13.6.1 Kepler’s Third Law

A photo of a satellite orbiting the Earth. — The International Space Station appears to be floating, but it’s actually traveling at nearly $8000 m/s$ as it orbits the earth.

An important parameter of circular motion is the period. Recall that the period T is the time to complete one full orbit. The relationship among speed, radius, and period is

v = \frac{circumference}{period} = \frac{2 π r}{T}

(13.23)

We can find a relationship between a satellite’s period and the radius of its orbit by using Equation 13.22 for v:

v = \frac{2 π r}{T} = \sqrt{\frac{G M}{r}}

(13.24)

Squaring both sides and solving for T give

T^{2} = (\frac{4 π^{2}}{G M}) r^{3}

(13.25)

In other words, the square of the period is proportional to the cube of the radius. This is Kepler’s third law. You can see that Kepler’s third law is a direct consequence of Newton’s law of gravity.

TABLE 13.2 contains astronomical information about the solar system. We can use these data to check the validity of Equation 13.25. FIGURE 13.17 is a graph of log T versus $log r$ for all the planets in Table 13.2 except Mercury. Notice that the scales on each axis are increasing logarithmically—by factors of 10—rather than linearly. (Also, the vertical axis has converted T to the SI units of s.) As you can see, the graph is a straight line with a best-fit equation

log T = 1.500 log r - 9.264

Taking the logarithm of both sides of Equation 13.25, and using the logarithm properties ${log a}^{n} = n log a$ $and log (a b) = log a + log b$ , we have

log T = \frac{3}{2} log r + \frac{1}{2} log (\frac{4 π^{2}}{G M})

In other words, theory predicts that the slope of a $log T -versus-log r$ graph should be exactly $\frac{3}{2}$ . As Figure 13.17 shows, the solar-system data agree to an impressive four significant figures. A homework problem will let you use the y-intercept of the graph to determine the mass of the sun.

The figure illustrates a log T versus log r graph for the planetary data from Table 13.2.

FIGURE 13.17 Full Alternative Text

TABLE 13.2 Useful astronomical data

Planetary body	Mean distance from sun (m)	Period (years)	Mass (kg)	Mean radius (m)
Sun	—	—	$1.99 \times 10^{30}$	$6.96 \times 10^{8}$
Moon	$3.84 \times 10^{8}$ *	27.3 days	$7.34 \times 10^{22}$	$1.74 \times 10^{6}$
Mercury	$5.79 \times 10^{10}$	0.241	$3.30 \times 10^{23}$	$2.44 \times 10^{6}$
Venus	$1.08 \times 10^{11}$	0.615	$4.87 \times 10^{24}$	$6.05 \times 10^{6}$
Earth	$1.50 \times 10^{11}$	1.00	$5.97 \times 10^{24}$	$6.37 \times 10^{6}$
Mars	$2.28 \times 10^{11}$	1.88	$6.42 \times 10^{23}$	$3.39 \times 10^{6}$
Jupiter	$7.78 \times 10^{11}$	11.9	$1.90 \times 10^{27}$	$7.15 \times 10^{7}$
Saturn	$1.43 \times 10^{12}$	29.5	$5.68 \times 10^{26}$	$6.02 \times 10^{7}$
Uranus	$2.87 \times 10^{12}$	84.0	$8.68 \times 10^{25}$	$2.54 \times 10^{7}$
Neptune	$4.50 \times 10^{12}$	165	$1.03 \times 10^{26}$	$2.46 \times 10^{7}$

*Distance from earth.

A particularly interesting application of Equation 13.25 is to communications satellites that are in geosynchronous orbits above the earth. These satellites have a period of $24 h = 86, 400 s$ , making their orbital motion synchronous with the earth’s rotation. As a result, a satellite in such an orbit appears to remain stationary over one point on the earth’s equator. Equation 13.25 allows us to compute the radius of an orbit with this period:

\begin{array}{l} r_{geo} & = & R_{e} + h_{geo} = {[(\frac{G M}{4 π^{2}}) T^{2}]}^{1 / 3} \\ = & {[(\frac{(6.67 \times 10^{- 11} {Nm}^{2} {/kg}^{2}) (5.97 \times 10^{24} kg)}{4 π^{2}}) {(86, 400 s)}^{2}]}^{1 / 3} \\ = & 4.22 \times 10^{7} m \end{array}

The height of the orbit is

h_{geo} = r_{geo} - R_{e} = 3.59 \times 10^{7} m = 36, 000 km \approx 22, 000 mi

NOTE When you use Equation 13.25, the period must be in SI units of s.

Geosynchronous orbits are much higher than the low earth orbits used by the space station and remote-sensing satellites, where $h \approx 300 km$ . Communications satellites in geosynchronous orbits were first proposed in 1945 by science fiction writer Arthur C. Clarke, more than 10 years before the first artificial satellite of any type!

EXAMPLE 13.5 Extrasolar planets

In recent years, astronomers have discovered thousands of planets orbiting nearby stars. These are called extrasolar planets. Suppose a planet is observed to have a 1200 day period as it orbits a star at the same distance that Jupiter is from the sun. What is the mass of the star in solar masses? (1 solar mass is defined to be the mass of the sun.)

SOLVE Here “day” means earth days, as used by astronomers to measure the period. Thus the planet’s period in SI units is $T = 1200 days = 1.037 \times 10^{8} s$ . The orbital radius is that of Jupiter, which we can find in Table 13.2 to be $r = 7.78 \times 10^{11} m$ . Solving Equation 13.25 for the mass of the star gives

\begin{array}{l} M & = \frac{4 π^{2} r^{3}}{G T^{2}} = 2.59 \times 10^{31} kg \times \frac{1 solar mass}{1.99 \times 10^{30} kg} \\ = 13 solar masses \end{array}

REVIEW This is a large, but not extraordinary, star.