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: Fix the position.
The particle rests at a height $3R$ above the surface. So its distance from the Earth's centre is the radius plus that height, $r = R + 3R = 4R$.

Step 2: Decide the meaning of 'not return'.
For the particle never to fall back, it must just barely escape, that is reach very far away with almost no speed left. That is the escape speed at that spot.

Step 3: Recall the escape speed formula.
At a distance $r$ from the centre, the escape speed is $v_e = \sqrt{\frac{2GM}{r}}$.

Step 4: Put in the distance.
Here $r = 4R$, so $v_e = \sqrt{\frac{2GM}{4R}}$.

Step 5: Simplify inside the root.
The two in the top and the four in the bottom reduce to one half: $v_e = \sqrt{\frac{GM}{2R}}$.

Step 6: State the answer.
So the smallest launch speed needed is $\left[\frac{GM}{2R}\right]^{1/2}$.
\[ \boxed{v_e = \left[\frac{GM}{2R}\right]^{1/2}} \]