Question

A planet having mass \(9M_e\) and radius \(4R_e\), where \(M_e\) and \(R_e\) are mass and radius of earth respectively, has escape velocity in km/s given by: (Given escape velocity on earth \(V_e= 11.2 × 10^3 m/s\))

• A. 67.2
• B. 33.6
• C. 16.8
• D. 11.2

Hint:

{Escape Velocity}: The escape velocity is the minimum speed needed for an object to "break free" from the gravitational attraction of a massive body without further propulsion. It depends on the mass and radius of the celestial body.

Answer 1

The Correct Option is C

To determine the escape velocity of a planet with mass \(9M_e\) and radius \(4R_e\), we can use the formula for escape velocity:

Escape velocity, \(V = \sqrt{\frac{2GM}{R}}\)

Here, \(G\) is the universal gravitational constant, \(M\) is the mass of the planet, and \(R\) is the radius of the planet. The escape velocity for Earth is given as \(V_e = 11.2 \times 10^3 \text{ m/s}\).

For Earth, the escape velocity can be expressed as:

\(V_e = \sqrt{\frac{2GM_e}{R_e}}\).

For the new planet, the escape velocity, \(V_{\text{planet}}\), is given by:

\(V_{\text{planet}} = \sqrt{\frac{2G \times 9M_e}{4R_e}} = \sqrt{\frac{9}{4}} \times \sqrt{\frac{2GM_e}{R_e}}\).

Therefore, \(V_{\text{planet}} = \frac{3}{2} \times V_e\).

Substituting the given escape velocity of Earth, \(V_e = 11.2 \times 10^3 \text{ m/s}\), we get:

\(V_{\text{planet}} = \frac{3}{2} \times 11.2 \text{ km/s} = 16.8 \text{ km/s}\).

Therefore, the escape velocity of the planet is 16.8 km/s.

Thus, the correct answer is 16.8.