Question

If the escape velocity at the earth is 11.2 km/s, then what will be the escape velocity of planet having mass 4 times of earth and gravitational acceleration equal to earth?

• A. 7.9 m/s
• B. 11.2 km/s
• C. 15.7 km/s
• D. None of these

Answer 1

Answer: (C)

15.7 km/s

Answer 2

${{v}_{e}}=\sqrt{2gR}$ $=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2GM}{{{(GM/g)}^{1/2}}}}$ $=\sqrt{2}.{{(GM)}^{1/4}}{{g}^{1/4}}$ $\left[ \because g=\frac{GM}{{{R}^{2}}} \right]$ ${{v}_{e}}\propto {{M}^{1/4}}$ $\therefore$ $\frac{{{v}_{p}}}{{{v}_{e}}}={{\left( \frac{4M}{M} \right)}^{1/4}}=\sqrt{2}$ ${{v}_{p}}=\sqrt{2}.{{v}_{e}}=\sqrt{2}\times 11.2=15.7\,km/s$