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I was thinking about what happens when two objects fall toward each other in space. The farther apart they are when they begin falling, the faster they will be traveling when they hit each other due to gravity's acceleration. So, if they are far enough apart, what prevents them from falling towards each other faster then the speed of light?

Can someone explain where in my thought process I am going wrong?

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  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/24319/2451 , physics.stackexchange.com/q/170502/2451 and links therein. $\endgroup$ – Qmechanic Jan 28 '18 at 16:35
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    $\begingroup$ The only thing wrong with your thought process is that you are thinking about it in terms of classical mechanics. Classical mechanics predicts no maximum speed. I am not a physicist, and I do not know how to set up and solve the relativistic equations that describe the situation, but I trust that the speed would be limited if you solved those equations. $\endgroup$ – Solomon Slow Jan 28 '18 at 16:39
  • $\begingroup$ And even just using Newtonian gravity, there will be a maximum speed, aka terminal velocity, assuming the masses are real objects, not point-like. $\endgroup$ – PM 2Ring Jan 28 '18 at 16:57
  • $\begingroup$ The gamma factor prevents superluminal motion. For your interest, see the case for charge in constant electric field here $\endgroup$ – Ng Chung Tak Jan 28 '18 at 20:01
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To simplify the calculation let's consider a light object of mass $m$ falling towards a heavy object of mass $M$. The reason this simplifies things is that we can treat the large object as fixed and just work put what happens to the light object as it falls into the gravitational field of the large object.

We'll also assume that the light object starts its fall with a negligible velocity i.e. with a kinetic energy of approximately zero.

Now we can work out the speed of the falling object just from conservation of energy. The object starts out with zero kinetic energy and zero potential energy so its total energy is zero, and conservation of energy means the total energy has to remain zero. When the object has fallen to a distance $r$ from the large object Newton's law of gravity tells us that its potential energy has become:

$$ V = -\frac{GMm}{r} $$

Since the total energy has to remain constant at zero that means the kinetic energy, $T$, has to be equal to $-V$ so that $T$ and $V$ add up to zero. That means:

$$ T = +\frac{GMm}{r} $$

And we know that kinetic energy is $\tfrac{1}{2}mv^2$ so we get:

$$ \tfrac{1}{2}mv^2 = \frac{GMm}{r} $$

Your question asks if there is some distance where the speed $v$ becomes greater than the speed of light $c$. If we set $v=c$ in our equation and rearrange it to find $r$ then we get:

$$ r = \frac{2GM}{c^2} $$

But ...

This distance $r$ has a special significance in general relativity. It is the radius of a black hole of mass $M$ and it is called the Schwarzschild radius. This means two things:

  1. this could only happen if our large mass $M$ is a black hole

  2. because relativity is involved we can't use Newtonian mechanics to analyse the motion of the falling object

You probably know that for fast moving objects time is dilated and length is contracted. Both effects become important as an object accelerates towards a black hole and as a result it does something very strange. Initially it accelerates towards the black hole, but then near to the event horizon it starts slowing down and eventually (given infinite time) comes to a halt at the horizon. At no point does it ever exceed the speed of light.

If you're interested in the details I do the calculation in my answer to Will an object always fall at an infinite speed in a black hole? The maximum speed the object attains is about $0.38c$.

So the answer is that no, an object cannot fall towards another object faster than the speed of light.

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  • $\begingroup$ Why is the small object's potential energy zero to start? If it is far above the surface of the large object, wouldn't it have a lot of gravitational potential energy? $\endgroup$ – TrumpetDude Jan 28 '18 at 22:58
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    $\begingroup$ @TrumpetDude it's just a convenient convention. We take $V=0$ at infinity then the potential becomes negative as we approach the massive object. $\endgroup$ – John Rennie Jan 29 '18 at 5:31
  • $\begingroup$ Is it fine to plug c in the T = -V ? Shouldn't the kinetic term be already modified to account for SR? I see the value of the treatment especially when combined with the other answer to "on falling towards a black hole". It is just me that I would have inserted SR already. At least conceptually. $\endgroup$ – Alchimista Jan 30 '18 at 12:08
  • $\begingroup$ @Alchimista no, potential energy is an evasive quantity in GR. You need to do a fully relativistic calculation to calculate the speed of an object falling into a black hole. This is a standard derivation that you'll find in any introductory GR book. $\endgroup$ – John Rennie Jan 30 '18 at 12:11

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