[CuriousOne (OP)]

My understanding is that time goes faster with altitude, but when I work the time dilation formula for gravity, T=T₀/√((1-(2GM/Rc²)), the dilation factor gets smaller as R increases, so time appears to go slower. I get less s/s difference at Neptune than I do Mercury. For Mercury I get 0.9999999745 and the #seconds is 1.0000000254 and Neptune is 0.9999999996 and 1.0000000003. Shouldn't I be getting a bigger s/s difference if time accelerates with R?

[Janus]

My understanding is that time goes faster with altitude, but when I work the time dilation formula for gravity, T=T₀/√((1-(2GM/Rc²)), the dilation factor gets smaller as R increases, so time appears to go slower. I get less s/s difference at Neptune than I do Mercury. For Mercury I get 0.9999999745 and the #seconds is 1.0000000254 and Neptune is 0.9999999996 and 1.0000000003. Shouldn't I be getting a bigger s/s difference if time accelerates with R?

In this equation, T is the amount of time it takes for a far removed observer to measure a time of T₀ pass for the clock deeper in the gravity field.  So for example if T₀ is one sec ticking off on an clock sitting on the surface of the Earth, then T is how much time it takes for our observer to see the Earth clock tick off that second.
So using the equation for a Earth surface clock :
T = 1.00000000069 sec if T₀ = 1 sec
It takes that much more than 1 sec for our observer to measure the Earth clock tick off just one sec, ergo, the Earth clock ticks slower than his.
He looks at a clock higher above the Earth's surface(let's say 2 Earth radii above it)
then  T = 1.00000000023 sec if T₀ = 1 sec.
This means it takes him not quite as long to measure this clock tick off 1 sec than it does for the the surface clock.
He measures the higher clock tick off secs faster than the lower clock, and the higher clock ticks faster than the lower one.
For Mercury and Neptune (computing time dilation due to orbital distance from the Sun), I get
1.000000051 for Mercury
1.00000000065 for Neptune.
It takes a longer time for our observer to measure Mercury's clock tick off one sec than it does for him to measure Neptune's clock tick off one sec, thus Neptune's clock ticks faster at its greater distance from the Sun.
All is in order.

[CuriousOne (OP)]

Thanks. I understand what you are saying, but am not getting the same answers. When I use T = T₀ / √1-(r_s / r), I get 1.0000000215855072659 for Mercury. I am using km in that equation for r_s and r. In the GM equation I am using m for r_s and r as G uses m. Is that correct?

OK. I see what I was looking at wrong. To the distant observer, Mercury is .0000000215855072659 slower. It relates to the distant observer, not the Sun.

But, why am I getting different answers than you? I am using 2.5 for r_s and 5.7909227*10⁷  for r for Mercury.

[Janus]

Thanks. I understand what you are saying, but am not getting the same answers. When I use T = T₀ / √1-(r_s / r), I get 1.0000000215855072659 for Mercury. I am using km in that equation for r_s and r. In the GM equation I am using m for r_s and r as G uses m. Is that correct?

OK. I see what I was looking at wrong. To the distant observer, Mercury is .0000000215855072659 slower. It relates to the distant observer, not the Sun.

But, why am I getting different answers than you? I am using 2.5 for r_s and 5.7909227*10⁷  for r for Mercury.

5.7909227 e7 is the orbital radius in km

My mistake.  I was plugging the numbers in by hand into a calculator, and neglected to take the square root before inverting.

[CuriousOne (OP)]

Well, thanks for helping me see what I was seeing wrong. It was driving me nuts. Pretty dumb as it is stated that T is the clock of the observer and not the sun. I guess I need a proof reader.