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November 24, 2014

Homework Help: Physics

Posted by Frederico Barreto on Monday, April 30, 2007 at 6:12am.

I really need you answered this to my email. It's very important, for this may decide my result in the school exams, and I really need a good mark.
Unfortunately, Portuguese system of education is full of problems...

The question is...

The altitude of a geostationary satellite depends on:

a) … the satellite’s mass.

b) … the absolute value of its speed
(Considering this is a circular motion)

c) … the Earth’s mass.

d) … the launching speed of the satellite (speed which he was thrown)


Consider while in orbit, there are one force operating on the satellite: Gravity. To be in orbit, gravity must equal exactly the "force" expected in centripetal acceleration.

Gravity Force= centripetalforce
GMeM/r^2= M v^2/r

dividing out M, the mass of the satellite, and r , one has

GMe/r= v^2 or

r= v^2/GMe so the factors are velocity in orbit, and the mass of the Earth.

The altitude of a geostationary satellite depends on:

a) … the satellite’s mass.

b) … the absolute value of its speed
(Considering this is a circular motion)

c) … the Earth’s mass.

d) … the launching speed of the satellite (speed which he was thrown)


How do you determine the altitude at which a satellite must fly in order to complete one orbit in the same time period that it takes the earth to make one complete rotation?

The force exerted by the earth on the satellite derives from

...................................................F = GMm/r^2

where G = the universal gravitational constant, M = the mass of the earth, m = the mass of the satellite and r = the radius of the satellite from the center of the earth.

GM = µ = 1.407974x10^16 = the earth's gravitational constant.

The centripetal force required to hold the satellite in orbit derives from F = mV^2/r.

Since the two forces must be equal, mV^2/r = µm/r^2 or V^2 = µ/r.

The circumference of the orbit is C = 2Pir.

A geosycnchronous orbit is one with a period equal to the earth's rotational period, which, contrary to popular belief, requires 23hr-56min-4.09sec. to rotate 360ş, not 24 hours. Therefore, the time to complete one orbit is 23.93446944 hours or 86,164 seconds

Squaring both sides, 4Pi^2r^2 = 86164^2

But V^2 = µ/r

Therefore, 4Pi^2r^2/(µ/r) = 86164^2 or r^3 = 86164^2µ/4Pi^2

Thus, r^3 = 86164^2(1.407974x10^16)/4Pi^2 = 2.647808686x10^24

Therefore, r = 138,344,596 feet. = 26,201.6 miles.

Subtracting the earth's radius of 3963 miles, the altitude for a geosynchronous satellite is ~22,238 miles.

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