Answer:
27h
Explanation:
We can answer this problem with Kepler's third law:
[tex]\frac{T^2}{a^3} =constant[/tex]
where [tex]T[/tex] is the period and [tex]a[/tex] 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 [tex]a[/tex] becomes the radius [tex]r[/tex] of a circle.
[tex]\frac{T^2}{r^3} =constant[/tex]
what this tells us is that the relationship between the first period with radius [tex]r[/tex] must be equal to the second period when the radius is now 9 times the original (the second radius is [tex]9r[/tex] because the orbit is nine times larger):
We will have the following:
[tex]\frac{T_{1}^2}{r^3} =\frac{T_{2}^2}{(9r)^3} [/tex]
[tex]T_{1}[/tex] is the original period: 1 hour
[tex]T_{2}[/tex] is the period of the second satellite.
and [tex](9r)^3=729r^3[/tex]
thus:
[tex]\frac{(1h)^2}{r^3} =\frac{T_{2}^2}{729r^3} [/tex]
Clearing for [tex]T_{2}[/tex]
[tex]\frac{(1h)^2}{r^3}(729r^3) ={T_{2}^2}[/tex]
[tex]729=T_{2}^2[/tex]
[tex]\sqrt{729}=T_{2}[/tex]
[tex]27=T_{2}[/tex]
The period of the second satellite is 27 hours
