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Suppose there are two masses on the ground $M$ and $m$. The distance is $d$. So according to newton's law the gravitational force between them is $F=\frac{GMm}{d^2}$. Now suppose that I am going upwards at a constant velocity $v$. Then by special relativity I would measure an increase in their mass relative to me. So according to newton's law, I would also measure an increase in the gravitational force between them. So if I go fast enough, then I would see that their gravitational force would overcome their friction and they would accelerate toward each other.

But according to another stationary observer, he would not see any increase in their masses and thus their gravitational force would be same to him and so they won't be able to overcome friction.

So how can two masses accelerate and stay still at the same time? So now I am confused does this mass increasing really happens?

Theoretical
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    This is a classic example of how relativistic mass can be misleading, and why modern treatments of relativity do not use relativistic mass. – PM 2Ring Nov 19 '18 at 15:52
  • Is this really a "classic example"? I think it's easily fixed by assuming increased friction due to increased normal force as well. – Jasper Nov 19 '18 at 15:54
  • @Jasper Fair point. OTOH, if the observer goes fast enough, the bodies (and the Earth) ought to acquire enough mass to become black holes. Obviously that doesn't happen. ;) – PM 2Ring Nov 19 '18 at 16:08
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    I think this is basically a duplicate of If a mass moves close to the speed of light, does it turn into a black hole? since that also asks if a fast moving object has an increased gravitational field. – John Rennie Nov 19 '18 at 16:19
  • Instead supposing that one goes horizontally in the direction pointed out of the two masses, the distance between the two masses would be contracted with a factor $\gamma^{-1},$ which would result in a factor $\gamma^4$ in Newton's law (twice from the masses and twice from the distance). – md2perpe Nov 19 '18 at 16:52
  • @John Rennie Apart from going really fast(forget the black hole part), will the gravitational force increase between them? If not why? Can you please give me a clear answer? – Theoretical Nov 19 '18 at 16:56
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    It is energy that creates gravity, so objections that relativistic mass is not mass are irrelevant. The question really is if the kinetic energy here contributes to gravity or not and why. None of the answers linked by John Rennie explain why or provide calculations in the moving frame of reference. So this is not a duplicate. – safesphere Nov 19 '18 at 17:13
  • @AsifIqubal OK, I've had a go at explaining why the speed doesn't affect the gravitational force. – John Rennie Nov 19 '18 at 18:13
  • This seems to be of a similar flavor to this supposed paradox, proposed by Tesla: https://physics.stackexchange.com/q/343556/4552 –  Nov 19 '18 at 22:26
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    Possible duplicate of If a mass moves close to the speed of light, does it turn into a black hole? –  Nov 22 '18 at 05:01

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Let's do a quick thought experiment to see why the speed does not affect the gravitational force between the masses. I'm going to modify your experiment slightly. We'll put a clock on each mass, and we'll start with the masses in contact. We'll give the give the two masses some initial small speed away from each other - this speed is small enough that relativistic effects can be ignored - so the masses separate, then slow to a halt then fall back together again.

Now, we're going to work in the centre of mass frame of our two masses. At the moment the masses separate we set both clocks to zero. We've started with the two masses together so that the two clocks are at the same point in spacetime, and we can therefore unambiguously synchronise both clocks to zero at the same time. All observers will agree that at the moment of separation the two clocks read zero.

The masses move away from each other until their mutual gravitational fields bring them to a halt, then they move towards each other and collide again at time $T$. That is at the moment of collision both clocks show the time to be $T$. Again both clocks are back in the same place so they can be directly compared. Again all observers will agree that the clocks show time $T$ at the moment of collision. So overall the process looks like this:

Clocks

But now suppose that you're in a rocket zooming past our experiment at $0.999c$. There will be various time dilation effects associated with your motion, but you are still going to agree that the two clocks showed $t=0$ when the masses separated and $t=T$ when they collided again. So you're going to agree that in the COM frame of the masses the process took $T$ seconds.

But all velocities are relative. Having the masses stationary and you zooming past at $0.999c$ is the same as having you stationary and the masses moving past you at $0.999c$. But that means with the masses moving past you at $0.999c$ they still separate when the clocks say zero and recollide when the clocks say $T$, and therefore even when the masses are moving at $0.999$ the process still takes $T$ seconds.

But of course the time taken for the masses to separate and recollide depends on the gravitational force, and since the time hasn't changed that means the gravitational force can't have changed either. Hence we have to conclude that having the masses move at $0.999c$, or indeed any speed, does not change the gravitational force between them.

John Rennie
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  • You said,"you're going to agree that the clock showed t=0 when the masses separate and T=T when they collide again." I couldn't understand how that would happen? What logic did you use? – Theoretical Nov 20 '18 at 06:31
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    @AsifIqubal the clock is just a mechanical device. If its hands point to zero when the masses first separate then all observers will see the two masses separating with the clock hand pointing to zero. Likewise if the clock hand points to $T$ when the masses recollide then all observers will see the clock hand pointing to $T$ when the masses recollide. – John Rennie Nov 20 '18 at 06:39
  • Now if I don't measure the time shown by the clock in the masses but measure the time by a clock which is in my hand (because of my curiosity) moving with a velocity equal to me, then what would I see? Will I see time $T$ after the complete process or see some else time in my clock? – Theoretical Nov 20 '18 at 07:56
  • @AsifIqubal if the masses are moving relative to you their time is dilated so you observe them to be moving more slowly. In your frame the separation and recollision takes longer than $T$ - longer not shorter. But this is just regular time dilation and you can easily correct for it. – John Rennie Nov 20 '18 at 08:48
  • Sorry for all the questions. But I couldn't help asking what would I feel if I were inside one of the masses? Will I measure the mass increased and how will I measure it from inside the ship? – Theoretical Nov 20 '18 at 10:21
  • @AsifIqubal The whole point is that the mass does not increase. The concept of a relativistic mass that increases with speed originated in the early days of special relativity before the theory was thoroughly understood and it is not used these days. It is a highly misleading concept that has confused generations of students. – John Rennie Nov 20 '18 at 10:32