Earlier, we used the expression “mgh” to compute potential energy or change in potential energy. We need to correct this formula for determining change in potential energy by referring calculation of potential energy to infinity. Using formula of potential energy with infinity as reference, we determine the potential difference between Earth’s surface and a point above it, as :

Figure 1: An object at a height "h"

Gravitational potential difference

⇒ Δ U = - G M m R + h − − G M m R ⇒ Δ U = - G M m R + h − − G M m R

⇒ Δ U = G M m 1 R − 1 R + h ⇒ Δ U = G M m 1 R − 1 R + h

We can eliminate reference to gravitational constant and mass of Earth by using relation of gravitational acceleration at Earth’s surface ( g = g 0 g = g 0 ),

g = G M R 2 g = G M R 2

⇒ G M = g R 2 ⇒ G M = g R 2

Substituting in the equation of change in potential energy, we have :

⇒ Δ U = m g R 2 1 R − 1 R + h ⇒ Δ U = m g R 2 1 R − 1 R + h

⇒ Δ U = m g R 2 X h R R + h ⇒ Δ U = m g R 2 X h R R + h

⇒ Δ U = m g h 1 + h R ⇒ Δ U = m g h 1 + h R

It is expected that this general formulation for the change in potential energy should be reduced to approximate form. For h<<R, we can neglect “h/R” term and,

⇒ Δ U = m g h ⇒ Δ U = m g h

For velocity less than escape velocity (the velocity at which projectile escapes the gravitation field of Earth), the projected particle reaches a maximum height and then returns to the surface of Earth.

When we consider that acceleration due to gravity is constant near Earth’s surface, then applying conservation of mechanical energy yields :

1 2 m v 2 + 0 = 0 + m g h 1 2 m v 2 + 0 = 0 + m g h

⇒ h = v 2 2 g ⇒ h = v 2 2 g

However, we have seen that “mgh” is not true measure of change in potential energy. Like in the case of change in potential energy, we come around the problem of variable acceleration by applying conservation of mechanical energy with reference to infinity.

Figure 2: The velocity is zero at maximum height, "h".

Maximum height

K i + U i = K f + U f K i + U i = K f + U f

⇒ − G M m R + 1 2 m v 2 = 0 + − G M m R + h ⇒ − G M m R + 1 2 m v 2 = 0 + − G M m R + h

⇒ G M R + h = G M R − v 2 2 ⇒ G M R + h = G M R − v 2 2

⇒ R + h = G M G M R − v 2 2 ⇒ R + h = G M G M R − v 2 2

⇒ h = G M G M R − v 2 2 − R ⇒ h = G M G M R − v 2 2 − R

⇒ h = G M − G M + v 2 R 2 G M R − v 2 2 ⇒ h = G M − G M + v 2 R 2 G M R − v 2 2

⇒ h = v 2 R 2 G M R − v 2 ⇒ h = v 2 R 2 G M R − v 2

We can also write the expression of maximum height in terms of acceleration at Earth’s surface using the relation :

⇒ G M = g R 2 ⇒ G M = g R 2

Substituting in the equation and rearranging,

⇒ h = v 2 2 g − v 2 R ⇒ h = v 2 2 g − v 2 R

This is the maximum height attained by a projection, which is thrown up from the surface of Earth.

Problem 1: A particle is projected vertically at 5 km/s from the surface Earth. Find the maximum height attained by the particle. Given, radius of Earth = 6400 km and g = 10 m / s 2 m / s 2 .

