JEE Mains · Physics · STD 11 - 7. gravitation
A body is projected vertically upwards from the surface of earth with a velocity sufficient enough to carry it to infinity. The time taken by it to reach height \(h\) is \(....\,S.\)
- A \(\frac{1}{3} \sqrt{\frac{2 R_{e}}{g}}\left[\left(1+\frac{h}{R_{e}}\right)^{3 / 2}-1\right]\)
- B \(\sqrt{\frac{2 R_{e}}{g}}\left[\left(1+\frac{h}{R_{e}}\right)^{3 / 2}-1\right]\)
- C \(\frac{1}{3} \sqrt{\frac{R_{e}}{g}}\left[\left(1+\frac{h}{R_{e}}\right)^{3 / 2}-1\right]\)
- D \(\sqrt{\frac{R_{e}}{g}}\left[\left(1+\frac{h}{R_{e}}\right)^{3 / 2}-1\right]\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{3} \sqrt{\frac{2 R_{e}}{g}}\left[\left(1+\frac{h}{R_{e}}\right)^{3 / 2}-1\right]\)
Step-by-step Solution
Detailed explanation
Applying energy conservation from \((1)\) to \((2)\) \(\frac{1}{2} m_{.}\left(\frac{2 G M}{R_{e}}\right)-\frac{G M m}{R_{e}}=\frac{1}{2} m v^{2}-\frac{G M m}{R+r}\) \(\Rightarrow \frac{1}{2} m v^{2}=\frac{G M m}{R+r}\) \(\Rightarrow v=\sqrt{\frac{2 G M}{R+r}}=\frac{d r}{d t}\)…
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