Question

A satellite of earth of mass '$m$' taken from orbital radius $2R$ to $3R$, then minimum work done is :-

• A. $\frac{GMm}{6R}$
• B. $\frac{GMm}{12R}$
• C. $\frac{GMm}{24R}$
• D. $\frac{GMm}{3R}$

Answer 1

Answer: (B)

$\frac{GMm}{12R}$

Centripetal acceleration of the satellite is provided by the gravitational force exerted by earth,

$⟹\frac{m{v}^2}{r}=\frac{GMm}{r^2}$

$⟹v^2=\frac{GM}{r}$

$\text{Total energy of the system}=\text{K.E.+ G.P.E.}=\frac{1}{2}m{v}^2-\frac{GMm}{r}$

$=-\frac{GMm}{2r}$

$\text{Therefore energy required to jump from orbital radius of}2R\text{to orbital radius of}3R\text{is given by,}$

$E=\frac{GMm}{2}(\frac{1}{2R}-\frac{1}{3R})$

$E=\frac{GMm}{12R}$