Question

Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential energy (final - initial) of an object of mass m, when taken to a heighth from the surface of earth (of radius R), is given by,​

Answer 1

Answer: see the image

Explanation:

Answer 2

Answer:

mgh

Explanation:

We are given that

Gravitational potential energy of an object at infinity=0

Mass of an object=m

We have to find the change in potential energy when the object from the surface of earth to a height .

Let h be the height from the surface of earth and R be  radius of earth

Distance of object from the center of earth to h=R+h

Potential energy=$-\frac{GmM}{r}$

Potential energy at  R=$-\frac{GmM}{R}$

Potential energy at R+h=$-\frac{GmM}{R+h}$

Change in potential energy= Final potential energy-Initial energy

$\Delta U=-\frac{GmM}{R+h}+\frac{GmM}{R}=GmM(\frac{R+h-R}{R(R+h)}$

$\Delta U=\frac{GmMh}{R^2(1+\frac{h}{R})}$

h < <R

$\frac{h}{R} <<1$

Substitute the value then we get

$\Delta U=\frac{GmMh}{R^2}=mgh$

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