Test Index

Gravitation Part 1

Section: Physics

Question : 15 of 100

Marks: +1, -0

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

Solution:

Applying energy conservation from (1) to (2)

\frac{1}{2} m \cdot \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}

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

\sqrt{2 G M} \int\limits_{0}^{t} d t = \int\limits_{R_{e}}^{R_{e}+h} \left(\sqrt{R+r}\right) 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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