A satellite of mass 5500 kg orbits the Earth and has a period of 6600 s.Determine (a) the radius of its circular orbit, (b) the magnitude of theEarth’s gravitational force on the satellite, and (c) the altitude of thesatellite.

(a) If the radius of the orbit is\(r\)and the period of the orbit is\(T = 6600\;{\rm{s}}\), then the orbiting speed is given by:

\(v = \frac{{2\pi r}}{T}\)

Let\({m_{\rm{E}}} = 5.98 \cdot {10^{24}}\;{\rm{kg}}\)be the mass of the Earth and\(m\)be the mass of the satellite.The universal law of gravity gives the gravitational force that the Earth exerts on this satellite as:

\(F = \;G\frac{{m{m_E}}}{{{r^2}}}\)

Here G is the universal gravitational constant.

The gravitational force provides the necessary centripetal force for executing the circular motion.Therefore,

\(\begin{aligned}G\frac{{m{m_{\rm{E}}}}}{{{r^2}}} &= \;\frac{{m{v^2}}}{r}\\G\frac{{m{m_{\rm{E}}}}}{{{r^2}}}\; &= \;m\frac{{4{\pi ^2}r}}{{{T^2}}}\end{aligned}\)

Now, the radius is given by:

\(r = \;\sqrt(3){{\frac{{G{m_{\rm{E}}}{T^2}}}{{4{\pi ^2}}}}} = 7600\;\;{\rm{km}}\)

So, the radius of the orbit is \(7600\;{\rm{km}}\).

(b) As explained in the previous step, the gravitational force on the satellite is given by:

\(\begin{aligned}F &= G\frac{{m{m_{\rm{E}}}}}{{{r^2}}}\\ &= Gm{m_{\rm{E}}}{\left( {\frac{{4{\pi ^2}}}{{G{m_{\rm{E}}}{T^2}}}} \right)^{\frac{2}{3}}}\\ &= 38\;{\rm{kN}}\end{aligned}\)

So, the gravitational force the Earth exerts on the satellite is \(38\;{\rm{kN}}\).

(c)The altitude of the satellite is the distance from the satellite to the surface of the Earth. This is the difference between the radius of the orbit and the radius of the Earth. Therefore,

\(\begin{aligned}h &= r - {r_{\rm{E}}}\\ &= 7600\;{\rm{km}} - 6380\;{\rm{km}}\\ &= 1200\;\;{\rm{km}}\end{aligned}\)

You have used \({r_{\rm{E}}} = 6380\;\;{\rm{km}}\).

So, the altitude of the satellite is \(1200\;{\rm{km}}\).