Question

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

• A. $mgR$
• B. $\frac{mgR}{3}$
• C. $\frac{mgR}{2}$
• D. $\frac{mgR}{12}$
• E. $\frac{mgR}{9}$

Answer 1

The correct option is C $\frac{mgR}{2}$

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
$E=U+K$
$\frac{G{M}_{e}m}{R}+\frac{G{N}_{e}m}{2R}$
$\frac{2G{M}_{e}m+G{M}_{e}m}{2R}=-\frac{G{M}_{e}m}{2R}$
But, acceleration due to gravity (g) in terms of gravitational constant (G) is
$\frac{G{M}_e}{R^2}$ ...(iii)
$\therefore \frac{G{M}_e}{R^2}\times R^2\times \frac{me}{2R}$ [from eq. (iii)]
$=g\times R^2\times \frac{m}{2R}$
(Cancelation of negative sign, because energy can never be negative)
$=\frac{gmR}{2}=\frac{mgR}{2}$