A Revision to my previous post
The equation is 4pi^2/T^2=GM/R^3 not 4pi/T^2=GM/R^3
I changed my answer to 5.46 periods
39.48/T^2=(3.98866E14)/(3.00763E20)
T^2=39.48/1.33x10^-6
sqrt of T=sqrt of 39.48/1.33x10^-6
T=5.45×10^3 periods/earth years
Is my answer correct?
4∏/T^2=GM/R^3
4∏/T^2=(6.67×10^-11)(5.98×10^24)/(6.7×10^6)^3
12.57/T^2=(3.98866E14)/(3.00763E20)
12.57/T^2=1.326180414E-6
T^2=1.66700878E-5
T=.0040829019
Is this answer correct?
Homework Statement
Find the period of a satellite that is in orbit 6.7×10^6 meters from the center of the earth?
Homework Equations
P^2=R^3 P=period and R=average distance
The Attempt at a Solution
so far I have tried P^2=(6.7×10^6)^3
then sqrt of (P^2)=sqrt of (3.00763×10^20)...