MHT CET · Physics · Gravitation
A body is projected vertically upwards from earth's surface. If velocity of projection is \(\left(\frac{1}{3}\right)^{\text {rd }}\) of escape velocity, then the height upto which a body rises is
\((\mathrm{R}=\) radius of earth \()\)
- A \(2 \mathrm{R}\)
- B \(\frac{\mathrm{R}}{4}\)
- C \(\frac{\mathrm{R}}{2}\)
- D \(\mathrm{R}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{R}}{2}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{V}=\frac{1}{3} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}\)
\(\mathrm{TE}_{\mathrm{At} \text { surface }}=\mathrm{TE}_{\text {at height } \mathrm{h}}\)
\(-\frac{G M m}{R}+\frac{1}{2} m v^{2}=-\frac{G M m}{R+h}+0\)
\(-\frac{G M m}{R}+\frac{1}{2} m \frac{1}{9} \times \frac{2 G M}{R}=-\frac{G M m}{R+h}\)
\(\left(\frac{1}{9}-1\right) \frac{G M m}{R}=-\frac{G M m}{R+h}\)
\(-\frac{8}{9} \frac{G M m}{R}=-\frac{G M m}{R+h}\)
\(\frac{8}{9} \frac{1}{R}=\frac{1}{R+h}\)
\(\Rightarrow 8 \mathrm{R}+8 \mathrm{h}=9 \mathrm{R}\)
\(8 h=R \quad \therefore h=\frac{R}{8}\)
\(\mathrm{TE}_{\mathrm{At} \text { surface }}=\mathrm{TE}_{\text {at height } \mathrm{h}}\)
\(-\frac{G M m}{R}+\frac{1}{2} m v^{2}=-\frac{G M m}{R+h}+0\)
\(-\frac{G M m}{R}+\frac{1}{2} m \frac{1}{9} \times \frac{2 G M}{R}=-\frac{G M m}{R+h}\)
\(\left(\frac{1}{9}-1\right) \frac{G M m}{R}=-\frac{G M m}{R+h}\)
\(-\frac{8}{9} \frac{G M m}{R}=-\frac{G M m}{R+h}\)
\(\frac{8}{9} \frac{1}{R}=\frac{1}{R+h}\)
\(\Rightarrow 8 \mathrm{R}+8 \mathrm{h}=9 \mathrm{R}\)
\(8 h=R \quad \therefore h=\frac{R}{8}\)
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