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hangontoyourego

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Hello, I had a bit of trouble figuring out this problem:

Given the following, determine the distance in miles above the Earth's surface of a geosynchronous satellite.

M

R

1 mile=1604 m

F

F

((6.67E-11m^3/kg x s^2)(m

For v of the satellite, I said the velocity is equal to the distance of the orbit divided by 86,400 seconds, so I have:

(3.98866E14)/(4.1165056E13+12832000x+x^2) = (((6.2832x)/(86400))^2)/(6416000+x)

and then

http://www4b.wolframalpha.com/Calculate/MSP/MSP94291ci3c60ee01fbifd00000i565236dhh0d08b?MSPStoreType=image/gif&s=2&w=476.&h=66.

Then, I found that x equaled 40,216,400 meters or

1. Homework Statement1. Homework Statement

Given the following, determine the distance in miles above the Earth's surface of a geosynchronous satellite.

M

_{Earth}=5.98E24 kgR

_{Earth}=4,000 miles1 mile=1604 m

## Homework Equations

F

_{g}=(Gm_{1}m_{2})/r^{2}F

_{C}=(mv^{2})/r## The Attempt at a Solution

((6.67E-11m^3/kg x s^2)(m

_{Satellite})(5.98E24kg))/(6416000m+x)^2 = (m_{Satellite}v_{Satellite}^2)/(6416000m+x)For v of the satellite, I said the velocity is equal to the distance of the orbit divided by 86,400 seconds, so I have:

(3.98866E14)/(4.1165056E13+12832000x+x^2) = (((6.2832x)/(86400))^2)/(6416000+x)

and then

http://www4b.wolframalpha.com/Calculate/MSP/MSP94291ci3c60ee01fbifd00000i565236dhh0d08b?MSPStoreType=image/gif&s=2&w=476.&h=66.

Then, I found that x equaled 40,216,400 meters or

**11,658 miles**. Is this correct?
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