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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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

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