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Gravitational potential difference

An object at a height "h"

Δ U = - G M m R + h G M m R

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 M 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 X h R R + h

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

Maximum height

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

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.

Maximum height

The velocity is zero at maximum height, "h".

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

Substituting in the equation and rearranging,

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 .

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

Putting values,

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 = 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

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

Escape velocity

In general, when a body is projected up, it returns to Earth after achieving a certain height. The height of the vertical flight depends on the speed of projection. Greater the initial velocity greater is the height attained.

Here, we seek to know the velocity of projection for which body does not return to Earth. In other words, the body escapes the gravitational influence of Earth and moves into interstellar space. We can know this velocity in verities of ways. The methods are equivalent, but intuitive in approach. Hence, we shall present here all such considerations :

1: binding energy :

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.

Questions & Answers

List the application of projectile
Luther Reply
How can we take advantage of our knowledge about motion?
Kenneth Reply
pls explain what is dimension of 1in length and -1 in time ,what's is there difference between them
Mercy Reply
what are scalars
Abdhool Reply
show that 1w= 10^7ergs^-1
Lawrence Reply
what's lamin's theorems and it's mathematics representative
Yusuf Reply
if the wavelength is double,what is the frequency of the wave
Ekanem Reply
What are the system of units
Jonah Reply
A stone propelled from a catapult with a speed of 50ms-1 attains a height of 100m. Calculate the time of flight, calculate the angle of projection, calculate the range attained
Samson Reply
58asagravitasnal firce
water boil at 100 and why
isaac Reply
what is upper limit of speed
Riya Reply
what temperature is 0 k
0k is the lower limit of the themordynamic scale which is equalt to -273 In celcius scale
How MKS system is the subset of SI system?
Clash Reply
which colour has the shortest wavelength in the white light spectrum
Mustapha Reply
how do we add
Jennifer Reply
if x=a-b, a=5.8cm b=3.22 cm find percentage error in x
Abhyanshu Reply
x=5.8-3.22 x=2.58

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Source:  OpenStax, Physics for k-12. OpenStax CNX. Sep 07, 2009 Download for free at http://cnx.org/content/col10322/1.175
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