Example 6.7 from College Physics, 6.6 Satellites and Kepler’s Laws: An Argument for Simplicity

Find the Time for One Orbit of an Earth Satellite

Given that the Moon orbits Earth each 27.3 d and that it is an average distance of $3.84 \times 10^{8} m$ from the center of Earth, calculate the period of an artificial satellite orbiting at an average altitude of 1500 km above Earth’s surface.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
The period, or time for one orbit, is related to the radius of the orbit by Kepler’s third law, given in mathematical form in $\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}$ . Let us use the subscript 1 for the Moon and the subscript 2 for the satellite. We are asked to find $T_{2}$ . The given information tells us that the orbital radius of the Moon is $r_{1} = 3 . 84 \times 10^{8} m$ , and that the period of the Moon is $T_{1} = 27.3 d$ . The height of the artificial satellite above Earth’s surface is given, and so we must add the radius of Earth (6380 km) to get $r_{2} = (1500 + 6380) km = 7880 km$ . Now all quantities are known, and so $T_{2}$ can be found.
Solution
Kepler’s third law is

$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}} .$
6.56

To solve for $T_{2}$ , we cross-multiply and take the square root, yielding

$T_{2}^{2} = T_{1}^{2} {(\frac{r_{2}}{r_{1}})}^{3}$
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$T_{2} = T_{1} {(\frac{r_{2}}{r_{1}})}^{3 / 2} .$
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Substituting known values yields

$\begin{array}{l} T_{2} & = & 27.3 d \times \frac{24.0 h}{d} \times {(\frac{7880 km}{3.84 \times 10^{5} km})}^{3 / 2} \\ = & 1.93 h. \end{array}$
6.59
Discussion
This is a reasonable period for a satellite in a fairly low orbit. It is interesting that any satellite at this altitude will orbit in the same amount of time. This fact is related to the condition that the satellite’s mass is small compared with that of Earth.

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