Question: Kepler's third law of planetary motion

[Jess048]

The mean distance between the Earth and the moon is 3.84x10^8 m, and the moon has an orbital period of 27.3 days. Find the distance from Earth of an artificial satellite that has an orbital period of 8.5 days.
a. 1.76x10^8 m
b. 1.76x10^4 m
c. 1.76x10^10 m
d. 5.24x10^8 m

So far I got:
I used the formula: (Ta/Tb)^2 = (rA/rB)^3
Plugged in these values:
Ta= 8.5 days
Tb= 27.3 days
rA= ?
rB= 3.84x10^8 m

rA^3=(3.84x10^8)^3(8.5/27.3)^2
=(5.489164304x10^24)^1/3
rA^3=1.829721435x10^24
This obviously is incorrect. Please help thanx!

 

[Dick]

Is (10^24)^(1/3)=10^24? Something is going seriously wrong with your cube root.

 

[m2003]

Your approach to using Kepler's third law is correct, but there seems to be a mistake in your calculation. The correct equation should be (Ta/Tb)^2 = (rA/rB)^3, where Ta and Tb are the orbital periods, and rA and rB are the distances from the central body (in this case, the Earth). Plugging in the given values, we get:

(8.5/27.3)^2 = (rA/3.84x10^8)^3

Solving for rA, we get:

rA = (8.5/27.3)^2 x (3.84x10^8)^3
= 1.74x10^8 m

Therefore, the correct answer would be option a. 1.76x10^8 m. It is important to double check your calculations to avoid any errors.