I don\'t know how to do this problem...can any1 please help meout??!?!?!?!?!? A

Question

I don't know how to do this problem...can any1 please help meout??!?!?!?!?!?


A satellite operates at an altitude of 20,200 km and uses communication frequencies of1575.42 MHz. Find the gravitationalfrequency change with respect to Earth.



The solution is .834 Hz but I keep getting this wrong,please help. Will rate lifesaver!! A satellite operates at an altitude of 20,200 km and uses communication frequencies of1575.42 MHz. Find the gravitationalfrequency change with respect to Earth.

The solution is .834 Hz but I keep getting this wrong,please help. Will rate lifesaver!!

Explanation / Answer

the gravitational force acting between the earth and the spaceshuttle orbitter is F = (GMm/r^2) -----------(1) the gravitational force acting between the earth and the spaceshuttle orbitter is balanced by the centripetal force acting on thespace shuttle orbitter therefore we get F = (m * v^2/r) ----------(2) from (1) and (2) we get (GMm/r^2) = (m * v^2/r) or v = (GM/r)^(1/2) = (GM/R + h)^(1/2) or (2r/T) = (GM/R + h)^(1/2) or 2r * f = (GM/R + h)^(1/2) or f = (1/2r) * (GM/R + h)^(1/2) = (1/2 * (R + h))* (GM/R + h)^(1/2) or f = (1/2) * [(GM)^(1/2)/(R + h)^(3/2)]----------(3) G = 6.67 * 10^-11 Nm^2/kg^2,M = 5.9742 * 10^24 kg,R = 6378.2km = 6378.2 * 10^3 m and h = 20200 km = 20200 * 10^3 m the gravitational frequency change with respect to earth whenthe orbiter is at an altitude of h is f = f' - f f' = 1575.42 MHz = 1575.42* 10^6 Hz and the value of f isobtained from equation (3).

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