Question

If g is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass m raised from surface of the earth to a height equal to radius R of the earth is

Answer 1

Please see the image for answer.

Answer 2

Answer:

$\Delta U=\dfrac{mgR}{2}$

Explanation:

The acceleration due to gravity is g

Let the mass of the earth be M

Let the mass of the object is m

The radius of the earth be R

The potential energy on the earth's surface is

$U_{i}=-\dfrac{GMm}{R^2}$

The final potential energy at the height h=R from the earth's surface

$U_{f}=-\dfrac{GMm}{R+h}$

The change in potential energy to raise the body at height R from the earth surface.

$\Delta U= U_{f}-U_{i}\\\Delta U=-\dfrac{GMm}{R+h}-\left(-\dfrac{GMm}{R^2}\right )\\\\\Delta U=-GMm\left (  \dfrac{1}{R+h}-\dfrac{1}{R} \right )\\\\\Delta U=-GMm \left(  \dfrac{R-R-h}{R(R+h)} \right )\\\\\Delta U=GMm \left(  \dfrac{h}{R(R+h)} \right )\\\\\Delta U=GMm \left(   \dfrac{h}{R^2\left(  1+\dfrac{h}{R} \right )}   \right )\\\\\Delta U=\dfrac{GMm}{R^2} \left(   \dfrac{h}{\left(  1+\dfrac{h}{R} \right )}   \right )$

We know that

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

So,

$\Delta U=\dfrac{mgh}{1+\dfrac{h}{R}}\\h=R\\\Delta U=\dfrac{mgR}{1+\dfrac{R}{R}}\\\Delta U=\dfrac{mgR}{1+1}\\\Delta U=\dfrac{mgR}{2}$