Gravitational Time Dilation?

9.8 m/s is not the velocity at the surface of the Earth. 9.8 m/s^2 is the ACCELERATION of gravity, not the velocity.

The basic formula for gravitational time dilation is t0/tf = sqrt[ 1 - 2GM/(rc^2) ]. So you need these numbers:

G = gravitational constant = 6.67384 x 1-^11 m^3 kg^-1 s^-2

M = mass of the object creating the gravitational field

r = distance from center of mass

c = speed of light = 299,792,458 m/s

t0 is the time between events A and B, as measured by a slow-ticking observer inside the gravity field

tf is the time between events for a fast-ticking observer at a large distance from the gravity field

On the surface of the Earth, this boils down to t0/tf = 1 - 0.695 x 10^-9. That means the the dilation is about 21.9 milliseconds per year.

This gets too complicated to explain on Yahoo answers. There is a good discussion of gravitational time dilation on Wikipedia.

You are confusing velocity (the "v" in your first equation) with gravitational acceleration. The gravitational acceleration at the surface of the earth is 9.8 m/s^2 - notice the fact that it is per second-squared. So NO you can't substitute in 9.8 in your formula for the time dilation.

The relativistic time dilation formula is

1. t/tₒ = √[1 - (v/c)²]

2. For a gravitational body, the escape velocity (when its KE = gravitational PE) is given by

½ mv² = GMm/r ,

v² = 2GM/r.

3. By substituting this in the first equation we get

t/tₒ = √[1 - (v/c)²]

= √[1 - (2GM/rc²) ].

4. Since g = GM/r² = acceleration at r ,

t/tₒ = √[1 - (2gr/c²) ].

For Earth (surface) it is

t/tₒ = √[1 - 1.4 X 10‾⁹] ≈ 1.

Gravitational time dilation has nothing to do with velocity. You are confusing Special Relativity Time Dilation with General Relativity Time Dilation.