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 $${v_e} = 11.2\,km\,{s^{ - 1}}.$$ 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.
$${v_s} = 22\,km\,{s^{ - 1}}$$
B.
$${v_s} = 42\,km\,{s^{ - 1}}$$
C.
$${v_s} = 62\,km\,{s^{ - 1}}$$
D.
$${v_s} = 72\,km\,{s^{ - 1}}$$
Answer :
$${v_s} = 42\,km\,{s^{ - 1}}$$
Solution :
$$\eqalign{
& \frac{1}{2}mV_e^2 - \frac{{G{M_e}m}}{{{R_e}}} - \frac{{G{M_e}m \times 3 \times {{10}^5}}}{{2.5 \times {{10}^4}{R_e}}} = 0 \cr
& \frac{{V_e^2}}{2} = \frac{{G{M_e}}}{{{R_e}}}\left[ {1 + \frac{{3 \times {{10}^5}}}{{2.5 \times {{10}^4}}}} \right] \cr
& {V_e} = \sqrt {13\left( {\frac{{2G{M_e}}}{{{R_e}}}} \right)} = \sqrt {13} \times 11.2 \approx 42 \cr} $$