Question

A satellite was placed in an orbit above the earth's atmosphere, at a distance of 15,000 km from the center of the earth. What is the orbital period (in hours) of the satellite?

Answer 1

The orbital period of this satellite is 5.08 hours (5 hours, 5 minutes, 3 seconds to be precise!) .

Explanation:

We need Kepler's third law for this problem, but in the form that is often referred to as "Newton's derivation"

`T^2=((4pi^2)/(GM))r^3`

where G is the gravitational constant `6.67xx10^(-11)` and `M` is the mass of the body that is being orbited (Earth in this case).

Also, care must be taken to use base units for `T, r and M`, as the value stated for `G` includes base units only.

Here is the solution:

`T^2=((4pi^2)/((6.67xx10^(-11))(5.97xx10^(24))))(1.5xx10^7)^3`

`T^2=3.35xx10^8s^2`

`T=18 292 s`

Divide by 3600 (seconds in each hour) and you get

`T=5.08` hours

Answer 2

The orbital period is `=5.077h`

Explanation:

Let the mass of the earth `=Mkg`

Let the mass of the satellite be `=mkg`

Let the radius of the earth be `=Rm`

Let the radius of the satellite's orbit be `=rm`

Let the gravitational constant be `=G`

There are `2` forces acting on the satellite, the gravitational force `F_G` anf the centripetal force `F_C`

`F_G=F_C`

`GMm/r^2=mv^2/r`

The velocity of the satellite is

`v^2=(GM)/r`

`v=sqrt((GM)/r)`

The orbital period is `=Ts`

The distance is `d=2pir`

Therefore,

`vT=2pir`

`T=(2pir)/v=2pirsqrt(r/(GM))`

`T=2pisqrt(r^3/(GM))`

`r=15000km=15*10^6m`

`G=6.67*10^-11Nm^2kg^-2`

`M=5.9810^24kg`

So,

`T=2pisqrt((15*10^6)^3/(6.67*10^-11*5.98*10^24))`

`=18276.9s`

`=5.077h`