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. $v_{es}=\sqrt{2GM{R}_e}$
• D. $v_{es}=\sqrt{\frac{2GM}{R_e^2}}$

Answer 1

Answer: (A)

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