Solution : We note here that velocity of projectile is less than escape velocity 11.2 km/s. The maximum height attained by the particle is given by:

h = v 2 2 g − v 2 R h = v 2 2 g − v 2 R

Putting values,

⇒ h = 5 X 10 3 2 2 X 10 − 5 X 10 3 2 6.4 X 10 6 ⇒ h = 5 X 10 3 2 2 X 10 − 5 X 10 3 2 6.4 X 10 6

⇒ h = 25 X 10 6 2 X 10 − 25 X 10 6 6.4 X 10 6 ⇒ h = 25 X 10 6 2 X 10 − 25 X 10 6 6.4 X 10 6

⇒ h = 1.55 x 10 6 = 1550000 m = 1550 k m ⇒ h = 1.55 x 10 6 = 1550000 m = 1550 k m

It would be interesting to compare the result, if we consider acceleration to be constant. The height attained is :

h = v 2 2 g = 25 X 10 6 20 = 1.25 X 10 6 = 1250000 m = 1250 k m h = v 2 2 g = 25 X 10 6 20 = 1.25 X 10 6 = 1250000 m = 1250 k m

As we can see, approximation of constant acceleration due to gravity, results in huge discrepancy in the result.

Gravitational binding energy represents the energy required to eject a body out of the influence of a gravitational field. It is equal to the energy of the system, but opposite in sign. In the absence of friction, this energy is the mechanical energy (sum of potential and kinetic energy) in gravitational field.

Now, it is clear from the definition of binding energy itself that the initial kinetic energy of the projection should be equal to the binding energy of the body in order that it moves out of the gravitational influence. Now, the body to escape is at rest before being initiated in projection. Thus, its binding energy is equal to potential energy only.

Therefore, kinetic energy of the projection should be equal to the magnitude of potential energy on the surface of Earth,

1 2 m v e 2 = G M m R 1 2 m v e 2 = G M m R

where “ v e v e ” is the escape velocity. Note that we have used “R” to denote Earth’s radius, which is the distance between the center of Earth and projectile on the surface. Solving above equation for escape velocity, we have :

⇒ v e = 2 G M R ⇒ v e = 2 G M R

The act of putting a body into interstellar space is equivalent to taking the body to infinity i.e. at a very large distance. Infinity, as we know, has been used as zero potential energy reference. The reference is also said to represent zero kinetic energy.

From conservation of mechanical energy, it follows that total mechanical energy on Earth’s should be equal to mechanical energy at infinity i.e. should be equal to zero. But, we know that potential energy at the surface is given by :

U = - G M m R U = - G M m R

On the other hand, for body to escape gravitational field,

K + U = 0 K + U = 0

Therefore, kinetic energy required by the projectile to escape is :

⇒ K = - U = G M m R ⇒ K = - U = G M m R

Now, putting expression for kinetic energy and proceeding as in the earlier derivation :

⇒ v e = 2 G M R ⇒ v e = 2 G M R

We again use conservation of mechanical energy, but with a difference. Let us consider that projected body of mass, “m” has initial velocity “u” and an intermediate velocity, “v”, at a height “h”. The idea here is to find condition for which intermediate velocity ,”v”, never becomes zero and hence escape Earth’s influence. Applying conservation of mechanical energy, we have :

E i = E f E i = E f

K i + U i = K f + U f K i + U i = K f + U f

1 2 m u 2 − G M m R = 1 2 m v 2 − G M m R + h 1 2 m u 2 − G M m R = 1 2 m v 2 − G M m R + h

Rearranging,

⇒ 1 2 m v 2 = 1 2 m u 2 − G M m R + G M m R + h ⇒ 1 2 m v 2 = 1 2 m u 2 − G M m R + G M m R + h

In order that, final velocity (“v”) is positive, the expressions on the right should evaluate to a positive value. For this,

⇒ 1 2 m u 2 ≥ G M m R ⇒ 1 2 m u 2 ≥ G M m R

For the limiting case, u = v e v e ,

⇒ v e = 2 G M R ⇒ v e = 2 G M R

In the case of Earth,

M = 5.98 X 10 24 k g M = 5.98 X 10 24 k g

R = 6.37 X 10 6 m R = 6.37 X 10 6 m

⇒ v e = 2 X 6.67 X 10 - 11 X 5.98 X 10 24 6.37 X 10 6 ⇒ v e = 2 X 6.67 X 10 - 11 X 5.98 X 10 24 6.37 X 10 6

