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Question

A body of mass is placed on the earth s surface. It is taken from the earth s surface to a height h=3R, R is the radius of earth. The change in gravitational potential energy of the body is


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Solution

Step1: Given data

  1. The radius of the earth is R.
  2. The height of the body from the earth's surface is h=3R.

Step2: Gravitational potential energy

  1. The gravitational potential energy at any point in a gravitational field of a system of masses is the amount of work done in bringing a unit mass from infinity to that point.
  2. Gravitational potential energy due to a point mass m at a distance r is, V=-∫∞rGMmx2dx=-GMmr, where G is the universal gravitational constant and M is the mass of the earth.

Step3: Finding the change in Gravitational potential energy

We know that the gravitational potential energy of a body of mass m placed on the earth's surface is

V=-GMmR ……………(1)

where M and R are the mass of the earth and the radius of the earth.

Now, gravitational potential energy at a height h=3R from the earth's surface is

V'=-GmMR+3R …………(2)

So, the change in gravitational potential energy is

∆V=V-V'=-GMmR--GmMR+3R. (From equations 1 and 2 )

or ∆V=-GMmR+GMm4R=-G3Mm4R

or ∆V=-G3mM4R

Therefore, the change in gravitational potential energy is ∆V=-G3mM4R.


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