Question #a2b13

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Question #a2b13

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Answer 1 · 2017-02-09T04:14:42

Please review for mathematical derivation here.

We see that velocity $v$ $v$ of a satellite is connected to orbital radius $R_{O}$ $R_O$ as

$v^{2} \propto \frac{1}{R_{O}}$ $v^2prop1/R_O$
where $R_{O} = R_{p} + h$ $R_O=R_p+h$
In this expression $R_{p}$ $R_p$ is the radius of planet and $h$ $h$ is the height of the satellite above planet's surface

Above equation implies that

$v^{2} \times R_{O} = Constant$ $v^2xxR_O="Constant"$ ......(1)

Let $R_{O N e w}$ $R_(ONew)$ be new orbital Radius when velocity of moon is increased by $42 %$ $42%$
Equation (1) becomes
${(1.42 v)}^{2} \times R_{O N e w} = Constant$ $(1.42v)^2xxR_(ONew)="Constant"$ .....(2)

Dividing (2) with (1) we get
$\frac{{(1.42 v)}^{2} \times R_{O N e w}}{v^{2} \times R_{O}} = 1$ $((1.42v)^2xxR_(ONew))/(v^2xxR_O)=1$
$\Rightarrow R_{O N e w} = \frac{1}{{1.42}^{2}} R_{O}$ $=>R_(ONew)=1/1.42^2 R_O$
$\Rightarrow R_{O N e w} = 0.496 R_{O}$ $=>R_(ONew)=0.496 R_O$

We see that new orbit shrinks by about $49.6 %$ $49.6%$ . That is, moon comes nearer to earth to maintain equilibrium.

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Question #a2b13

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