Astronomy 105

Kepler's Third Law

10. Kepler's Third Law (Pages 42,48) (Problem 2-3)
    

    This formula is explained well in the text. It relates the orbital period of a planet in years (how long it takes for the planet to orbit the Sun) to the distance from the planet to the Sun in astronomical units (AU)

    

    It is a simple formula to use, but it took astronomers over a century to figure out why it is true.

    Two examples. Simple. The Earth is 1.0 AU from the Sun. What is the period of the Earth's orbit? Since the Sun weighs one solar mass, . Thus P = 1.0 y. But you already knew that, so we'll try a harder one.

    Saturn's orbital period is 29.5 y. How far is it from the Sun? Here, P = 29.5, and once again, M = 1.0, so:
    

    
    

    Now we need to find a. Since a³ = 870, a = or a = (870)^1/3 (the two forms are equivalent). On your calculator, punch in 870, then , then 3. The answer should be a = 9.5 AU. (Which can be checked on page 511).
    

    In some cases, the distances are not given in AU nor the periods in years. To solve a problem like this you will need to convert to the appropriate units.
    

    Example. How much does the Earth weigh in solar masses? Use the orbital period and radius of the Moon's orbit to figure this out.
    

    
    

    
    

    Now solve Kepler's Equation for M:

    

    You can check this answer by dividing the mass of the Earth by that of the Sun.

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