Answer:
The mass of the Earth is [tex]6.02x10^{24}Kg[/tex].
Explanation:
The Universal law of gravitation shows the interaction of gravity between two bodies:
[tex]F = G\frac{Mm}{r^{2}}[/tex] (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:
[tex]a = \frac{v^{2}}{r}[/tex] (2)
Then Newton's second law ([tex]F = ma[/tex]) will be replaced in equation (1):
[tex]ma = G\frac{Mm}{r^{2}}[/tex]
By replacing (2) in equation (1) it is gotten:
[tex]m\frac{v^{2}}{r} = G\frac{Mm}{r^{2}}[/tex]
[tex]v^{2} = G\frac{Mmr}{mr^{2}}[/tex]
[tex]v^{2} = \frac{GM}{r}[/tex] (3)
Since it is a circular motion, the orbital velocity can be represented by:
[tex]v = \frac{2\pi r}{T}[/tex] (4)
Where r is the orbital radius and T is the period.
Then, equation 4 can be replaced in equation 3.
[tex](\frac{2\pi r}{T})^{2} = \frac{GM}{r}[/tex]
[tex]\frac{4\pi^{2} r^{2}}{T^{2}} = \frac{GM}{r}[/tex]
Notice that r can be represented by the semi-major axis (a)
[tex]\frac{4\pi^{2} a^{2}}{T^{2}} = \frac{GM}{a}[/tex]
[tex]T^{2}GM = 4\pi^{2} a^{2} a[/tex]
[tex]T^{2} = \frac{4\pi^{2} a^{3}}{GM}[/tex] (5)
Therefore, the mass of the Earth can be determined if M is isolated from equation (5):
[tex]M = \frac{4\pi^{2} a^{3}}{GT^{2}}[/tex] (6)
Notice that it is necessary to express a in units of meters and T in units of seconds.
[tex]a = 384000Km . \frac{1000m}{1Km}[/tex] --- [tex]3.84x10^{8}m[/tex]
[tex]T = 27.3days . \frac{24hrs}{1day}[/tex] --- [tex]655.2hours.\frac{3600s}{1hour}[/tex] --- [tex]2358720s[/tex]
[tex]M = \frac{4\pi^{2} (3.84x10^{8}m)^{3}}{(6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})(2358720s)^{2}}[/tex]
[tex]M = 6.02x10^{24}Kg[/tex]
Hence, the mass of the Earth is [tex]6.02x10^{24}Kg[/tex].
