Originally:
Earth mass = M
Moon mass = m
distance between center of earth and center of moon = r
G = gravitational constant
Let’s start with our equation for the speed of an object in a circular orbit:
v^2 = G(M)/r
We are looking for an expression with Period T in it, and we can find that in the velocity. Velocity is distance / time, and the period is the time of 1 orbit.
v = distance of orbit/ period T
the distance of one orbit is the circumference of the circle that the moon follows. C = 2 pi r
v = 2*pi*r/T
v^2 = G(M)/r square root both sides
v = sqrt(GM/r) plug in our expression for v
2*pi*r/T = sqrt (GM/r) get T by itself: multiply by T
2*pi*r= T * sqrt(GM/r) divide by the sqrt(GM/r)
when we divide by a square root that has a fraction inside, we can apply Keep Change Flip to write the square root as if it is multiplying with 2 pi r while flipping the inside
2 * pi * r * sqrt(r/GM) = T
We are going to ignore all constants and compare the original scenario to the new scenario where we plug in double the quantity for each value.
Original scenario: I see r and M in this equation (no mass of moon), so am going to choose M = 1 and r = 2 to see what coefficient I will be comparing. It doesn’t matter what you choose for M and r but you want easy numbers to work with.
2 * pi * 2 * sqrt( 2 /1G) = T I plugged in 2 for r and 1 for Me
4 pi sqrt(2/G) = T
New scenario: I will double both quantities. Me = 2 and r = 4
2 * pi * 4 * sqrt (4/2G) = T
8 pi sqrt (2/G) = T
Old period: 4 pi sqrt (2/G)
New period: 8 pi sqrt (2/G)
Ignore everything that exactly matches. The only number that changed was the 4, which became an 8. The 4 doubled to become an 8.
So, if we double all the physical values in the solar system, THE PERIOD OF THE MOON WILL ALSO DOUBLE compared to what it is now. :) (currently the moon takes about 27.3 days (.075 years) to go around the earth so it would take about 54.6 days (0.15 years))
