Why does time pass slow in an orbiting satellite relative to an observer on the ground? I know that according special relativity, if an object travels at a very high constant velocity relative to an observer, the observer will see the object's clock running slow. I also know that according to general relativity, gravity affects the flow of time (gravitational time dilation). So what is the main reason time passes slow on the satellite? is it because it is moving relative to us or is it because it is at a higher point in the earths gravitational field.
Also since the satellite is in a constant free fall, should the tangential velocity be the same velocity which will cause a time dilation equal to the gravitational time dilation?
 A: People commonly talk about time dilation due to motion and time dilation due to gravity as though they were separate effects, but in general there is no natural way to make such distinctions. Questions about time dilation in curved spacetime can be handled by applying the general principles directly to calculate the total effect, and that is the only thing that matters in the real world. By comparing the total effects in special situations, the spirit of the question can still be addressed.
For simplicity, suppose that the earth is non-rotating and has a perfectly spherically symmetric distribution of mass. Then the question can be addressed using the Schwarzschild metric, with "time" coordinate $t$, "radial" coordinate $r$, and angular coordinate $\phi$. We only need three coordinates here because we'll only be considering motion in a single plane. 
It's important to remember that these are just coordinates. The time coordinate is not necessarily the proper time experienced by a given object, though one is a monotonically increasing function of the other (ignoring what happens inside the event horizon if the earth were to be compressed to a black hole). Similarly, the radial coordinate is not necessarily the proper distance from the origin, though one is again a monotonically increasing function of the other (outside the horizon).
Now, consider two different objects:


*

*Object $1$ is described by $(r,\,\phi) = (R_1,\,0)$ where $R_1$ is independent of $t$. This corresponds to an object hovering over some fixed point on the (non-rotating) earth. If we take $R_1$ to be the earth's radius, then the object is resting on the surface of the earth.

*Object $2$ is described by $(r,\,\phi) = (R_2,\,\omega t)$ where $R_2$ is independent of $t$. This corresponds to an object moving in a circular path centered on the origin, with period $\Delta t=2\pi/\omega$ in coordinate-time.
In both cases, we can use the Schwarzschild metric to deduce equations that say how the object's elapsed proper time $\Delta\tau$ is related to the elapsed coordinate time $\Delta t$. I won't show the derivations here, but I'll show the results. For object $1$, the result is
$$
 \Delta \tau_1 = \sqrt{1-\frac{2GM}{c^2R_1}}\,\Delta t.
\tag{1}
$$
For object $2$, the result is
$$
 \Delta \tau_2 
 = \sqrt{1-\frac{2GM}{c^2R_2}-\frac{R_2^2\omega^2}{c^2}}\,\Delta t.
\tag{2}
$$
The constants $G$, $M$, and $c$ are Newton's gravitational constant, the mass of the earth, and the speed of light, respectively, so 
$$
 \frac{2GM}{c^2}\approx 9\text{ millimeters}.
\tag{3}
$$
Equation (2) illustrates the statement made at the beginning of this answer: because of the square root, there is no natural way to say how much of the total time-dilation effect is due to gravity (the $G$ term) versus due to motion (the $\omega$ term). This is analogous to the relationship between the length of the hypotenuse of a right triangle and the lengths of its sides: because of the square root in that relationship, there is no natural way to say how much of the length of the hypotenuse is due to one side versus due to the other.
However, we can still make some interesting comparisons, very much in the spirit of the original question. For example, suppose that objects $1$ and $2$ are at the same radius, so that $R_1=R_2$. Then the objects meet each other once each period, and they can compare their elapsed proper times between any two consecutive meetings. Using $R\equiv R_1=R_2$ to denote their common radial coordinate, equations (1) and (2) give
$$
 \frac{\Delta \tau_2}{\Delta \tau_1}
 = \frac{\sqrt{1-\frac{2GM}{c^2R}-\frac{R^2\omega^2}{c^2}}}{
  \sqrt{1-\frac{2GM}{c^2R}}} < 1
\hskip2cm
 \text{for all }\omega\neq 0,\text{ if }R_1=R_2.
\tag{4}
$$
This says that when the two objects are at the same distance from the earth, the object moving around the circular path ages less between visits, compared to the object that is hovering over a fixed point on the earth. This is true for any value of the angular speed $\omega$, whether or not it corresponds to an orbit. So, in this particular situation, we can say that time passes more slowly on the satellite because the satellite is moving.
Equation (2) is valid for any circular motion, whether or not the object is in orbit. For an object in orbit, the Schwarzschild metric implies
$$
 \omega = \pm\sqrt{\frac{GM}{R_2^3}}
\tag{5}
$$
in the given coordinate system. Use this in equation (2) to get the proper time for an object in orbit:
$$
 \Delta \tau_2 
 = \sqrt{1-\frac{3GM}{c^2R_2}}\,\Delta t
\hskip2cm
 \text{(in orbit)}.
\tag{6}
$$
Now we can make another interesting comparison. For $R_1\neq R_2$, the two objects never meet. However, we can still compare the proper-time intervals that they each assign to a single orbital period, because equations (1) and (6) express both of their proper times in terms of the same coordinate time $t$. Therefore, the ratio
$$
 \frac{\Delta \tau_2}{\Delta \tau_1}
 = \frac{\sqrt{1-\frac{3GM}{c^2R_2}}}
        {\sqrt{1-\frac{2GM}{c^2R_1}}}
\hskip2cm
 \text{(in orbit)}
\tag{7}
$$
describes the period of the orbit as measured by object $2$ (the orbiting object) compared to the period of the orbit measured by object $1$ (the fixed object). Equation (7) implies the following interesting conclusion:
\begin{align}
\Delta\tau_2>\Delta\tau_1 &\hskip1cm\text{if }R_2>3R_1\,/2 \\
\Delta\tau_2=\Delta\tau_1 &\hskip1cm\text{if }R_2=3R_1\,/2 \\
\Delta\tau_2<\Delta\tau_1 &\hskip1cm\text{if }R_2<3R_1\,/2.
\end{align}
This is interesting, because it says that if the radial coordinate of the orbiting object is exactly $50\%$ larger than the radial coordinate of the fixed object, then their respective clocks agree with each other regarding the orbital period. So in this special case, we can say that the effects of gravity and motion balance each other. The other two cases are:


*

*If the orbit is higher than this special case, then the orbiting object's clock runs faster than the fixed object's clock. (The orbiting object ages more.)

*If the orbit is lower than this special case, then the orbiting object's clock runs slower than the fixed object's clock. (The orbiting object ages less.)
Trying to interpret these last results in terms of the relative magnitudes of gravity-induced and motion-induced effects would be an exercise of doubtful value, and it isn't necessary. The equations shown here capture the total effect, just like the pythagorean theorem captures the total length of the hypotenuse, without trying to separate it into an amount due to one side and another amount due to the other.
