Explanation:
The acceleration due gravity (free-fall acceleration) [tex]g=[/tex] of a body is given by the following formula:
[tex]g=\frac{GM}{r^{2}}[/tex] (1)
Where:
[tex]G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the gravitational constant
[tex]M[/tex] the mass of the body (the moon in this case)
[tex]r[/tex] is the distance from the center of mass of the body to its surface. Assuming the moon is a sphere with a diameter [tex]d=3500km[/tex], its radius is [tex]r=\frac{d}{2}=1750km=1750(10)^{3}m [/tex]
If the value of [tex]g[/tex] is given: [tex]g=1.60m/s^{2}[/tex] we can find the mass of the moon with equation (1):
[tex]M=\frac{gr^{2}}{G}[/tex] (2)
[tex]M=\frac{(1.60m/s^{2})(1750(10)^{3}m)^{2}}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}[/tex] (3)
[tex]M=7.34(10)^{22}kg[/tex] (4)
Now, according to the 3rd Kepler's Law, there is a relation between the orbital period [tex]T[/tex] of a body (the spacecraft in this case) orbiting a greater body (the moon) in space with the size [tex]r[/tex] of its orbit.
[tex]T^{2}=\frac{4\pi^{2}}{GM}r^{3}[/tex] (5)
Substituting the known values and the calculated mass of the moon in (6), we can find the period of the orbit of the spacecraft around the moon:
[tex]T=\sqrt{\frac{4\pi^{2}}{GM}r^{3}}[/tex] (6)
[tex]T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(7.34(10)^{22}kg)}(1750(10)^{3}m)^{3}}[/tex] (7)
Finally:
[tex]T=6571.37619s=109.522 min[/tex]
