Q115 / 150 · MHT CET 2026 (16 April Shift 1) · Physics
The masses and radii of the earth and moon are $M_1$, $R_1$ and $M_2$, $R_2$ and respectively, Their centres are at a distance 'd' apart. The minimum speed with which body of mass 'm' should be projected from a distance $2d/3$ from the centre of $M_1$ so as to escape to infinity is
- A$\sqrt{\dfrac{6G}{d}(2M_1 + M_2)}$
- B$\sqrt{\dfrac{6G}{d}(M_1 - 2M_2)}$
- Correct answerC$\sqrt{\dfrac{3G}{d}(M_1 + 2M_2)}$
- D$\sqrt{\dfrac{8G}{d}(M_1 - 2M_2)}$
Solution
Let the body of mass $m$ be projected from a point $P$ on the line joining the centres of the two bodies.
The distance of point $P$ from the centre of $M_1$ is $r_1 = \dfrac{2d}{3}$.
The distance of point $P$ from the centre of $M_2$ is $r_2 = d - \dfrac{2d}{3} = \dfrac{d}{3}$.
The total gravitational potential energy of the body at point $P$ is the sum of the potential energies due to $M_1$ and $M_2$:
$U = -\dfrac{G M_1 m}{r_1} - \dfrac{G M_2 m}{r_2}$
$U = -\dfrac{G M_1 m}{\dfrac{2d}{3}} - \dfrac{G M_2 m}{\dfrac{d}{3}}$
$U = -\dfrac{3G M_1 m}{2d} - \dfrac{3G M_2 m}{d}$
$U = -\dfrac{3G m}{2d} (M_1 + 2M_2)$
For the body to just escape to infinity, its total mechanical energy must be zero. By conservation of energy:
$K + U = 0$
$\dfrac{1}{2} m v^2 - \dfrac{3G m}{2d} (M_1 + 2M_2) = 0$
$\dfrac{1}{2} m v^2 = \dfrac{3G m}{2d} (M_1 + 2M_2)$
$v^2 = \dfrac{3G}{d} (M_1 + 2M_2)$
$v = \sqrt{\dfrac{3G}{d} (M_1 + 2M_2)}$
Answer: $\sqrt{\dfrac{3G}{d}(M_1 + 2M_2)}$
The distance of point $P$ from the centre of $M_1$ is $r_1 = \dfrac{2d}{3}$.
The distance of point $P$ from the centre of $M_2$ is $r_2 = d - \dfrac{2d}{3} = \dfrac{d}{3}$.
The total gravitational potential energy of the body at point $P$ is the sum of the potential energies due to $M_1$ and $M_2$:
$U = -\dfrac{G M_1 m}{r_1} - \dfrac{G M_2 m}{r_2}$
$U = -\dfrac{G M_1 m}{\dfrac{2d}{3}} - \dfrac{G M_2 m}{\dfrac{d}{3}}$
$U = -\dfrac{3G M_1 m}{2d} - \dfrac{3G M_2 m}{d}$
$U = -\dfrac{3G m}{2d} (M_1 + 2M_2)$
For the body to just escape to infinity, its total mechanical energy must be zero. By conservation of energy:
$K + U = 0$
$\dfrac{1}{2} m v^2 - \dfrac{3G m}{2d} (M_1 + 2M_2) = 0$
$\dfrac{1}{2} m v^2 = \dfrac{3G m}{2d} (M_1 + 2M_2)$
$v^2 = \dfrac{3G}{d} (M_1 + 2M_2)$
$v = \sqrt{\dfrac{3G}{d} (M_1 + 2M_2)}$
Answer: $\sqrt{\dfrac{3G}{d}(M_1 + 2M_2)}$
Hence, the correct answer is option: 'C'.