Question

A satellite orbits the earth in circular orbits. Find the period if:
a) The satellite is just above the surface of the earth.
b) The satellite is at an altitude of 300 km above the earth.
(Assume that air resistance can be neglected)
Provide your answers in minutes.

Answer 1

a) The gravitational acceleration just above the surface of the earth is $a_1= 9.8m/s^2$. For a circular orbit the following holds:

$$\dfrac{v^2}{R_1} = a_1$$

where $v$ is the speed of the satellite on this orbit and $R_1 = 6.4\times 10^6m$ is the radius of the Earth. By defintion, the speed is:

$$v = \dfrac{2\pi R_1}{T_1}$$

where $T_1$ is the period (time, required to cover the full orbit of length $2\pi R_1$ once). Substituting this into the first equation, obtain:

$$\dfrac{4\pi^2 R_1}{T_1^2} = a_1$$

Expressing $T_1$, get:

$$T_1 = 2\pi\sqrt{\dfrac{R_1}{a_1}}\\ T_1 = 2\pi\sqrt{\dfrac{6.4\times 10^6}{9.8}}\approx 5077s \approx 84.6\space min$$

b) The acceleration at an altitude of 300 km above the earth is:

$$a_2 = G\dfrac{M_E}{R_2^2}$$

where $G = 6.67\times 10^{-11}m^3kg^{-1}s^{-2}$ it the gravitational constant, $M_E = 5.97\times 10^{24}kg$ is the mass of the Earth, and $R_2 = R_1 + 300km = 6.7\times 10^{6}m$ it the radius of the orbit. The period can be found similarly to the part (a):

$$T_2 = 2\pi\sqrt{\dfrac{R_2}{a_2}} = 2\pi\sqrt{\dfrac{R_2^3}{GM_E}}\\ T_2 = 2\pi\sqrt{\dfrac{(6.7\times 10^{6})^3}{6.67\times 10^{-11}\cdot 5.97\times 10^{24}}} \approx 5445s\approx 90.8\space min$$

Answer. 84.6 min, and 90.8 min.