Question

The moon takes about 27.3 days to revolve round the earth in a nearly circular orbit of radius 3.84 × 105 km/ Calculate the mass of the earth from these data.

Answer 1

From this data, it is found that the mass of the earth is $6.02 \times 10^2^4 \mathrm{kg}$.

Explanation:

The third law of Kepler describes the period of celestial body’s rotation around a larger celestial body using the orbital mass and radius of the larger body. By using the below equation, the rotation’s orbital period is explained:

$T=2 \pi \sqrt{r 3 G M T}=2 \pi r 3 G M$

Where,

M is the star’s mass orbiting around

r is the radius

$G=6.67 \times 10^{-11} \mathrm{m}^ 3 / \mathrm{kgs}^{2}$

where G is the gravitational force.

T = 27.3 d, where ‘d’ is the moon’ period.

$r=3.84 \times 10^{5} \mathrm{km}$ is the orbital radius of the moon

Using Kepler’s third law, the period is calculated as:

$T=2 \pi \sqrt{r 3 G M T}=2 \pi r ^3 G M$

We square both sides:

$T 2=4 \pi 2(r ^3 G M) T 2=4 \pi ^2(r ^3 G M)$

When the squared period, the mass is isolated here.

$M=4 \pi ^2(r^ 3 G T 2) M=4 \pi^ 2(r 3 G T 2)$

We will get the below after evaluation as $M = 6.02 \times 10^2^4 kg$