JEE Mains · Physics · STD 11 - 7. gravitation
If earth has a mass nine times and radius twice to the of a planet \(P\). Then \(\frac{v_e}{3} \sqrt{x}\; ms ^{-1}\) will be the minimum velocity required by a rocket to pull out of gravitational force of \(P\), where \(v_e\) is escape velocity on earth. The value of \(x\) is
- A \(2\)
- B \(3\)
- C \(18\)
- D \(1\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
\(v_{\text {(escape) plant }}=\sqrt{\frac{2 G M_P}{R_P}}\) \(=\sqrt{\frac{2 G\left(\frac{M_e}{9}\right)}{\left(\frac{R_e}{2}\right)}}=\frac{v_e \sqrt{2}}{3} \therefore x=2\)
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