Question

The moon orbits the earth at a distance of 3.85 x 10^8 m. Assume that this distance is between the centers of the earth and the moon and that the mass of the earth is 5.98 x 10^24 kg. Find the period for the moon's motion around the earth. Express the answer in days and compare it to the length of a month.

Answer 1

Answer:

27.5 days

0.92 month

Explanation:

$r$ = radius of the orbit of moon around the earth = $3.85\times10^{8} m$

$M$ = Mass of earth = $5.98\times10^{24} m$

$T$ = Time period of moon's motion

According to Kepler's third law, Time period is related to radius of orbit as

$T^{2} = \frac{4\pi ^{2} r^{3} }{GM}$

inserting the values, we get

$T^{2} = \frac{4(3.14)^{2} (3.85\times10^{8})^{3} }{(6.67\times10^{-11})(5.98\times10^{24})}\\T = 2.3754\times10^{6} sec$

we know that

1 day = 24 hours = 24 x 3600 sec = 86400 s

$T = 2.3754\times10^{6} sec \frac{1 day}{86400 sec} \\T = 27.5 days$

1 month = 30 days

$T = 27.5 days \frac{1 month}{30 days} \\T = 0.92 month$