Question

A body is at a point \(P\), at some height above the surface of a planet of mass \(M\) and radius \(R\). If the potential energy of the body at point \(P\) is half of its potential energy on the surface of the planet, then the difference between the escape velocity of the body from point \(P\) and its escape velocity from the surface of the planet is \[ (G=\text{Universal Gravitational Constant}) \]

• A. \(\sqrt{\dfrac{GM}{R}}(\sqrt3-\sqrt2)\)
• B. \(\sqrt{\dfrac{GM}{R}}(\sqrt3-1)\)
• C. \(\sqrt{\dfrac{GM}{R}}(\sqrt2-1)\)
• D. \(\sqrt{\dfrac{GM}{R}}(2-\sqrt2)\)

Hint:

Escape velocity at distance \(r\) from the centre is \[ \boxed{ v_e=\sqrt{\frac{2GM}{r}} } \] and gravitational potential energy is \[ \boxed{ U=-\frac{GMm}{r}. } \]

Answer 1

The Correct Option is C

Step 1: Determine the distance of point \(P\) from the planet's centre. Gravitational potential energy is \[ U=-\frac{GMm}{r}. \] Given, \[ U_P=\frac12 U_{\text{surface}}. \] Thus, \[ -\frac{GMm}{r} = \frac12\left(-\frac{GMm}{R}\right), \] which gives \[ \boxed{r=2R.} \]

Step 2: Find the escape velocities. Escape velocity at the surface: \[ v_s=\sqrt{\frac{2GM}{R}}. \] Escape velocity from point \(P\): \[ v_P=\sqrt{\frac{2GM}{2R}} =\sqrt{\frac{GM}{R}}. \]

Step 3: Calculate the difference. \[ v_s-v_P = \sqrt{\frac{GM}{R}} (\sqrt2-1). \] Hence, \[ \boxed{ \sqrt{\frac{GM}{R}}(\sqrt2-1) } \] Therefore, \[ \boxed{(C)} \] is the correct answer.