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.

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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 – 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.

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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.