MHT CET · Physics · Gravitation
A body is projected vertically from earth's surface with velocity equal to half the escape velocity. The maximum height reached by the satellite is ( \(R=\) radius of earth)
- A \(\mathrm{R}\)
- B \(\frac{\mathrm{R}}{2}\)
- C \(\frac{\mathrm{R}}{3}\)
- D \(\frac{\mathrm{R}}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{R}}{3}\)
Step-by-step Solution
Detailed explanation
Given: \(\mathrm{v}=\frac{\mathrm{v}_{\mathrm{c}}}{2}\)
If body is projected with velocity \(\mathrm{v}\left(\mathrm{v} < \mathrm{v}_e\right)\) then height up to which it will rise, \(h=\frac{R}{\left(\frac{v_c^2}{v^2}-1\right)}\)
\(\therefore \quad \mathrm{h}=\frac{\mathrm{R}}{\left(\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}} / 2}\right)^2-1}=\frac{\mathrm{R}}{4-1}=\frac{\mathrm{R}}{3}\)
If body is projected with velocity \(\mathrm{v}\left(\mathrm{v} < \mathrm{v}_e\right)\) then height up to which it will rise, \(h=\frac{R}{\left(\frac{v_c^2}{v^2}-1\right)}\)
\(\therefore \quad \mathrm{h}=\frac{\mathrm{R}}{\left(\frac{\mathrm{v}_{\mathrm{e}}}{\mathrm{v}_{\mathrm{e}} / 2}\right)^2-1}=\frac{\mathrm{R}}{4-1}=\frac{\mathrm{R}}{3}\)
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