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No equivalence for GPS
The fact that SR & GR cancel across the Earth’s surface is a mixed blessing. On one hand it’s good because we don’t need to worry about unsynchronized clocks at different latitudes. On the flip side it means that we can’t use them to verify relativity. But there is a reason for pointing this out and that has to do with the Equivalence Principle. We know that gravity at the equator is made of two components: actual gravity from the Earth (a downward pull), and centripetal (upward) force which lessens the effect. We can easily calculate the strength of the centripetal component to be 0.034 m/s2. In above equation (2) the quoted value of gravity at the equator, 9.78 m/s2, is the measured value. The true equatorial gravity must be 9.780+0.034=9.814 m/s2. However we don’t use this true value of gravity in our calculation of GR time-dilation, only the net (real gravity minus centripetal force) value. The reason for this has to do with the Equivalence Principle which states that gravity and acceleration are indistinguishable and should be treated alike. If we used the number 9.814 we would calculate a gravitational dilation at the equator that is higher than at the pole – the opposite of what we need to counteract SR at the equator. Now cast your eye to our earlier calculation of GPS time dilation shown in equation (1). Here we determined the difference between dilation at the equator and dilation at the satellite. The first term, with R in denominator, represents time dilation at the equator. The second, with R+A, represents dilation at the satellite. Let’s split these apart: Time dilation at equator is: 1 + 6.9774x10-10 Time dilation at satellite is: 1 + 1.6702x10-10 The first term represents the gravity at the equator. The second represents gravity at the satellite. Is something amiss here? Hang on a tick... gravity at the satellite? There’s no ‘gravity’ on man-made satellites! The satellite is in orbit and experiences an outward force exactly equal to the inward real-gravitational pull. The net gravity, according to the Equivalence Principle, is therefore zero and the satellite shouldn’t experience gravitational time dilation. So what should the net dilation between the equator and satellite be? Using the above equation (2) with the second term set to 1 (for zero dilation), we get: \left(1\left/\sqrt{1-\frac{2 9.78 6378000}{c^2}}\right.-1\right)*86400 = 59960 ns This value is 31% higher than the quoted value of 45900ns. |