Expert: Philip Stahl - 11/7/2009

Question
Can you please show me step by step how to use Kepler’s Law of Periods to find the period of the earth’s orbit, recorded in seconds? Also, how would it be expressed in years?

I know the distance from the earth to the sun is 1.5x10^11m, and the mass of the sun is 1.9x10^30kg.

Thanks!


Answer
Hello,

Actually - the mass of the Sun is 1.99 x 10^30 kg

Anyway, we begin by setting the gravitational force of attraction equal to the centripetal force:

where M denotes the Sun's mass, m is Earth's mass = 6.0 x 10^24 kg.

G is the gravitational constant: G = 6.7 x 10^-11 N-m^2/kg^2 and (r1 + r2) is the distance to the Sun where r1, r2 refer to the respective distances of the Earth, solar centers to their common center of mass (which, of course, is inside the Sun since it has ~ 330,000x the mass of Earth)

Then (get set for lots of algebra!):



GMm/ (r1 + r2)^2  = mv^2/ r1

but v = r1 (omega) where omega is the angular velocity of Earth

so: mv^2/ r1  = m[r1 (omega)]^2/ r1   

but omega = (2 pi) / P

so (omega)^2 = (4 pi^2)/P^2



Then:

GMm/ (r1 + r2)^2  = m (4 pi^2) r1/P^2   (for body m)

and

GMm/ (r1 + r2)^2  = M (4 pi^2) r2/P^2        (for body M)


Now cancel out the masses common to each side of the equation, and add the result:


e.g.   G(M + m) / (r1 + r2)^2  = (4 pi^2)[r1 + r2] /P^2


call (r1 + r2) the semi-major axis, and we get:


G(M + m) / (a)^2  = 4 pi^2 a /P^2

Or:

G (M + m) =  4 pi^2 a^3 /P^2



Now, solve for P, the period:

P  = [4 pi^2 a^3 / G (M + m)]^1/2



Substituting the values given earlier, one computes:

P = 3.16 x 10^7 secs

Now there are 86,400 seconds in one day so divide this into P:


P(d) =   (3.16 x 10^7 s)/ (86400) =  365.8 days or very close to the accepted value.

Expressed in years it is simply "1 year".

It *can't* be anything else for the Sun-Earth system since the semi-major axis of 1 AU is tied to the formal definition of "1 year" for the period.

I obtained P in *seconds* to show you how it can be done in SI units, as you inquired into it. (But that quantity of seconds, again, is equivalent to 1 year- taking into account round off errors etc.)