Posted by **John** on Saturday, March 17, 2012 at 10:59pm.

In a computer simulation, a satellite orbits around Earth at a distance from the Earth's surface of 2.1 X 104 miles. The orbit is circular, and one revolution around Earth takes 10.5 days. Assuming the radius of the Earth is 3960 miles, find the linear speed (velocity) of the satellite.

- trig -
**Steve**, Sunday, March 18, 2012 at 3:14pm
speed is distance/time

for a circle, C=2pi*r

r = 3960+104 = 4064

C = 25534

speed = 25534mi/10.5day = 2432mi/day

- trig -
**tchrwill**, Monday, March 19, 2012 at 1:48pm
/r] sqrt(sqrt[(1.407974x10^16)/(3960+218.4)5280]

= 25,262 feet per second.

The orbital period is

T = 2(3.14)sqrt[22,061,952^3/1.407974x10^16] = 5487 seconds or 91.45 minutes.

- trig -
**tchrwill**, Monday, March 19, 2012 at 2:04pm
Part of my previous reply was lost in the posting process.

The orbital radius is 3960 + 2.1(104) = 4178.4 miles or 22,061,952 feet.

The alleged time to complete one orbit is 10.5(24)3600 = 907,200 seconds making the derived orbital velocity

Vc = 152.7fps.

Unfortunately, the real orbital velocity required to remain in a circular orbit derives from Vc=sqrt(µ/r)=

where µ = the earth's gravitational constant and r = the orbital radius in feet =

sqrt[(1.407974x10^16)/(3960+218.4)5280]

= 25,262 feet per second.

The orbital period is

T = 2(Pi)sqrt[r^3/µ]=

2(3.14)sqrt[22,061,952^3/1.407974x10^16] = 5487 seconds or 91.45 minutes.

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