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Gravitational time dilation

by Seedling
Tags: dilation, gravitational, time
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Seedling
#1
Feb15-10, 08:41 PM
P: 4
1. The problem statement, all variables and given/known data

Calculate the difference in time after one year between a clock at Earth's surface and a clock on a satellite orbiting at 300 km above the surface


2. Relevant equations

T = T0 / (1 - 2gR/c^2)^.5

That is, this:
http://hyperphysics.phy-astr.gsu.edu...grel/gtim3.gif


3. The attempt at a solution

I don't understand how to use this equation to get the difference between the clock on the satellite and the clock on the surface. Do I just take the value of T with R = Earth's radius, and again with R = Earth's radius + 300 km, and take the difference?
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tiny-tim
#2
Feb16-10, 05:13 AM
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tiny-tim's Avatar
P: 26,148
Hi Seedling!

(try using the X2 and X2 tags just above the Reply box )
Quote Quote by Seedling View Post
T = T0 / (1 - 2gR/c^2)^.5

I don't understand how to use this equation to get the difference between the clock on the satellite and the clock on the surface. Do I just take the value of T with R = Earth's radius, and again with R = Earth's radius + 300 km, and take the difference?
The ratio, rather than the difference

T0 is the time on a clock "at infinity", and TR is the time on a clock at radius R, so TR/T0 is the ratio of their "speeds", and TR=h/TR is the ratio you want.

(But remember that this formula only gives you the general relativity time dilation there'll also be an ordinary Lorentz time dilation, in the opposite direction ).


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