Question

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. \(\sqrt{\frac{6G}{d}(2M_1+M_2)}\)
• B. \(\sqrt{\frac{6G}{d}(M_1-2M_2)}\)
• C. \(\sqrt{\frac{3G}{d}(M_1+2M_2)}\)
• D. \(\sqrt{\frac{8G}{d}(M_1-2M_2)}\)

Hint:

Add the potentials of both bodies at the launch point and set total energy to zero.

Answer 1

The Correct Option is C

Step 1: Energy per unit mass:
Escape needs $\tfrac12v^2 = |V|$.

Step 2: Evaluate |V|:
$|V| = \dfrac{3GM_1}{2d} + \dfrac{3GM_2}{d} = \dfrac{3G}{2d}(M_1 + 2M_2)$.

Step 3: Solve:
$v^2 = 2|V| = \dfrac{3G}{d}(M_1 + 2M_2)$, so $v = \sqrt{\dfrac{3G}{d}(M_1+2M_2)}$, option (C).

Final Answer:
The escape speed is root of 3G(M1 + 2M2)/d. \[ \boxed{\text{(C) }\sqrt{\dfrac{3G}{d}(M_1+2M_2)}} \]