Question

An artificial satellite is moving in a circular orbit around the earth with a speed equal to half of the magnitude of escape velocity from the earth. If the satellite is stopped suddenly in its orbit and allowed to fall freely onto the earth, find the speed $(\text{km/sec)}$ with which it hits the surface of the earth. $(g=9.8{\text{ms}}^{-2}$ and Radius of earth $=6400\text{km})$

• A. $10$
• B. $6$
• C. $8$
• D. $4$

Answer 1

The correct option is C $8$

Let us say height of the satellite is $h$.


As centripetal force is provided by gravitation pull,
$\frac{m{v}^2}{(R+h)}=\frac{GMm}{(R+h{)}^2}$

Given, $V=\frac{V_e}{2}=\left(\frac{1}{2}\sqrt{\frac{2GM}{R}}\right)$

So, $m(\frac{1}{2}\sqrt{\frac{2GM}{R}}{)}^2=\frac{GMm}{(R+h)}$
Hence, $h=R$

If satellite is stopped then using conservation of mechanical energy,
Total energy at $A$, $\left(E_A\right)=\frac{-GMm}{R+h}+0$

Total energy at $B$, $\left(E_B\right)=\frac{-GMm}{R}+\frac{1}{2}m{v}^2$

Now, $E_A=E_B$

So,$v=\sqrt{\frac{GM}{R}}$

or, $v=\sqrt{gR}$

$v=\sqrt{10\times 6400\times {10}^3}$

$v=8\text{km/sec}$