Kepler's laws form the basis for most of the calculations that will be made in this chapter. Our first example deals with his law of equal areas.
1. The radius vector of a certain planet sweeps out an area of 36 million square miles in a
10-day period. How many square miles are swept out in a month?
Kepler's second law states that equal areas are swept out in equal time periods, so since a
month is about 30 days long, the area swept out in one month should be about (30/10) or 3 times
that swept out in 10 days:
3 (36 million sq. mi.) = 108 million sq. mi.
2. Show that Kepler's third law holds (approximately) for Earth and Venus.
We can use the data found in the Spotlight features on the terrestrial and Jovian planets to set
up this problem. We will use data in days and miles.
T2(Earth) / R3(Earth) = T2(Venus) / R 3(Venus)
(365.25 days)2 / (93 x 106 mi)3 = (687 days)2 / (142 x 106 mi)3
1.66 x 10-17 = 1.65 x 10-17
Since these numbers are very nearly equal, Kepler's third law is shown to hold for Earth
and Venus to very good agreement.
If we consider the units in years and astronomical units, we have a convenient value for the constant
k used in Kepler's third law. In this case " R " for Earth is 1 AU and " T " is
1 year, so the constant is simply:
k = T2 / R3 = (1 year)2 / (1 AU)3 = 1 year2/AU3
Let's apply this conversion factor to another problem.
3. What would be the period of a planet located at an average distance of 4.0 astronomical units from
the Sun?
Again k = T2 / R3
where k has the value calculated above, but the R (in AU) and T (in years) can now apply to any planet.
T2 / R3 = k
orT2 = k R3
T2 = (1 year2/AU3) x (4.0 AU)3
T2 = 64 years2
We must take the square root of both sides of this equation to get our final answer.
T = 8.0 years
Thus we have found that a planet 4 times as far as Earth is from the Sun will take 8 times as
long to complete one orbit.
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