Question: Kepler's third law of planetary motion

  • Thread starter Jess048
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  • #1
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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!
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Is (10^24)^(1/3)=10^24? Something is going seriously wrong with your cube root.
 
Last edited:

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