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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

sarat.kant
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Matthew Barber
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  • You would need to read up on Einsteins theory of special relativity for this. When a object nears the speed of light, its mass starts increasing in relation to its speed. So the acceleration produced due to the gravitational force on that body will decrease due to its increasing mass. This acceleration keeps decreasing at such a rate that the particle will never reach the speed of light. And by the way, the mass of such a planet exceeds the mass of the largest supermassive black hole found till date. So I don't think it will remain a planet anymore. – sarat.kant May 11 '16 at 13:01
  • and whoever edited the question to include mathjax forgot a little thing that the OP forgot too. The $c^2/20$ in the denominator should be squared. – sarat.kant May 11 '16 at 13:03
  • The answer is that it can: there are very famous objects where things falling into them reach the speed of light, which are black holes. Caveats: the field is not uniform, and you necessarily have to treat the whole thing relativistically which means that simple Newtonian gravitation approach does not work at all well. –  May 11 '16 at 13:08
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    @sarat.kant The use of "relativistic mass" (gamma m) is considered old-fashioned and avoided these days. Only the invariant, rest mass is of interest in SR. As an example of the problems relativistic mass causes: if the falling object's mass increases shouldn't the gravitational force increase as well and compensate? – Paul T. May 11 '16 at 13:37

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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.

Paul T.
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