Question

A rocket is launched normal to the surface of the Earth, away from the Sun, along the line, joining the Sun and the Earth. The Sun is $3\times {10}^5$ times heavier than the Earth and is at a distance $2.5\times {10}^4$ times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is $v_e=11.2km{s}^{-1}$. The minimum initial velocity $\left(v_s\right)$ required for the rocket to be able to leave the Sun-Earth system is closest to (Ignore the rotation and revolution of the Earth and the presence of any other planet)

• A. $v_s=62km{s}^{-1}$
• B. $v_s=42km{s}^{-1}$
• C. $v_s=72km{s}^{-1}$
• D. $v_s=22km{s}^{-1}$

Answer 1

Answer: (B)

$v_s=42km{s}^{-1}$

Given:

$M_s=3\times {10}^5{M}_e$

$d=2.5\times {10}^4{R}_e$

Also $V_e=11.2Km/s$ = $\sqrt{\frac{2G{M}_e}{R_e}}$

Now for the sun earth system,

$\frac{1}{2}m{V}_s^2-\frac{G{M}_{e}m}{R_e}-\frac{G{M}_{s}m}{R_e+d}=0$

Assuming $R_e<<d$

and taking the value of $\frac{G{M}_e}{R}=\frac{{11.2}^2}{2}$

$V_s\approx 42Km/s$