Physics
posted by Katie on .
A GPS satellite orbits at an altitude of 2.0 x 10^7 m and a speed of 3.9 x 10^3 m/s, Earthâ€™s radius is 6.4 x 10^6 m, and a point at its equator has tangential speed of 460 m/s, By approximately what fraction must the time be adjusted to account for both regular/speeddependent and gravitational time dilation?

t(Earth) = t(satellite) / (square root) 1  (v/c)^2
= t(satellite) /
(square root) 1  [ (3.9 x 10^3) / (3x 10^8)^2 ]
= (1  1.7 x 10^10)^  1/2 [ t(satellite) ]
t(Earth) = (1 + 8.5 x 10^11) [ t(satellite) ]
GMEarth = (6.67 x 10^11 N x m^2/kg^2)
(5.98 x 10^24 kg)
= 4.0 x 10^14 m^3/s^2
But how did you get (5.98 x 10^24 kg)??
To plug in
t(Earth) = t(satellite) [1  (1/c^2) [ ( GM/r, Earth )
 ( GM/r, satellite )