Question

The escape velocity from the earth's surface 11 km/sec. A certain planet has a radius twice that of the earth but its mean density is the same as that of the earth. The value of escape velocity from this planet would be

• A. 11 km/ sec
• B. 22 km/sec
• C. 33 km/sec
• D. 16.5km/sec

Answer 1

The correct option is B 22 km/sec

escape velocity, $v_e=\sqrt{\frac{2GM}{R}}where,M=\frac{4}{3}{\pi R}^{2}d\therefore v_e=\sqrt{\frac{2G\times \frac{4}{3}{\pi R}^{3}d}{R}=}\sqrt{\frac{8G\times {\pi R}^{2}d}{3}}\Rightarrow v_e\propto RNow,\frac{\left(v_e\right)earth}{\left(v_e\right)planet}=\frac{R_{earth}}{R_{planet}}\Rightarrow \frac{11}{v_e}=\frac{R}{2R}\Rightarrow v_e=22km/sec.$