Question

A satellite in orbit around the earth has a period of one hour. An identical satellite is placed in an orbit having a radius that is nine times larger than that of the first satellite. What is the period of the second satellite?

Answer 1

27h

Step-by-step explanation:

We can answer this problem with Kepler's third law:

`(T^2)/(a^3) =constant`

where
`T` is the period and
`a` is the semi-major axis of an ellipse, but because the satellite's orbit is around the earth it must be a circular orbit, so that
`a` becomes the radius
`r` of a circle.

`(T^2)/(r^3) =constant`

what this tells us is that the relationship between the first period with radius
`r` must be equal to the second period when the radius is now 9 times the original (the second radius is
`9r` because the orbit is nine times larger):

We will have the following:

`(T_(1)^2)/(r^3) =(T_(2)^2)/((9r)^3)`

`T_(1)` is the original period: 1 hour

`T_(2)` is the period of the second satellite.

and
`(9r)^3=729r^3`

thus:

`((1h)^2)/(r^3) =(T_(2)^2)/(729r^3)`

Clearing for
`T_(2)`

`((1h)^2)/(r^3)(729r^3) ={T_(2)^2}`

`729=T_(2)^2`

`√(729)=T_(2)`

`27=T_(2)`

The period of the second satellite is 27 hours