Circular Orbits and Motion - Satellites

AI Thread Summary
A spy satellite orbits Earth with a period of 6.02 hours, and the radius of Earth is 6.371 million meters. To determine the satellite's height above Earth's surface, it's essential to convert the orbital period from hours to seconds. The satellite's acceleration can be calculated using the formula v = sqrt(G*M Earth / r), where r includes both the Earth's radius and the satellite's altitude. The discussion highlights confusion regarding the application of Newton's form of Kepler's 3rd Law and the need for unit conversion. Accurate calculations require careful attention to these details to avoid incorrect results.
BlueSkyy
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Homework Statement



A spy satellite is in circular orbit around Earth. It makes one revolution in 6.02 hours. (Radius of the Earth=6.371 times 106 m)

(a) How high above Earth's surface is the satellite?
(b) What is the satellite's acceleration?

Homework Equations



v = sqrt(G*M Earth / r)
Kepler's 3rd Law

The Attempt at a Solution



I found the angular velocity to be 2.899 rad/sec...
I really don't know where to start after that
 
Physics news on Phys.org
What is Kepler's 3rd law? Have you learned Newton's form of Kepler's 3rd law for circular orbits?
 
i just found Newton's form of Kepler's 3rd law on a different website - i will see if i can get it to work...
 
i'm still getting a ridiculous number for the distance...i assume i use Newton's form of Kepler's 3rd law but do i keep the period in hours or seconds? argh...
 
BlueSkyy said:
i'm still getting a ridiculous number for the distance...i assume i use Newton's form of Kepler's 3rd law but do i keep the period in hours or seconds? argh...

The SI units are seconds, so you will have to convert. Remember the distance will be the radius of the Earth plus the satellites orbital height.
 
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