Question

The Moon orbits Earth in an average of p = 27.3 days at an average distance of a =384,000 kilometers. Using Newton’s version of Kepler’s third law determine the mass of Earth. You may neglect the mass of the Moon in comparison to the mass of Earth.

Please no pictures or handwriting, it is always difficult to read.

Answer 1

The values given in the question are the following :

Average period,p = 27.3 days

Average distance, a = 384,000 km

Newton's version of Kepler 's third law is given by the equation:

M₁ + M₂ = a³ / p²

where, p - period

a- distance travelled/semi major axis

M_{1 ,} M₂ - masses of the body

In using this equation masses should in the units of solar mass, period in years and distance in astronomical units.

First we convert the given data into these units,

Average period, p = 27.3 days = (27.3 / 365) years = 0.07479452 years

Average distance, a = 384,000 km = (384,000/149,600,000) AU = 2.5668 × 10^-3 AT

Since 1 AU = 149,600,000 km

It is said in the question to neglect the mass of the moon.

So the required equation reduces to:

Mass of the earth , M₁ = ( a³/ p²)

= [ ( 2.5668 × 10^-3)³​​​^​ / ( 0.07479452 )² ]

= [ ( 1.69 × 10 ^-8 ) / ( 5.594 × 10^-3) ]

= 3.02 × 10^-6 solar masses

1 solar mass = 1.99 × 10³⁰ kg

The mass of the earth = 3.02 × 10^-6 × 1.99 × 10³⁰ Kg

= 6.0098 × 10²⁴ Kg