 10²⁴ kg, Radius of earth = 6400 km, Height of telescope above the surface of earth = h = 598 km, G = 6.67 x 10^-11 N m²/kg²

To find: Weight of telescope = W =?

Solution:

r = R + h = 6400 + 598 = 6998 km = 6.698  x 10⁶ m

The mass is always constant, hence mass at an altitude of 598 km = 11600 kg

Ans: The weight of Hubble telescope in its orbit is 9.447 x 10⁴ N and mass at an altitude of 598 km = 11600 kg

Example – 14:

Find the value of Universal gravitational Constant G from the following data: M = 6 x 10²⁴ kg, R = 6400 km, g = 9.774 m/s²,
Given: M = 6 x 10²⁴ kg, R = 6400 km = 6.4 x 10⁶ m, g = 9.774 m/s²,
To find: G = ?

Solution:

Ans: The value of G is 6.672 x 10^-11 N m²/kg².

Example – 15: 

Assuming the earth to be homogeneous sphere, find the density of material of the earth from the following data.g = 9.8 m/s², G = 6.673 x 10^-11 Nm²/kg² ,    R = 6400 km,

Given: g = 9.8 m/s², Universal gravitational Constant = G = 6.673 x 10^-11 Nm²/kg² ,R = 6400 km = 6.4 x 10⁶ m,

To find:  Density = r = ?

Solution:

Ans: Density of material of the earth = 5483 kg/m³

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