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As a particle's velocity increases, its mass increases(gamma times). Therefore, if a particle is in a gravitational field, the gravitational force it experiences must also increase(gamma times). The net acceleration of the particle, i.e.(gamma*force)/(gamma*mass) should therefore remain constant for small distance traversed by the particle and increase as the particle travels closer and closer to the object generating the gravitational field(due to inverse square relation). Hence. is it possible to accelerate a particle indefinitely using gravitational field?

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What do you mean by indefinitely? Eventually either the particle hits the object or it misses and flies away. That article uses Newtonian mechanics but the picture is qualitatively the same in general relativity. –  Michael Brown Jun 26 '13 at 5:30

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No. It is not possible.. The misconception you have is because you are thinking of the gravitational field as being constant. However, this is not true. Ignoring air resistance for the time being, consider a projectile launched off the earth. Now, if it is launched at the escape velocity from the earth, then it will take an infinite amount of time to reach 0 speed, and by then it will be infinitely far away.

Now, consider the opposite situation. If the projectile is dropped from infinitely far away, it will take an infinite time to fall, but when it falls, its velocity would be the escape velocity, i.e. it would be exactly $$v=\sqrt{\frac{2GM}{r}}$$ Now, we know that this escape velocity cannot be greater than the speed of light. It is the speed of light only when $$r=r_s=\frac{2GM}{c_0^2}$$ Now, consider the situation where the object had some initial velocity towards the planet (not due to the gravity) when it was infinitely far away? Naive Newtonian Mechanics (NNM) would tell you that it would be $v=v_0+\sqrt{\frac{2GM}{r}}\mbox{ could be greater than } c_0$. However, Special Relativity (SR) is required at such velocities. It tells you that the actual final velocity would be: $$v=\frac{v_0+\sqrt{\frac{2GM}{r}}}{\sqrt{1+\frac{v_0\sqrt{\frac{2GM}{r}}}{c_0^2}}}=c_0\frac{v_0+\sqrt{\frac{2GM}{r}}}{\sqrt{c_0^2+{v_0\sqrt{\frac{2GM}{r}}}}}\leq c_0$$ As required.

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