Question

Suppose the gravitational attraction on a satellite varies inversely as the distance from centre of the earth. The orbital velocity of a satellite in such a case varies as $nth$ power of distance where $n$ is equal to

• A. $-1$
• B. zero
• C. $+1$
• D. $+2$

Answer 1

The correct option is B zero

Let, $r$ is the distance from the centre of the earth.

According to problem, the gravitational force, $$F\propto \frac{1}{r}$$
$\Rightarrow F=\frac{k}{r}$, where $k$ is a constant.

Gravitation force provides the necessary centripetal force to satellite to revolve around the earth. So,

$F=\frac{k}{r}=\frac{m{v}^2}{r}$

$\Rightarrow v=\sqrt{\frac{k}{m}}$

$\Rightarrow v\propto r^0$

Given, $v\propto r^n$

$\therefore n=0$

Hence, option (b) is the correct answer.

Why this question: Caution: The result $v=\sqrt{\frac{GM}{r}}$ is derived by the fact that gravitational force follows inverse square law. If the inverse square law is not followed, the result for orbital velocity will not be valid.