Question

. A 2000-kg communication satellite is placed in the geosynchronous orbit around the Earth. Find its speed and altitude. Give the mass of the earth is 5.98x1024kg, the radius of the Earth is 6.38x106
meter.

Answer 1

According to the https://en.wikipedia.org/wiki/Geosynchronous_orbit#Period, the altitude of the satellite is given by:

$$h = \sqrt[3]{GM_{Earth}\cdot\dfrac{T^2}{4\pi^2}}$$

where $G = 6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant, $M_{Earth} = 5.98\cdot 10^{24} kg$ is the mass of Earth, $T$ is the period of satellite revolution. On the geosynchronous orbit $T = 24 h = 86400 s$.

Thus:

$$h = \sqrt[3]{6.67\cdot 10^{-11}\cdot 5.98\cdot 10^{24}\cdot\dfrac{86400^2}{4\pi^2}} = 42\space227 km$$

The acceleration on this height is:

$$a = G\dfrac{M_{Earth}}{(R + h)^2} = 6.67\cdot 10^{-11} \dfrac{5.98\cdot 10^{24}}{(6.38\cdot 10^6 + 42.2 \cdot10^{6})^2} = 0.17 m/s^2$$

where $R = 6.38\cdot10^6$ is the radius of the Earth

Assuming the circular orbit, the speed will be:

$$v = \sqrt{a(R + h)} = \sqrt{0.17(6.38\cdot 10^6 + 42.2 \cdot10^{6})} = 2.87\cdot10^3 m/s = 2.87 km/s$$

Answer. v = 2.87 km/s, h = 42.2 km.