⇒ v e = 11.2 k m / s ⇒ v e = 11.2 k m / s

We should understand that this small numerical value is deceptive. Actually, it is almost impossible to impart such magnitude of speed. Let us have a look at the magnitude in terms of “km/hr”,

v e = 11.2 k m / s = 11.2 X 60 x 60 = 40320 k m / h r v e = 11.2 k m / s = 11.2 X 60 x 60 = 40320 k m / h r

If we compare this value with the speed of modern jet (which moves at 1000 km/hr), this value is nearly 40 times! It would be generally a good idea to project object from an artificial satellite instead, which itself may move at great speed of the order of about 8-9 km/s. The projectile would need only additional 2 or 3 km/hr of speed to escape, if projected in the tangential direction of the motion of the satellite.

This mechanism is actually in operation in multistage rockets. Each stage acquires the speed of previous stage. The object (probe or vehicle) can, then, be let move on its own in the final stage to escape Earth’s gravity. The mechanism as outlined here is actually the manner an interstellar probe or vehicle is sent out of the Earth’s gravitational field. We can also appreciate that projection, in this manner, has better chance to negotiate friction effectively as air resistance at higher altitudes is significantly less or almost negligible.

This discussion of escape velocity also underlines that the concept of escape velocity is related to object, which is not propelled by any mechanical device. An object, if propelled, can escape gravitational field at any speed.

Escape velocity of Moon :

In the case of Earth’s moon,

M = 7.4 X 10 20 k g M = 7.4 X 10 20 k g

R = 1.74 X 10 6 m R = 1.74 X 10 6 m

⇒ v e = 2 X 6.67 X 10 - 11 X 7.4 X 10 20 1.74 X 10 6 ⇒ v e = 2 X 6.67 X 10 - 11 X 7.4 X 10 20 1.74 X 10 6

⇒ v e = 2.4 k m / s ⇒ v e = 2.4 k m / s

The root mean square velocity of gas is greater than this value. This is the reason, our moon has no atmosphere. Since sound requires a medium to propagate, we are unable to talk directly there as a consequence of the absence of atmosphere.

It may appear that we may need to fire projectile vertically to let it escape in interstellar space. This is not so. The spherical symmetry of Earth indicates that we can project body in any direction with the velocity as determined such that it clears physical obstructions in its path. From this point of view, the term “velocity” is a misnomer as direction of motion is not involved. It would have been more appropriate to call it “speed”.

The direction, however, makes a difference in escape velocity for some other reason. The Earth rotates in particular direction – it rotates from East to west at a linear speed of 465 m/s. So if we project the body in the tangential direction east-ward, then Earth rotation helps body’s escape. The effective escape velocity is 11200 – 465 = 10735 m/s. On the other hand, if we project west-ward, then escape velocity is 11200 + 465 = 11635 m/s.

Black holes are extremely high density mass. This represents the final stage of evolution of a massive star, which collapses due to its own gravitational force. Since mass remains to be very large while radius is reduced (in few kms), the gravitational force becomes extremely large. Such great is the gravitational force that it does not even allow light to escape.

Lesser massive star becomes neutron star instead of black hole. Even neutron star has very high gravitational field. We can realize this by calculating escape velocity for one such neutron star,

M = 3 X 10 30 k g M = 3 X 10 30 k g

R = 3 X 10 4 m R = 3 X 10 4 m

⇒ v e = 2 X 6.67 X 10 - 11 X 3 X 10 30 3 X 10 4 ⇒ v e = 2 X 6.67 X 10 - 11 X 3 X 10 30 3 X 10 4

⇒ v e = 1.1 X 10 5 m / s ⇒ v e = 1.1 X 10 5 m / s

It is quite a speed comparable with that of light. Interstellar Black hole is suggested to be 5 times the mass of neutron star and 10 times the mass of sun! On the other hand, its dimension is in few kilometers. For this reason, following is possible :

⇒ v e = 2 G M R > c ⇒ v e = 2 G M R > c

where “c” is the speed of light. Hence even light will not escape the gravitational force of a black hole as the required velocity for escape is greater than speed of light.