If potential energy of a body of mass m on the surface of earth is taken as zero then its potential energy at height h above the surface of earth is [R is radius of earth and M is mass of earth]

Question

If potential energy of a body of mass m on the surface of earth is taken as zero then its potential energy at height h above the surface of earth is [R is radius of earth and M is mass of earth] (A) (- GMm /R+h) (B) (- GMm /h) (C) (G M m h/R(R+h)) (D) ( GMmh /h+2 R)

Tardigrade Admin · Accepted Answer

The concept involved here is that,
Gravitational potential energy difference between any two points in a gravitational field is independent of the choice of reference.
When potential at infinity is assigned zero value,
Potential energy of a body of mass

m

on the surface of earth

= \frac{- GM m}{R} = U_{s}

Potential energy at height,

h = \frac{- GM m}{R + h} = U_{n}

U_{s} - U_{h} = + GM m (- \frac{1}{R} + \frac{1}{R + h})

= GM m (\frac{- R - h + R}{R ( R + h )})

= \frac{- GM mh}{R ( R + h )}

Now, when potential at the surface is taken zero,
Let

U_{s}^{'}, U_{h}^{'}

be the new values of potential energy at the surface and height

h

respectively,
And,

U_{s} - U_{h} = U_{s}^{'} - U_{h}^{'}

\Rightarrow \frac{- GM mh}{R ( R + h )} = 0 - U_{h}^{'}

\Rightarrow U_{h}^{'} = \frac{GM mh}{R ( R + h )}

Q. If potential energy of a body of mass $m$ on the surface of earth is taken as zero then its potential energy at height $h$ above the surface of earth is [R is radius of earth and $M$ is mass of earth]

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

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

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

$\frac{GM mh}{h + 2 R}$

Q. If potential energy of a body of mass m on the surface of earth is taken as zero then its potential energy at height h above the surface of earth is [R is radius of earth and M is mass of earth]

R+h−GMm​

h−GMm​

R(R+h)GMmh​

h+2RGMmh​

Q. If potential energy of a body of mass $m$ on the surface of earth is taken as zero then its potential energy at height $h$ above the surface of earth is [R is radius of earth and $M$ is mass of earth]

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

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

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

$\frac{GM mh}{h + 2 R}$