Mass of the Earth Using Kepler's Third Law

Kepler's Third Law (Newton's Formulation)

Mtotal [units of solar masses] = D[AU]3 / t[yr]2

Where D = distance between the orbiting bodies, (in Astronomical units = mean Earth-Sun distance), and

t = period of the orbit (with respect to the stars) (in years).

Mean Distance of Moon from Earth

D = 3.8 x 105 km

Period of Moon (with respect to the stars)

t = 27.3 days

Convert units from km to AU

                      1 AU = 1.5 x 108 km 

                      D(km) x (AU/km) = D (AU) 

                      D (AU) = 3.8 x 105 [km] x 1 / ( 1.5 x 108 km/AU )

                               3.8   105   [km] 
                             = --- x --- x ------   
                               1.5   108   [km/AU]  

                             = 2.5 x 105-8 [AU] = 2.5 x 10-3 [AU]
Convert units from days to years

1 yr = 365.25 days

t (days) x (years/day) = t (years)

t (years) = 27.3 days x 1/ ( 365.25 days/year)

= 7.5 x 10-2 years

Now can use Newton's version of Kepler's Thrid Law to calculate the combined Earth + Moon mass.

                          (2.5 x 10-3)3 
       M (solar masses) = -------------- 
                          (7.5x 10-2)2 

                          1.56 x 10-8
                        = -------------
                          5.6 x 10-3

                        = 2.8 x 10-6 (solar masses)
Now convert solar masses to kg. 
           Mass of Sun = 2 x 1030 kg
Then 
           M (Earth+Moon) = 2.8 x 10-6 (solar masses) x 2 x 1030 kg/Msun

                          = 5.6 x 1024 kg

Since the Mass of the Moon is much less than the mass of the Earth, we can neglect it and take M(Earth+Moon) = M (Earth).

For information on arithmetic with numbers in scientific notation, see Scientific Arithmetic at the University of Oregon.

Updated:

This page has been accessed times.

Bob Stein's home page, email: steinr@pilot.msu.edu