Question

The change in potential energy when a body of mass $m$ is raised to height $nR$ from the earth's surface is ($R$ is radius of earth)

• A. $\Delta U=\left(\frac{n+1}{n}\right)\frac{GMm}{R}$
• B. $\frac{GMm}{R}$
• C. $\Delta U=\left(\frac{n}{n+1}\right)\frac{GMm}{R}$
• D. $\Delta U=-\left(\frac{n}{n+1}\right)\frac{GMm}{R}$

Answer 1

The correct option is C $\Delta U=\left(\frac{n}{n+1}\right)\frac{GMm}{R}$

The potential energy at a distance $r$ from the center of the Earth at a point outside is given by $$U=-\frac{GMm}{r}$$ where, $M$ is the mass of the Earth and $m$ is the mass of the object.

Hence,
Potential energy at the surface where $r=R$ is,

$Ui=-\frac{GMm}{R}....\left(1\right)$

Potential energy at height $nR$ where $(r=R+nR=(n+1\left)R\right)$,

$Uf=-\frac{GMm}{(n+1)R}....\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, change in potential energy

$\Delta U=Uf-Ui$

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

$\Rightarrow \Delta U=\left(\frac{n}{n+1}\right)\frac{GMm}{R}....\left(3\right)$

Hence, option (c) is correct answer.

The gravitational potential energy for two point masses is given by $$U=-\frac{GMm}{r}$$ A uniform spherical mass can be considered to be a point mass concentrated at its center to calculate gravitational potential energy, if the other mass is outside the sphere.