Answer:
Mass of the earth is [tex]6.03 \times 10^{24} kg[/tex]
Explanation:
The relationship between the orbital period, orbital radius and Mass is given by Kepler's third law
[tex]T^2=\frac{4\pi^2r^3 }{GM}[/tex]
here G is gravitational constant =[tex]6.67\times 10^{-11} Nm^2/kg^2[/tex]
M is mass of the earth
T orbital period =[tex]27.32 days= 27.32\times 24\times 3600=2360448 sec[/tex]
r is the orbital radius =[tex]238910 miles=238910\times1609.34 m=38.448\times 10^7 m[/tex]
Mass of the earth is
[tex]M=\frac{4\pi^2r^3 }{GT^2}\\M=\frac{4\pi^2(38.44\times10^7 )^3 }{6.67\times 10^{-11}(2360448)^2}\\M=6.03 \times 10^{24} kg[/tex]
