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Classical Mechanics Level 1

The acceleration due to gravity on Moon( $g_m$ ) is $\frac16$ that of Earth( $g_e$ ) and the radius of the Moon( $R_m$ )is $\frac14$ that of Earth( $R_e$ ). Find the escape velocity of Moon in terms of escape velocity of Earth.

The answer is in the from $\dfrac{v_e}{\sqrt{\overline{ab}}}$ where $\overline{ab}$ is a two digit number, find $a^b$ .

16 9 36 1

1 solution

Sravanth C.
Sep 30, 2015

We have been given; $\dfrac{g_e}{6}=g_m;\dfrac{R_e}4=R_m$

We know that, $v_e=\sqrt{2g_eR_e}\\\therefore v_m= \sqrt{2\dfrac{g_e\times R_e}{6\times4}}\\ = \dfrac{\sqrt{2gR}}{\sqrt{24}} \\\boxed{v_e=\dfrac{v_e}{\sqrt{24}}}$

So, $a=2$ and $b=4$ , $a^b=2^4=\boxed{16}$ .

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