Question

The escape velcoity of earth is 11.2 Km/s. Find the escape velocity of planet with mass half of that earth and the radius is two times greater than that of earth
1. 5.1 Km/Sec
2. 5.4 Km/Sec
3. 5.6 Km/Sec
4. 6.2 Km/Sec

Answer 1

Correct Answer - Option 3 : 5.6 Km/Sec

CONCEPT: 

• Escape velocity: The minimum velocity required to escape the gravitational field of the earth is called escape velocity. It is denoted by V_e and is given by

\(\Rightarrow V _{e}= \sqrt{\frac{2GM_{e}}{R_e}}\)

Where Ve = Escape velocity, G = Universal gravitational constant, Me = Mass of earth, and R_e =  radius of earth

• The escape velocity of the earth is 11.2 Km/Sec

CALCULATION :

Given - Mass of the planet M_P = M_e/2 , The radius of planet R_p = 2R_e 

• The escape velocity of earth is given by

\(\Rightarrow V _{e}= \sqrt{\frac{2GM_e}{R_e}}\)

• The escape velocity of the planet is given by

\(\Rightarrow V _{p}= \sqrt{\frac{2GM_p}{R_p}}= \sqrt{\frac{2GM_e}{4R_e}}=\frac{1}{2}\sqrt{\frac{2GM_e}{R_e}}=\frac{V_e}{2}=\frac{11.2}{2}=5.6\, km/sec\)

• Thus the escape velocity of the planet is 5.6 Km/s
• Hence, option 3 is the answer