Question

A body of mass $m$ is situated on the earth in the gravitational field of sun. For the body to escape from the gravitation pull of the solar system the body must be imparted an escape velocity of (assume earth to be stationary)

• A. 11.2 km/s
• B. 22.4 km/s
• C. 33.6 km/s
• D. 42 km/s

Answer 1

The correct option is A 11.2 km/s

The potential energy of a mass $m$ in the Sun-Earth field is given by

$V=-\frac{G{M}_{s}m}{d_s}-\frac{G{M}_{e}m}{R_e}$

Let the velocity required to escape the solar system be $v$. Then Kinetic energy is $K=\frac{1}{2}m{v}^2$

To escape the solar system, $K+V=0$

$\Rightarrow v=\sqrt{2G(\frac{M_s}{d_s}+\frac{M_e}{R_e})}$

Using $M_s=1.98855\times {10}^{30}\text{kg,}{M}_e=5.97237\times {10}^{24}\text{kg,}{d}_s=1.496\times {10}^{13}\text{m,}{R}_e=6.371\times {10}^6\text{m}$

we get $v_{esc}=\sqrt{2\times 6.674\times {10}^{-11}(\frac{1.98855\times {10}^{30}}{1.496\times {10}^{13}}+\frac{5.97237\times {10}^{24}}{6.371\times {10}^6})}=11.24\times {10}^3\text{m/s}$