Orbital period questions continued

AI Thread Summary
To determine the orbital period of a satellite, both the altitude and mass are provided, but they primarily serve to contextualize the problem. The key equation to use is the centripetal acceleration formula, Ac = 4π²r/T², along with the velocity derived from gravitational force, v = (GMe/r)^(1/2). By calculating the satellite's velocity and using it to find acceleration, the orbital period can be derived. The final calculation yields an orbital period of approximately 15.79 hours. Understanding Kepler's third law is not essential for this specific problem, as the necessary calculations can be completed using the provided formulas.
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Homework Statement


A 9.0x10^3kg satellite with an orbital radius of 3.20x10^7m orbits the Earth at an altitude of 2.56x10^7m. What is the orbital period?

Homework Equations


ΣFc = mac
Kepler's third law equation

The Attempt at a Solution


For this question I know that it is asking me to find the orbital period, but it has also given me an altitude and mass for the satellite. Are they there just to through me off because you are supposed to use Kepler's third law equation to find the orbital period?

Or are they important and I am supposed to use the equation:

Fg = 4(pie)^2r/T^2

Basically what I am wondering is which equation do we use?
 
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What does Kepler's third law give you? In relation to what you are looking for? What are the inputs required to use Kepler's Law?

~Lyuokdea
 
Keplers third law: "The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit." i don't think it is necessary to use keplars law to solve this question, this is how i would figure it out;

ok so you can solve how fast the satellite is moving in relation the Earth: v=(GMe/r)^.5
(G=6.67x10^-11 gravitational constant) so plugging the numbers in gets you a velocity of 3536.4176...ms^-1, next we use the formula Ac=4\pi^2r/T^2 like you had before but you can solve the acceleration by using Ac=v^2/r now that you have the velocity. This gives you 0.39082m/s^2 now put this back into your original formula which gives you a period of 15.79 hours

hope this helps
 
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