Question

An artificial satellite is moving in a circular orbit around the earth with a speed equal to half 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 $\text{km/s}$ with which it hits surface of the earth. $(g=10{\text{m/s}}^2\text{and}{R}_E=6400\text{km})$

• A. $4$
• B. $10$
• C. $6$
• D. $8$

Answer 1

The correct option is D $8$



$\frac{m{V}^2}{(R+h)}=\frac{GMm}{(R+h{)}^2}$

Given, $V=\frac{V_e}{2}=\frac{1}{2}\sqrt{\frac{2GM}{R}}$

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

hence, $h=R$
If satellite is stopped, then using conservation of mechanical energy.

Total energy at $A$

${\epsilon }_A=\frac{-GMm}{(R+h)}+0$

Total energy at $B$

${\epsilon }_B=\frac{-GMm}{R}+\frac{1}{2}m{V}^2$

Now, ${\epsilon }_A={\epsilon }_B$

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

or, $V=\sqrt{gR}$

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

$V=8\text{km/s}$

Thus, the correct choice is option $\text{(C)}$ .