Question

If the potential energy of a body on a planet is numerically $U$ and the escape velocity for the same body be $v_{e}$ for the same planet then $U / v_{e}$ will be :

• A. $\frac{U}{v_{e}}=m \sqrt{\frac{G M}{2 R}}$
• B. $ \frac{U}{{{v}_{e}}}=m\sqrt{\frac{GM}{2R}} $
• C. $ \frac{U}{{{v}_{e}}}=m\sqrt{\frac{2GM}{R}} $
• D. $ \frac{U}{{{v}_{e}}}=m\frac{GM}{R} $

Answer 1

Answer: (B)

The work obtained in bringing a body from infinity to a point in a gravitational field is called the gravitational potential energy of the body at that point.
$U=-\frac{G M_{m}}{R}\,\,\,...(1)$
where $G$ is gravitational constant, $M$ is mass of earth $R$ its radius and $m$ is mass of body.
Also for a body projected upwards at a certain velocity of projection the body will go out of the gravitational field of the earth and will never return to the earth, this initial velocity is called escape velocity.
$v_{e}=\sqrt{\frac{2 G M}{R}}\,\,\,\,...(2)$
Since, work is required to take a body from earth's surface to infinity, we have
$U=+\frac{G M_{m}}{R}\,\,\,\,....(3)$
Dividing Eq. (3) by (2), we get
$\frac{U}{v_{e}}=m \sqrt{\frac{G M}{2 R}}$