Question

An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.225×105 km and the planet has a radius of 3.150×103 km and a mass of 7.90×1022 kg , how long (in days) does it take the moon to make one revolution around the planet? The gravitational constant is 6.67×10−11N·m2/kg2.

Answer 1

F=G*(m1*m2)/r^2 where G is uni. grav. const., m1 and m2 are masses, and r is radius from CENTERS. This question is tricky because they give you radius of planet and distance of moon from surface; you need to add these two to get distance between centers.

From there, you need to apply centripital force.

F=m2*w^2*r where w is ANGULAR velocity. (take m2 as moon)

sooo
F=G*(m1*m2)/r^2=m2*w^2*r
therefore, w^2=G*m1/r^3

what you want is the period, T. T=1/f=2(pi)/w

[2(pi)/T]^2=G*m1/r^3
1/T=sqrt(G*m1/r^3)/(2pi)
T=2(pi)*sqrt(r^3/[G*m1])

T=6.28*sqrt(2.180*10^12)

T=9272308.882 sec

t(days) = t(in seconds) / (86400 seconds per day)

T=9272308.882/86400

T=107.32 days