The satellites used in the Global Positioning System go around the earth in circ

Question

The satellites used in the Global Positioning System go around the earth in circular orbits whose radius is 26,600km and period is 12 hours exactly. Assume for the sake of simplicity, the earth is not rotating, so that a clock on its surface is in an inertial frame.

a) In unit N, we saw that the speed of an object in a circular orbit of radius R with mass M is v = (GM/R)^0.5, where G is the universal gravitational constant. Argue that in SR units, G = G(si)/c^3 = 2.475x10^-38.

b) Let event A be a certain GPS satellite passing a given position in space and event B be it passing that point again after one complete orbit. At each event, this satellite sends a radio signal directly below it to a clock directly below it on the (nonrotating) earth, which receives the signals at events C and D, respectively. What is the difference between the time an atomic clock on board the satellite registers between events A and B and the time a clock on the earth's surface registers between events C and D? Express your results symbollically in terms of G, M, and R (don't crunch numbers yet), though you can assume GM/R << 1. (Hint: Argue that the signal's travel time is the same in both cases.

c) Now calculate numerically how much less time a clock on the GPS satellite measures for a complete orbit than the clock on the ground does.

Explanation / Answer

Given radius of orbit, R = 26,600 km

Time period = T = 12 hours

Assuming earth is not rotating

a.

in standard relative units

1 sec = 3*10^8 m

So unit of G in SI = m^3/s^2 * kg

so G ( SR) = G(SI)/c^2 = 6.67*10^-11/(3*10^8)^2 = 7.411*10^-28 m / kg

b. for the satelite with fast ticking clock, time between events A and B = tf

for earth bound observer with slow ticking clocks, time between events C and D = to

then to = tf*sqroot(1 - 2G(SR)M/R)

but as 2GM/R < < 1

to = tf

c. time difference as compared by the satellite = to - tf = tf(1 - sqroot(1 - 2G(SR)M/R))

now tf = 12 hours

M = mass of earth = 6*10^24 kg

so, to - tf = 12*1.67165*10^-10 hours = 0.00722 mili second per orbit

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