Question

A particle of mass 'm' is kept at rest at a height \( 3R \) from the surface of earth, where 'R' is radius of earth and 'M' is the mass of earth. The minimum speed with which it should be projected, so that it does not return back is ( \( g = \) acceleration due to gravity on the earth's surface)

• A. \(\left[ \frac{GM}{2R} \right]^{1/2}\)
• B. \(\left[ \frac{gR}{4} \right]^{1/2}\)
• C. \(\left[ \frac{2g}{R} \right]^{1/2}\)
• D. \(\left[ \frac{GM}{R} \right]^{1/2}\)

Hint:

Escape velocity depends on distance from the center: \(v_e = \sqrt{2GM/r}\). At Earth’s surface \(r = R\), \(v_e = \sqrt{2gR}\). At height \(3R\), \(r = 4R\) and \(v_e\) becomes half of surface value? Check: \(\sqrt{GM/(2R)}\) vs \(\sqrt{2GM/R}\) ratio = 1/2, yes.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A particle is at height \(3R\) above Earth’s surface, so its distance from Earth’s center is \(r = R + 3R = 4R\). We need the minimum speed to escape (i.e., reach infinity with zero speed).

Step 2: Key Formula or Approach:
Escape speed from a distance \(r\) is \(v_e = \sqrt{\frac{2GM}{r}}\).

Step 3: Detailed Explanation:
Here \(r = 4R\). So: \[ v_e = \sqrt{\frac{2GM}{4R}} = \sqrt{\frac{GM}{2R}}. \] Option (A) matches exactly. (Note: \(GM = gR^2\) can be used to rewrite, but the given form is correct.)

Step 4: Final Answer:
The minimum speed is \(\left[ \frac{GM}{2R} \right]^{1/2}\), option (A).