Mechanics — Hard JEE physics MCQ
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 $v_{e}=11.2 \mathrm{~km} \mathrm{~s}^{-1}$. 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)
$[\mathrm{A}] \quad v_{S}=22 \mathrm{~km} \mathrm{~s}^{-1}$
$[\mathrm{B}] v_{S}=42 \mathrm{~km} \mathrm{~s}^{-1}$
$[\mathrm{C}] \quad v_{S}=62 \mathrm{~km} \mathrm{~s}^{-1}$
D. $v_{S}=72 \mathrm{~km} \mathrm{~s}^{-1}$
- A.
- B.
- C.
- D.
Solution
The correct option is **B**.
