Question

The masses and radii of the earth and moon are (\(M_1, R_1\)) and (\(M_2, R_2\)) respectively. Their centres are at a distance 'r' apart. Find the minimum escape velocity for a particle of mass 'm' to be projected from the middle of these two masses :

• A. \(V = \frac{\sqrt{2G(M_1 + M_2)}}{r}\)
• B. \(V = \frac{1}{2} \sqrt{\frac{2G(M_1 + M_2)}{r}}\)
• C. \(V = \sqrt{\frac{4G(M_1 + M_2)}{r}}\)
• D. \(V = \frac{1}{2} \sqrt{\frac{4G(M_1 + M_2)}{r}}\)

Hint:

Potential is a scalar. Sum the individual potentials due to all nearby masses before applying energy conservation for escape velocity.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Escape velocity is the minimum velocity required for an object to escape the gravitational field of a system, meaning its total mechanical energy at infinity becomes zero.
Step 2: Key Formula or Approach:
Total Energy = Kinetic Energy + Potential Energy = 0.
Potential Energy at midpoint (distance \(r/2\) from both): \( U = -\frac{GM_1m}{r/2} - \frac{GM_2m}{r/2} \).
Step 3: Detailed Explanation:
\[ U = -\frac{2Gm}{r}(M_1 + M_2) \] By conservation of energy:
\[ \frac{1}{2} m V^2 + U = 0 \] \[ \frac{1}{2} m V^2 - \frac{2Gm}{r}(M_1 + M_2) = 0 \] \[ V^2 = \frac{4G(M_1 + M_2)}{r} \] \[ V = \sqrt{\frac{4G(M_1 + M_2)}{r}} \] Step 4: Final Answer:
The minimum escape velocity is \(\sqrt{\frac{4G(M_1 + M_2)}{r}}\).