Question

How would I do this physics problem?

Suppose the earth had another moon which was 2.55 times as far from the center of the earth as our own moon. Determine the orbital period of this moon if our own moon has a period of 27.32 days.

Answer 1

You have to rely on Kepler's 3rd Law which states that
http://www.1728.org/kepler3.htm
T^2 = r^3
In this case the distance is being increased by a factor of 2.55 so
T^2 = (2.55)^3
T^2 = 16.581375
T = 4.0720 greater than its present period
T = 4.0720 * 27.32
= 111.25 Days

Answer 2

Here's a really fancy Kepler's 3rd Law calculator:
http://www.1728.org/kepler3a.htm
It says it's about 112.58 Days so I think I'm right!!

Answer 3

Ah yes, you are indeed correct. I was first confused at this problem because I was only given the equation \[\large \frac{ GM }{ 4 \pi^2 }=\frac{ R^3 }{ T^2}\]

Answer 4

Okay then you would rewrite the formula as
t = square root (4 * PI^2 * r^3 /G * m)

Answer 5

Yeah, I think I get it now, thanks wolf.

Answer 6

The units you would have to use are G = 6.674 x 10^-11
m would be Earth's mass in kilograms
r would be Moon's orbit in in meters (2.55 times Moon's current radius)
Let's see if I got that right.

Answer 7

Moon's orbit is presently 384,399,000 meters so 2.55 times that =
980,217,450 meters

t = square root (4 * PI^2 * 980,217,450^3 meters / 6.674x 10^-11 * 5.97237×10^24 kg)
t = square root (39.4784176044 * 9.418 x10^26 / 3.986x10^14)
t = square root (3.7182 x 10^28 / 3.986x10^14)
t = square root (9.328 x 10^13)
t = 9658170.4 seconds
t = 111.78 days