Asked by Katie
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, Earth’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/speed-dependent and gravitational time dilation?
Answers
Answered by
Katie
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 )
= 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 )
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