# More questions about time and gravity

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#### Fallingstar1971

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I will start by saying that time on the surface of the Earth passes at a *normal rate. (I use the term *normal just so we have a point of comparison)

Now we know that time passes at a different rate in orbit, say at the range of the GPS satellites. You are farther up the gravity well (generated by the mass of the Earth) then you would be standing on the surface.

We also know that the Earth itself is caught in the gravity well of the Sun.

What would be the differences in the rate that time passes in close orbit of the Sun (say Mercury, for example) verses the rate time passes on the Earth? What would the differences be at Jupiter, or the Oort cloud?

And lets not forget that the Sun is caught in the gravity well of the Milky Way. What would the differences in time be at one of the Stars whipping around the Galactic center? Or at the very edge of the spiral arms (In comparison with *normal time on the surface of the Earth?)

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#### derekmcd

##### Guest
Fallingstar1971":doqfcm9b said:

I will start by saying that time on the surface of the Earth passes at a *normal rate.

Now we know that time passes at a different rate in orbit, say at the range of the GPS satellites. You are farther up the gravity well (generated by the mass of the Earth) then you would be standing on the surface.

We also know that the Earth itself is caught in the gravity well of the Sun.

What would be the differences in the rate that time passes in close orbit of the Sun (say Mercury, for example) verses the rate time passes on the Earth? What would the differences be at Jupiter, or the Oort cloud?

And lets not forget that the Sun is caught in the gravity well of the Milky Way. What would the differences in time be at one of the Stars whipping around the Galactic center? Or at the very edge of the spiral arms (In comparison with *normal time on the surface of the Earth?)

In Special Relativity, the conversion factor is 1/sqrt [1-(v^2/c^2)]... aka the Lorentz factor. For ease of math, i'll use a geostationary orbit around the Earth which is just a bit over 3km per second. I'll use 3km per seconds because it makes for a simple calculation.

where v=velocity of the object and c=speed of light. If you make c=1 and v a ratio of c, it makes the calculation that much more simple.

You ultimately end up with a conversion factor of 5*10^-10... or 1.00000000005... or 50 billionths. In other words, 50 billion seconds would need to pass for you on Earth before you are 1 second older than that satellite.

Now, you need to factor in gravitational time dilation under General Relativity in which case your clock on Earth runs slower by a considerable amount compared to said satellite (in comparison to special relativity). It's only slightly more complicated, but the input values are such that doing it on the back of an envelope is a bit tougher... so I'll just point you to:

http://hyperphysics.phy-astr.gsu.edu/hb ... im.html#c4

IIRC, a while back, on a different forum, someone calculated the difference between Mars and Earth using gravitational time dilation and figured that the surface of the Earth would be only a couple years younger over the course of 4 billion years.

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