Why can an object not reach the speed of light by falling in a gravitational field with constant acceleration? I recently learned in school that the gravitational potential energy of an object is equal to $mgh$. I also learned that the kinetic energy of an object is $\frac12 mv^2$. If an object falls in a vacuum, it will convert all of its gravitational potential energy to kinetic energy before hitting the floor. Therefore, 
$$mgh = 1/2mv^2$$
$$gh = 1/2v^2$$
If $g = 10 N/kg$ on a planet then to reach the speed of light:
$h = 1/20c^2$
If an object were dropped from this height I assume it would not reach the speed of light. I think this is because $g$ would get smaller as you went further away. I found out that $F = \frac {Gm_1m_2}{r^2}$ and that the weight of something is $F = m_2g$. Therefore, $g =\frac {Gm_1}{r^2}$. If instead of having $g = 10 N/kg$ on the surface, if $g = 10 N/kg$ at that height then $10 =\frac {Gm_1}{(c^2/20)^2}$. Therefore, the mass of the planet would be $m_1 = c^4/(40G)$. 
So, if you dropped something from a height of $c^2/20$ in a vacuum (although unreasonably large) on a planet with mass $c^4/40G$, surely it would accelerate fast enough to reach the speed of light? As it fell, $g$ would only increase so why wouldn't it become faster than light? What have I done wrong? Thanks
 A: The short answer is special relativity.
classically
According to classical mechanics, there is nothing to prevent an object from accelerating faster than the speed of light.  By conservation of energy you can calculate the speed of a dropped object by looking at the change in gravitational potential.  As you note, the equation $PE=mgh$ needs to be modified to handle the changing gravitational field $g$ once the object is far from the surface of the Earth.
We can use the Newtonian gravitational potential $V(r) = - \frac{GM}{r}$, where $r$ is the separation between the central mass $M$ and the falling object.  The gravitational potential energy is $PE=mV=- \frac{GMm}{r}$.  Dropping an object from rest we find:
$$ E_\mathrm{initial} = E_\mathrm{final} $$
$$ -\frac{GMm}{r_\mathrm{initial}} = -\frac{GMm}{r_\mathrm{final}} + \frac{1}{2}\, m\, v^2 $$
$$ v = \sqrt{ 2\, GM \left( \frac{1}{r_\mathrm{final}} - \frac{1}{r_\mathrm{initial}} \right) } $$
By choosing $M$ and the $r$'s appropriately we can make $v$ whatever we want.
special relativity
If we want to use special relativity, we need to modify the expression for kinetic energy.  Special relativity also accounts for the rest energy of the falling particle.  The rest plus kinetic energy is $E = \gamma\, mc^2$, where
$$\gamma = \frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}$$
is the Lorentz factor for a moving particle.  If $v=0$, then $\gamma=1$ so the energy is just the rest energy $mc^2$.
By conservation of energy
$$ -\frac{GMm}{r_\mathrm{initial}} +  mc^2 =  -\frac{GMm}{r_\mathrm{final}} + \gamma\,mc^2 $$
$$ (\gamma - 1)\,c^2 = GM \left( \frac{1}{r_\mathrm{final}} - \frac{1}{r_\mathrm{initial}} \right)$$
The velocity of the dropped particle is hiding in the $\gamma$.  After some algebra we can find:
$$ v = c \sqrt{1 - \left[ \frac{1}{\frac{GM}{c^2} \left( \frac{1}{r_\mathrm{final}} - \frac{1}{r_\mathrm{initial}} \right) + 1} \right]^2}$$
You can check and see that the square root term is always less than 1, so the dropped particle does not go faster than the speed of light!
Better yet, you could use general relativity instead of Newtonian gravity... but that's beyond the scope of this question.
