Question

What is the escape velocity for a body on the surface of a planet on which the acceleration due to gravity is $(3.1)^2 ms^{-2}$ and whose radius is 8100 km?

• A. $2790\, km-s^{-1}$
• B. $27.9\,km-s^{-1}$
• C. $\frac{27.9}{\sqrt5}\,km^{-1}$
• D. $27.9\sqrt5 \,km-s^{-1}$

Answer 1

Answer: (C)

Escape velocity $v_e=\sqrt{2gR}$
Given, $g=(3.1)^2ms^{-2},$
$R = 8100km = 8 1 00 x 10^3 m.$ = $v_e=\sqrt{2\times(3.1)^2\times8100\times10^3}$
$v_e=12.5\times10^3ms^{-1}$
$\Rightarrow v_e=12.5 kms^{-1}$
$\approx\frac{27.9}{\sqrt5}kms^{-1}$