Question

The escape velocity of an object from the earth depends upon the mass of the earth $\left( M _{ e }\right)$, its mean density $(\rho)$, its radius $\left( R _{e}\right)$, and the gravitational constant $( G )$, Thus, the formula for escape velocity is

• A. $v_{es}=R_{e} \sqrt{\frac{8 \pi}{3} G \rho}$
• B. $v_{es}=M\sqrt{\frac{8 \pi}{3} G R_{e}}$
• C. $v_{es}=\sqrt{2 GMR_{e}}$
• D. $v_{e s}=\sqrt{\frac{2 G M}{R_{e}^{2}}}$

Answer 1

Answer: (A)

Escape velocity
$v_{es}=\sqrt{2 g R_{e}}=\sqrt{2 \frac{G M_{e}}{R_{e}^{2}} \times R_{e}}\left[g=\frac{G M_{e}}{R_{e}^{2}}\right]$
$=\sqrt{2 \frac{G M_{e}}{R_{e}}}$
[mass = volume $\times$ density]
$=\sqrt{\frac{2 G}{R_{e}} V \rho}=\sqrt{\frac{2 G \times \frac{4}{3} \pi R_{e}^{3} \rho}{R_{e}}}$
$=\sqrt{\frac{8}{3} \pi G R_{e}^{2} \rho}=R_{e} \sqrt{\frac{8 \pi}{3}} G \rho$