Question #a2b13

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Answer 1 · 2017-02-09T04:14:42

Please review for mathematical derivation here.

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

$v^2prop1/R_O$
where $R_O=R_p+h$
In this expression $R_p$ is the radius of planet and $h$ is the height of the satellite above planet's surface

Above equation implies that

$v^2xxR_O="Constant"$ ......(1)

Let $R_(ONew)$ be new orbital Radius when velocity of moon is increased by $42%$
Equation (1) becomes
$(1.42v)^2xxR_(ONew)="Constant"$ .....(2)

Dividing (2) with (1) we get
$((1.42v)^2xxR_(ONew))/(v^2xxR_O)=1$
$=>R_(ONew)=1/1.42^2 R_O$
$=>R_(ONew)=0.496 R_O$

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