Particle in uniform circular motion

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A satellite orbits 600 km above Earth's surface, where the free fall acceleration is 8.21 m/s² and Earth's radius is 6400 km. The calculations involve determining the satellite's speed and the orbital period using the equations for circular motion. The angular velocity (w) was calculated as 0.0346, leading to a period (T) of 0.76 seconds, which is incorrect. The correct orbital period should be approximately 7000 seconds or around 2 hours. Emphasizing the importance of using units in calculations could help identify errors in the process.
shreddinglicks
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Homework Statement



A satellite is orbiting 600km above the Earths surface. The free fall speed is 8.21 m/s^2. The radius of Earth is 6400km.

What is the satellites speed, and the time interval for one orbit around Earth.

Homework Equations



v = ((2)(PI)(r))/T
ac = rw^2
w = (2)(PI) / T

The Attempt at a Solution



8.21 = 7000w^2
w = .0346

.0346 = (2(PI)) / T
T = .76

(2)(PI)(7000) / .76 = 57871.44

I am unsure of what to do or where I went wrong.
 
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You should always use units in your calculations. In this case, units would have shown you where the error is.
(The result should be somewhere around 2 hours or ~7000 seconds, probably a bit less)
 
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