Speed of light in a gravitational field? How do I solve the speed of light in gravitational field?
Should I just add gravitational acceleration in speed of light?
$$c'=c_0+g(r)t~?$$
 A: This is a far more complicated question than you (probably) realise, as to answer it requires an understanding of general relativity.
In GR the speed of light is locally invariant, that is if you measure the speed of light at your location you'll always get the value $c$. However if you measure the speed of light at some distant location you may find it to be less than $c$. The obvious example of this is a black hole, where the speed of light falls as it approaches the event horizon and indeed slows to zero at the event horizon.
The reason we may measure the speed of light at a distant location to be less than $c$ is because, as alexo says in his answer, spacetime is curved by mass/energy. The co-ordinates that you use for measuring spacetime will not match the co-ordinates a distant observer uses, and that's why the two of you measure different values for the speed of light. To calculate the speed of light at some distant point you need to solve Einstein's equations to find out how spacetime curves relative to your co-ordinate system.
To show this let's take an example. If you solve Einstein's equations for a spherically symmetric mass you get the Schwarzschild metric:
$$ \mbox{d}s^2 = -\left(1-\frac{r_s}{r}\right)c^2~\mbox{d}t^2 + \frac{\mbox{d}r^2}{\left(1-\frac{r_s}{r}\right)} + r^2 (\mbox{d}\theta^2 + \sin^2\theta~ \mbox{d}\phi^2) $$
In this equation $r$ is the distance to the black hole (the radius) and $t$ is time (what you measure on your wristwatch). $\theta$ and $\phi$ are basically longitude and latitude measurements. The quantity $\mbox{d}s$ is called the interval. $r_s$ is the radius of the event horizon. The co-ordinate system strictly speaking is the one used by an observer at infinity, but it's a good approximation as long as you are well outside the event horizon.
For light rays $\mbox{d}s$ is always zero, and we can use this to calculate the velocity of the light ray. For simplicity let's take a ray headed directly towards the black hole, so the longitude and latitude are constant i.e. $\mbox{d}\theta$ and $\mbox{d}     \phi$ are both zero. This simplifies the above equation to:
$$ 0 = -\left(1-\frac{r_s}{r}\right)c^2~\mbox{d}t^2 + \frac{\mbox{d}r^2}{\left(1-\frac{r_s}{r}\right)} $$
The velocity of the light, $v$, is just the rate of change of radius with time, $\mathrm dr/\mathrm d t$, and we get this by a quick rearrangement:
$$ \frac{\mathrm{d}r}{\mathrm{d}t} = v = c \left(1-\frac{r_s}{r}\right) $$
The variation of the velocity of light with distance from the black hole looks like:

At large distances (large $r$) the velocity tends to 1 (i.e. $c$) but close to the black hole it decreases, and falls to zero at the event horizon.
So, to calculate the speed of light in your co-ordinate system solve the Einstein equations to get the metric, set $\mbox{d}       s$ to zero and solve the resulting equation - sounds easy, but it rarely is!
But, but, but, be absolutely clear what you're calculating. All you're calculating is the speed of light in your co-ordinate system i.e. the result you get applies only to you. Other observers in other places will calculate a different value, and every observer everywhere will find the local speed of light to have the same value of $c$.
A: The speed of light does not increase in a gravitational field. Its the space that is bending and so the light will follow that bending
A: The light is not attracted by mass (at least by a small mass such as Earth). Hence there is no gravitational pull or acceleration by earth on light. Hence the velocity will be "c", that is all, no need to do a correction to it.
A: If we consider that a second is slightly shorter for our heads than it is for our feet, which is the conclusion we must come to based on the SI definition of the second, then we are forced into the GR solution of John Rennie and Co. which is, I would presume, a complicated but valid solution.
Now consider that a day is a day for our heads in exactly the same way as it is for our feet, since they both start and finish one rotation of the planet together. We can then, justifiably, assume a constant rate of time and a variable speed of light to come up with the following solution:
$$\Delta E = m(c + \Delta c) ^ 2 - mc^2$$
and
$$\Delta E = mg\Delta h$$
Which, by substitution and cancellation of m, gives us
$$(c + \Delta c) ^ 2 - c^2= g\Delta h$$
expanding
$$c^2 + 2c\Delta c + (\Delta c)^2 - c^2 = g\Delta h$$
[neglecting $(\Delta c) ^ 2$ as it is normally insignificantly small] 
We have
$$\Delta c = \frac {g\Delta h}{2c}$$
