Question

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

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

Answer 1

Answer: (A)

$22.4\, km/s^{-1}$

Answer 2

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