Question

The escape velocity of a body on the surface of the earth is $11.2km/s^{-1}$. If the earth's mass increases to twice its present value and radius of the earth becomes half, the escape velocity becomes

• A. $22.4km/s^{-1}$
• B. $44.8km/s^{-1}$
• C. $5.6km/s^{-1}$
• D. $11.2km/s^{-1}$

Answer 1

Answer: (A)

Escape velocity of a body $(v_e)=11.2km/s$;
New mass of the earth ${M^{\prime }}_e=2M_e$ and new radius
of the earth ${R^{\prime }}_e=0.5R_e.$
Escape velocity $(v_e)=\sqrt{\frac{2G{M}_e}{R_e}}\propto \sqrt{\frac{M_e}{R_e}}$.
Therefore $\frac{v_e}{{v^{\prime }}_e}=\sqrt{\frac{M_e}{R_e}\times \frac{0.5R_e}{2M_e}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$
or, ${v^{\prime }}_e=2v_e=22.4km/sec$.