MHT CET · Physics · Gravitation
The ratio of the acceleration due to gravity on two planets \(\mathrm{P}_1\) and \(\mathrm{P}_2\) is \(\mathrm{K}_1\). The ratio of their respective radii is \(\mathrm{K}_2\). The ratio of their respective escape velocities is
- A \(\sqrt{\mathrm{K}_1 \mathrm{~K}_2}\)
- B \(\sqrt{2 \mathrm{~K}_1 \mathrm{~K}_2}\)
- C \(\sqrt{\frac{\mathrm{K}_1}{\mathrm{~K}_2}}\)
- D \(\sqrt{\frac{\mathrm{K}_2}{\mathrm{~K}_1}}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{\mathrm{K}_1 \mathrm{~K}_2}\)
Step-by-step Solution
Detailed explanation
Since escape velocity from the surface of a planet can be written as \(v_e=\sqrt{2 g \mathrm{R}}\)
\(\text { ratio } \frac{v_{e p_1}}{v_{e p_2}}=\sqrt{\frac{2 g_{p_1} \mathrm{R}_{p_1}}{2 g_{p_2} \mathrm{R}_{p_2}}}=\sqrt{\mathrm{K}_1 \mathrm{~K}_2}\)
\(\text { ratio } \frac{v_{e p_1}}{v_{e p_2}}=\sqrt{\frac{2 g_{p_1} \mathrm{R}_{p_1}}{2 g_{p_2} \mathrm{R}_{p_2}}}=\sqrt{\mathrm{K}_1 \mathrm{~K}_2}\)
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