Calculating Time for Cannon Ball to Orbit Earth at 4000mi Radius

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The discussion focuses on calculating the time it takes for a cannonball to orbit Earth at a height of 800 miles above a 4000-mile radius. Using gravitational formulas and orbital mechanics, the calculated time for the cannonball's orbit is approximately 5138 seconds, or 85.6 minutes. This result aligns closely with the International Space Station's orbit, which takes about 90 minutes at a lower altitude of 200 miles. The calculations involve using gravitational force equations and orbital velocity. Overall, the approach and result appear to be accurate and reasonable within the context of orbital mechanics.
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How long would it take a cannon ball to orbit the Earth given that the radius of the
earth is 4000 miles and the height of the (quite fictitious) mountain is 800 miles?

Me = mass of earth
Mc = mass of cannon ball
R = Earth's radius
v = 2piR/T
4000 mi = 6437200 meters
a = acceleration of cannon ball
G = 6.67e-11
Using an applet for a previous question, I found Vo to be 15468 miles/hr. But I didn't use Vo... Instead I used Fnet = GMeMc/r^2 as follows.

Fnet = GMeMc/r^2 = Mca = (Mc(2piR/T)^2)/R -->
GMe/r^2 = (4pi^2*R^2)/RT^2 -->
r^2/GMe = T^2/(4pi^2*R) -->
sqrt(4pi^2*R^3)/GMe = T

my result was 5138s, which is about 85.6 minutes.
Does my work and result look all right?? thanks in advance
 
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well, the International Space Station is in Low Earth Orbit and it takes about 90 minutes to orbit the Earth once, so you're in the right ballpark ;)

it's about 200 miles up for comparison, and has an orbital velocity of about 7,000 m/s
 
thanks!
 
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