Question

the change in potential energy when a body of mass M is raised to height and r from the Earth's surface is R is the radius of Earth​

Answer 1

The change in potential energy is $\boxed{mgR(\frac{n}{n+1})}$

Explanation:

I guess the question is :

The change in the gravitational potential energy when a body of mass m is raised to a height nR above the surface of the earth is (here, R is the radius of the earth)

We know that the gravitational potential energy of a body of mass m at a distance R from the centre of the Earth is given by

$U=-\frac{GMm}{R}$

Therefore, the potential energy at a height nR from the surface of the Earth

$U'=-\frac{GMm}{(R+nR)}$

$\implies U'=-\frac{GMm}{R(1+n)}$

Change in potential energy

$\Delta U=U'-U$

$\implies \Delta U=-\frac{GMm}{R(1+n)}+\frac{GMm}{R}$

$\implies \Delta U=\frac{GMm}{R}(1-\frac{1}{1+n})$

$\implies \Delta U=\frac{GMm}{R}(\frac{n}{n+1})$

From gravitation, we also know that

Acceleration due to gravity is given by

$g=\frac{GM}{R^2}$

$\implies gR=\frac{GM}{R}$

Thus,

$\Delta U=mgR(\frac{n}{n+1})$

Hope this answer is helpful.

Know More:

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Answer 2

Answer:

Explanation:

We known that

Potential energy = mgh/1+h/r

=mgnR/1+nR/R (h=nR)

=mgnR/R+nR/R

=(mgnR) R/R(1+n)

=mgR{n/1+n}