Question

Question #a2b13

Answer 1

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.