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

Answer:

27h

Explanation:

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

$\frac{T^2}{a^3} =constant$

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.

$\frac{T^2}{r^3} =constant$

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:

$\frac{T_{1}^2}{r^3} =\frac{T_{2}^2}{(9r)^3}$

$T_{1}$ is the original period: 1 hour

$T_{2}$ is the period of the second satellite.

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

thus:

$\frac{(1h)^2}{r^3} =\frac{T_{2}^2}{729r^3}$

Clearing for $T_{2}$

$\frac{(1h)^2}{r^3}(729r^3) ={T_{2}^2}$

$729=T_{2}^2$

$\sqrt{729}=T_{2}$

$27=T_{2}$

The period of the second satellite is 27 hours