Answer to Question #88943 in Mechanics | Relativity for Zebib

Question

You are in charge of a project that requires having a communications satellite orbiting the earth as fast as possible. A Canadian corporation has said that the best they can do is to provide a 50 kg satellite that will orbit the earth is 5 * 10^3 s. An American company said that because of advance lightweight materials they can build the same satellite with a mass of 12.5 kg and have it orbit at 7.94 * 10^3 m/s. which satellite will you order? Show calculation to justify your choice.

Answer 1

We have two satellites orbiting the Earth.

Given:

Canadian satellite: mass is 50 kg, period is 5000 s.

American satellite: mass is 12.5 kg, speed is 7940 m/s.

According to third law of Kepler with the provided information we can calculate the semi-major axis for the Canadian satellite:

$$a_1=\sqrt[3]{\frac{T_1^2G(M+m_1)}{4\pi^2}}=$$

$$=\sqrt[3]{\frac{(5\cdot10^3)^2\cdot6.673\cdot10^{-11}(5.974\cdot10^{24}+50)}{4\pi^2}}=6320075\text{ m},$$

or 6320 km. Very interesting, since the Earth's radius is 6371 km, and the satellite will dig the Earth 41 km below its surface. Indeed, just imagine the Canadian satellite orbiting the Earth in 1 hour 23 minutes 20 seconds (5000 seconds)! Just impossible. The least possible period for a satellite is 89 minutes - 6 minutes more than the Canadian corporation promises.

Let's perform some calculations for the American satellite. Since its speed is higher than 7.9 km/s (orbital velocity, rarely "first cosmic velocity") and less that the escape velocity of 11.2 km/s, it will have elliptical low Earth orbit. Otherwise (in case of circular orbit) the american satellite with its parameters will dig 48 km below the planet's surface, which is nonsense too.

Low Earth orbits range from 200 to 2000 km above the surface of the Earth, and since they're elliptical, the velocity must change: reach its maximum at perigee and its minimum at apogee. The American corporation promises velocity of 7.94 km/s, hence determine the parameter responsible for current distance between centers of the Earth and the satellite $r$ and semi-major axis $a_2$:

$$v_2^2=G(M+m_2)\Big(\frac{2}{r}-\frac{1}{a_2}\Big),$$

$$\frac{2}{r}-\frac{1}{a_2}=\frac{7940^2}{6.673\cdot10^{-11}(5.974\cdot10^{24}+12.5)}=1.58\cdot 10^{-7}\text{ m}^{-1},$$

$$a_2(r)=\frac{r}{2-r\cdot1.58\cdot 10^{-7}}\textbf{ meters}.$$

If we choose current location of the satellite 6,700,000 m from center of the Earth, the semi-major axis will be 7117 km, which is quite appropriate result. However, at such low distances from earth satellites' lifetime reduces quickly because of high air friction, and choosing the american Satellite is not reasonable also. Of course, perhaps Americans wanted to use their satellite at Molniya orbit, but periods there are about 12 hours.

The most effective communication satellites are geostationary ones, therefore neither of the two considered satellites is fine. So, if I were to choose, I'd rather ask Russians to design a satellite :)