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 height h from the surface of earth (of radius R), is given by,

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 height h from the surface of earth (of radius R), is given by, (A) -(GMm/R + h) (B) (GMmh/R(R + h)) (C) mgh (D) (GMm/R + h)

Tardigrade Admin · Accepted Answer

Δ U = -GMm [ (1/rf) - (1/ri)] = - GMm[(1/R+h) - (1/R)] = (GMmh/R(R+h))

Q. 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 height h from the surface of earth (of radius R), is given by,

$- \frac{GM m}{R + h}$

$\frac{GM mh}{R ( R + h )}$

mgh

$\frac{GM m}{R + h}$

$Δ U = - GM m [\frac{1}{r _{f}} - \frac{1}{r _{i}}]$
$= - GM m [\frac{1}{R + h} - \frac{1}{R}]$
$= \frac{GM mh}{R ( R + h )}$

Q. 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 height h from the surface of earth (of radius R), is given by,

−R+hGMm​

R(R+h)GMmh​

mgh

R+hGMm​

ΔU=−GMm[rf​1​−ri​1​] =−GMm[R+h1​−R1​] =R(R+h)GMmh​

$- \frac{GM m}{R + h}$

$\frac{GM mh}{R ( R + h )}$

$\frac{GM m}{R + h}$

$Δ U = - GM m [\frac{1}{r _{f}} - \frac{1}{r _{i}}]$
$= - GM m [\frac{1}{R + h} - \frac{1}{R}]$
$= \frac{GM mh}{R ( R + h )}$