Question

Escape velocity at earth's surface is $11.2km/s$. Escape velocity at the surface of a planet having mass $100$ times and radius $4$ times that of earth will be

• A. $56km/s$
• B. $112km/s$
• C. $305km/s$
• D. $11.2km/s$

Answer 1

The correct option is A $56km/s$

$V_e=\sqrt{2gR}$

$V_e=\sqrt{\frac{2GM}{R}}[\therefore g=\frac{GM}{R^2}]$

Now,

$11.1Km/hr=\sqrt{\frac{2GM}{R}}$

$V_e^1=\sqrt{\frac{2G\left(100M\right)}{4R}}$

$\frac{11.1}{V_e^1}=\frac{1}{5}$

$V_e^1=55.5$ Km/hr