Question

An artificial satellite is moving in a circular orbit around the earth with speed equal to half the magnitude of escape velocity from the earth. The height of the satellite above the earth’s surface is found to be $640\times n\text{km}$. The value of $n$ will be (Give integer value).

Answer 1

For the revolution of satellite $\text{S}$ in a circular path of radius $(R+h)$, where $R$ denotes radius of earth.

Centripetal force = Gravitational force.

$\Rightarrow \frac{m{v}^2}{(R+h)}=\frac{GmM}{(R+h{)}^2}$

$v^2=\frac{GM}{(R+h)}.......\left(i\right)$

But from the data given in the question, $v=\frac{v_e}{2}$

Or $v=\frac{1}{2}\sqrt{\frac{2GM}{R}}.......\left(ii\right)$

From $\left(1\right)$ and $\left(2\right)$,

$\frac{2GM}{4R}=\frac{GM}{R+h}$

$\Rightarrow 2R=h+R$

$\Rightarrow R=h$

Or $h=R=6400\text{km}$

But the given height is $640\times n\text{km}$.

Therefore, on comparison we get, $n=10$.

Accepted answer : $10$