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7. gravitation — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics7. gravitation1 Mark
MCQ
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 $....\,S.$
  • ✓
    $\frac{1}{3} \sqrt{\frac{2 R_{e}}{g}}\left[\left(1+\frac{h}{R_{e}}\right)^{3 / 2}-1\right]$
  • 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

Correct option: A.
$\frac{1}{3} \sqrt{\frac{2 R_{e}}{g}}\left[\left(1+\frac{h}{R_{e}}\right)^{3 / 2}-1\right]$
a
Applying energy conservation from $(1)$ to $(2)$

$\frac{1}{2} m_{.}\left(\frac{2 G M}{R_{e}}\right)-\frac{G M m}{R_{e}}=\frac{1}{2} m v^{2}-\frac{G M m}{R+r}$

$\Rightarrow \frac{1}{2} m v^{2}=\frac{G M m}{R+r}$

$\Rightarrow v=\sqrt{\frac{2 G M}{R+r}}=\frac{d r}{d t}$

$\Rightarrow \sqrt{2 G M} \int_{0}^{t} d t=\int_{R_{e}}^{R_{e}+h}(\sqrt{R+r}) d r$

$\sqrt{2 G M} \cdot t=\frac{2}{3}\left[(R+r)^{3 / 2}\right]_{R_{e}}^{R_{e}+h}$

$t=\frac{2}{3} \sqrt{\frac{R_{e}^{3}}{2 G M}}\left[\left(1+\frac{h}{R_{e}}\right)^{3 / 2}-1\right]$

$\frac{G M}{R_{e}^{2}}=g$

$t=\frac{1}{3} \sqrt{\frac{2 R_{e}}{g}}\left[\left(1+\frac{h}{R_{e}}\right)^{3 / 2}-1\right]$

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