MHT CET · Physics · Gravitation
A body of mass \(m\) is dropped from a height \(\frac{R}{2}\), from the surface of earth where \(R\) is radius of earth. Its speed when it will hit the earth's surface is \(\left(v_e=\right.\) escape \(v\) elocity from earth' surface)
- A \(\sqrt{2} V_e\)
- B \(\frac{V_e}{\sqrt{3}}\)
- C \(\frac{V_e}{\sqrt{2}}\)
- D \(\sqrt{3} V_e\)
Answer & Solution
Correct Answer
(B) \(\frac{V_e}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
Considering the total energy conservation:
\(\begin{aligned}
& -\frac{G M m}{R+\frac{R}{2}}=-\frac{G M m}{R}+\frac{1}{2} m v^2 \\
& \Rightarrow v=\sqrt{\frac{2 G M}{3 R}}=\frac{v_e}{\sqrt{3}}
\end{aligned}\)
\(\begin{aligned}
& -\frac{G M m}{R+\frac{R}{2}}=-\frac{G M m}{R}+\frac{1}{2} m v^2 \\
& \Rightarrow v=\sqrt{\frac{2 G M}{3 R}}=\frac{v_e}{\sqrt{3}}
\end{aligned}\)
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