Answer: [tex]1.893(10)^{27}kg [/tex]
Explanation:
This problem can be solved by the Third Kepler’s Law of Planetary motion, which states:
“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.
In other words, this law stablishes a relation between the orbital period [tex]T[/tex] of a body (moon, planet, satellite) orbiting a greater body in space with the size [tex]a[/tex] of its orbit.
This Law is originally expressed as follows:
[tex]T^{2}=\frac{4\pi^{2}}{GM}a^{3}[/tex] (1)
Where;
[tex]T=7.16days=618624s[/tex] is the period of the orbit Ganymede describes around Jupiter
[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]
[tex]M[/tex] is the mass of Jupiter (the value we need to find)
[tex]a=10.7(10)^{8}m[/tex] is the semimajor axis of the orbit Ganymede describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)
If we want to find [tex]M[/tex], we have to express equation (1) as written below and substitute all the values:
[tex]M=\frac{4\pi^{2}}{GT^{2}}a^{3}[/tex] (2)
[tex]M=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(618624s)^{2}}(10.7(10)^{8}m)^{3}[/tex] (3)
Finally:
[tex]M=1.8934(10)^{27}kg[/tex] This is the mass of Jupiter
