Question

the moon orbits the earth in an average time of 27.3 days at an average distance of 384000 km. use these facts and newtons version of keplers 3rd law to determine the mass of earth

Answer 1

Answer:

The mass of the Earth is $6.02x10^{24}Kg$.

Explanation:        

The Universal law of gravitation shows the interaction of gravity between two bodies:

$F = G\frac{Mm}{r^{2}}$  (1)

Where G is the gravitational constant, M and m are the masses of the two objects and r is the distance between them.

For this particular case, M is the mass of the Earth and m is the mass of the moon. Since it is a circular motion, the centripetal acceleration will be:

$a = \frac{v^{2}}{r}$  (2)

Then Newton's second law ($F = ma$) will be replaced in equation (1):

$ma = G\frac{Mm}{r^{2}}$

By replacing (2) in equation (1) it is gotten:

$m\frac{v^{2}}{r} = G\frac{Mm}{r^{2}}$  

$v^{2} = G\frac{Mmr}{mr^{2}}$

$v^{2} = \frac{GM}{r}$  (3)

Since it is a circular motion, the orbital velocity can be represented by:

$v = \frac{2\pi r}{T}$  (4)

Where r is the orbital radius and T is the period.

Then, equation 4 can be replaced in equation 3.

$(\frac{2\pi r}{T})^{2} = \frac{GM}{r}$  

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

Notice that r can be represented by the semi-major axis (a)

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

$T^{2}GM = 4\pi^{2} a^{2} a$  

$T^{2} = \frac{4\pi^{2} a^{3}}{GM}$  (5)

Therefore, the mass of the Earth can be determined if M is isolated from equation (5):

$M = \frac{4\pi^{2} a^{3}}{GT^{2}}$  (6)

Notice that it is necessary to express a in units of meters and T in units of seconds.

$a = 384000Km . \frac{1000m}{1Km}$ --- $3.84x10^{8}m$

$T = 27.3days . \frac{24hrs}{1day}$ --- $655.2hours.\frac{3600s}{1hour}$ --- $2358720s$

$M = \frac{4\pi^{2} (3.84x10^{8}m)^{3}}{(6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})(2358720s)^{2}}$  

$M = 6.02x10^{24}Kg$  

Hence, the mass of the Earth is $6.02x10^{24}Kg$.