## Satellites and Kepler's Laws

Read this text, which includes visual diagrams of Kepler's Laws of Planetary Motion, which describe the motion of planets around the sun. We can also apply these laws to explain the motion of satellites around planets.

1. Kepler's First Law of Planetary Motion states that planets move around the sun in an ellipse shaped orbit with the sun at the center of the ellipse (see Figure 6.29).
2. Kepler's Second Law of Planetary Motion states that planets move so that a point on the planet sweeps an equal area in equal times (see Figure 6.30).
3. Kepler's Third Law of Planetary Motion refers to the relationship between the time it takes for two planets to revolve around the sun, and their distances from the sun: $\frac{T_1^2}{T_2^2}=\frac{r_1^3}{r_2^3}$, where $T_{1}$ and $T_{2}$ are periods of orbit while $r_{1}$ and $r_{2}$ are radii for planets one and two.

We can use Kepler's Third Law to solve problems to determine the period for planetary or satellite orbits. See a worked example of using the equation from Kepler's Third Law to determine the period of a satellite in Example 6.7. Pay attention to the derivation of Kepler's Third Law using the concept of centripetal forces.

### Kepler's First Law

The orbit of each planet about the Sun is an ellipse with the Sun at one focus.

Figure 6.29 (a) An ellipse is a closed curve such that the sum of the distances from a point on the curve to the two foci $\left(f_{1}\right.$ and $\left.f_{2}\right)$ is a constant. You can draw an ellipse as shown by putting a pin at each focus, and then placing a string around a pencil and the pins, and tracing a line on paper. A circle is a special case of an ellipse in which the two foci coincide (thus any point on the circle is the same distance from the center). (b) For any closed gravitational orbit, $m$ follows an elliptical path with $M$ at one focus. Kepler's first law states this fact for planets orbiting the Sun.

### Kepler's Second Law

Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal times (see Figure 6.30).

### Kepler's Third Law

The ratio of the squares of the periods of any two planets about the Sun is equal to the ratio of the cubes of their average distances from the Sun. In equation form, this is

$\dfrac{T_{1}^{2}}{T_{2}^{2}}=\dfrac{r_{1}^{3}}{r_{2}^{3}}$

where $T$ is the period (time for one orbit) and $r$ is the average radius. This equation is valid only for comparing two small masses orbiting the same large one. Most importantly, this is a descriptive equation only, giving no information as to the cause of the equality.

Figure 6.30 The shaded regions have equal areas. It takes equal times for $m$ to go from A to B, from C to D, and from E to F. The mass $m$ moves fastest when it is closest to $M$. Kepler's second law was originally devised for planets orbiting the Sun, but it has broader validity.

Note again that while, for historical reasons, Kepler's laws are stated for planets orbiting the Sun, they are actually valid for all bodies satisfying the two previously stated conditions.

#### Example 6.7 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 $3.84 \times 10^{8} \mathrm{~m}$ from the center of Earth, calculate the period of an artificial satellite orbiting at an average altitude of 1500 km above Earth's surface.

#### 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 $\dfrac{T_{1}^{2}}{T_{2}^{2}}=\dfrac{r_{1}^{3}}{r_{2}^{3}}$.

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 $r_{1}=3.84 \times 10^{8} \mathrm{~m}$, and that the period of the Moon is $T_{1}=27.3 \mathrm{~d}$.

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 $r_{2}=(1500+6380) \mathrm{km}=7880 \mathrm{~km}$. Now all quantities are known, and so $T_{2}$ can be found.

#### Solution

Kepler's third law is

$\dfrac{T_{1}^{2}}{T_{2}^{2}}=\dfrac{r_{1}^{3}}{r_{2}^{3}}$

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

\begin{aligned} &T_{2}^{2}=T_{1}^{2}\left(\dfrac{r_{2}}{r_{1}}\right)^{3} \\&T_{2}=T_{1}\left(\dfrac{r_{2}}{r_{1}}\right)^{3 / 2}\end{aligned}

Substituting known values yields

\begin{aligned} T_{2} &=27.3 \mathrm{~d} \times \dfrac{24.0 \mathrm{~h}}{\mathrm{~d}} \times\left(\dfrac{7880 \mathrm{~km}}{3.84 \times 10^{5} \mathrm{~km}}\right)^{3 / 2} \\ &=1.93 \mathrm{~h} \end{aligned}

#### 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.

People immediately search for deeper meaning when broadly applicable laws, like Kepler’s, are discovered. It was Newton who took the next giant step when he proposed the law of universal gravitation. While Kepler was able to discover what was happening, Newton discovered that gravitational force was the cause.