Discussion Forum

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 3× 10⁵ times heavier than the earth and is at a distance 2.5×10⁴ times larger than the radius of the earth. The escape velocity from the earth's gravitational field is V_e=11.2 km s^-1. The minimum initial velocity (V_s) 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)

Answer:

Explanation:

Given V_e= 11.2 km/s = $\sqrt{\frac{2GM_{e}}{R_{e}}}$

From energy conservation

                            K_i+U_i=K_f+U_f

$\frac{1}{2}mv_s^2-\frac{GM_{s}m}{r}-\frac{GM_{e}m}{R_{e}}=0+0$

Here r= distance of rocket from the sun

$\Rightarrow$     $                           V_{s}=\sqrt{\frac{2GM_{e}}{R_{e}}+\frac{2GM_{s}}{r}}$

  Given M_s  = 3× 10⁵ M_e

and r=2.5× 10⁴   R_e

$\Rightarrow$    $ V_{s}=\sqrt{\frac{2GM_{e}}{R_{e}}+(2G)(\frac{3\times 10^{5}M_{e}}{2.5\times 10^{4}R_{e}})}$

                          =$\sqrt{\frac{2GM_{e}}{R_{e}}(1+\frac{3\times 10^{5}}{2.5\times 10^{4}})}$

    =  $\sqrt{\frac{2GM_{e}}{R_{e}}\times 13}$

    = 42 km/s