Question

If a body of mass m has to be taken from the surface of earth to a height $h = R$, then the amount of energy required is ($R$ = radius of earth)

• A. mgR
• B. $\frac{mgR}{3}$
• C. $\frac{mgR}{2}$
• D. $\frac{mgR}{12}$

Answer 1

Answer: (C)

$\frac{mgR}{2}$

Answer 2

We know that,
Gravitational potential energy
$U=-\frac{G M_{e} m}{R}\,\,\,...(i)$
and gravitational kinetic energy
$K=\frac{1}{2} \cdot \frac{G M_{e} m}{R}\,\,\,...(ii)$
$\therefore$ Total energy of a body is
$E =U+K$
$=-\frac{G M_{e} m}{R}+\frac{G M_{e} m}{2 R}$
$=-\frac{2 G M_{e} m+G M_{e} m}{2 R}=-\frac{G M_{e} m}{2 R}$
But, acceleration due to gravity $(g)$ in terms of aravitational constant $(G)$ is
$g=\frac{G M_{e}}{R^{2}}\,\,\,...(ii)$
$\therefore -\frac{G M_{e}}{R^{2}} \times R^{2} \times \frac{m}{2 R}$[From E (iii)]
$=g \times R^{2} \times \frac{m}{2 R}$
(Cancelation of negative sign, because energy can never be negative)
$=\frac{g m R}{2}=\frac{m g R}{2}$