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Take the most massive known star, R136a1, with a mass of $5.3 * 10^{32}$kg and a raidus of $2.5*10^{10}$m. Now place a $1$kg metal ball $10*10^{24}$m away (about a billion ly, while still being SI units). Now let everything go and wait until they collide.

All rounded numbers to hopefully make the math easier. Also ignoring effects like space dust and solar wind.

How would I calculate the impact velocity of this baseball? How do I take into account relativistic effects?

When looking around for solutions to this, I found answers for very earthy problems, like dropping a baseball from 3 meters up, or even asteroid impacts. I'm looking for how to factor in relativistic effects into it.

P.S. by "impact velocity" I mean the velocity the ball reaches when its distance from the star's core is equal to the star's radius. I'm aware there wouldn't be a real "impact".

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  • $\begingroup$ Hint : use potential energy values. I do not think relativistic effects are significant. $\endgroup$ – StephenG Oct 2 '18 at 23:32
  • $\begingroup$ You play baseball with metal balls? :) $\endgroup$ – Aaron Stevens Oct 2 '18 at 23:46
  • $\begingroup$ @StephenG Turns out they're not. I underestimated the speed of light. $\endgroup$ – Daffy Oct 3 '18 at 1:03
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    $\begingroup$ @AaronStevens lol, forgot to change that part. The question originally was about a 0.15kg baseball, but that just adds unnecessary confusion when a 1kg metal ball works just as well. $\endgroup$ – Daffy Oct 3 '18 at 1:04
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Earthy solutions suffice. Relativistic corrections only matter if the metal ball ends up moving near the speed of light. If you do the calculations for what the potential energy of the ball is at 1 billion light years away, $PE = -GMm/r$ with $M = 5.3 \times 10^{32}\mathrm{kg}$, $R = 1\times10^{25}\mathrm{m}$, and compare that for when $R = 2.5 \times 10^{10}m$, you'll find that the ball has gained only $\approx 1.4 \times 10^{12}$ Joules of energy. This sounds like a lot, but if you convert that amount of kinetic energy into velocity using the nonrelativistic formula $KE = 1/2 mv^2$, you'll find that the ball only moves at $10^6$m/s.

Since the gamma factor for relativistic corrections is $\gamma = 1/\sqrt{1-v^2/c^2}$, at this speed, the gamma factor is still very close to 1, and relativistic corrections are still too small to matter.

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    $\begingroup$ How did you get only 150 J?? $G$ is ~$10^{-11}$, so the answer should be somewhere around $10^{32-11-10}$ = $10^{11}$ J. $\endgroup$ – HiddenBabel Oct 2 '18 at 23:56
  • $\begingroup$ @HiddenBabel indeed, edited. $\endgroup$ – Allure Oct 3 '18 at 0:00
  • $\begingroup$ Good correction! My calculations show that the energy gained is $1.41\times10^{12}$, like you said. But the final velocity would be $1.68\times10^{6}$, which is about .0056 times the speed of light. So there is no need for relativistic calculations. $\endgroup$ – Stuart Van Horne Oct 3 '18 at 0:08
  • $\begingroup$ Very helpful answer! Turns out I underestimated the speed of light. This started because I read a comment on a Youtube video about the speed of light and why it can't be broken. The comment was about a large mass and a far away object, and the constant acceleration supposedly breaking the speed of light. I knew intuitively that relativity would prevent that but I didn't have the math behind it. Would it be alright to post another question with changed numbers? $\endgroup$ – Daffy Oct 3 '18 at 1:09
  • $\begingroup$ @Daffy absolutely, go ahead and ask. $\endgroup$ – Allure Oct 3 '18 at 1:21
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Very near the escape speed. This is because the particle is starting from what is in effect "near infinity". To get a particle all the way to infinity requires you to give it at least the escape velocity upon launch, by definition thereof. To get it near infinity you thus need a velocity just shy of that.

The situation for an object falling in from infinity or near infinity is simply the time reversal of the situation where the object is climbing out toward that remote point, and when you take the time reversal the final point of the motion becomes the initial point, thus representing an object starting out very if not infinitely far and finally ending at the surface with the same speed - thus near or at the escape speed.

So the object here will hit the surface with speed very close thereto.

That's the intuitive understanding. To make it rigorous, we can find it mathematically. We do this by considering the energy difference between the two points. In particular, the difference in potential energy between a radius $r_i$ and $r_f$ is

$$\Delta U = -GMm \left(\frac{1}{r_f} - \frac{1}{r_i}\right)$$

and if we set this to kinetic energy to clear the test mass $m$, we get

$$\frac{1}{2} mv^2 = -GMm \left(\frac{1}{r_f} - \frac{1}{r_i}\right)$$

so

$$v = \sqrt{2GM \left|\frac{1}{r_i} - \frac{1}{r_f}\right|}$$

(we use the absolute value to make it independent of the direction we are going) and if you take now $M = 5.3 \times 10^{32}\ \mathrm{kg}$ and $r_i = 10\ \mathrm{Ym} = 10^{25}\ \mathrm{m}$ while $r_f = 2.5 \times 10^{10}\ \mathrm{m}$, you get that the velocity is about 1682 km/s. For comparison, you can find the escape speed by

$$v_\mathrm{esc} = \sqrt{\frac{2GM}{r}}$$

and this gives also indistinguishably, 1682 km/s. Indeed, using these specific mass figures, which is going way beyond their level of precision, the difference is on the order of parts per trillion , that is, micrometers per second .

We can also find the time required for the ball to fall in from this distance. The fall, since we're outside the uniform-gravity approximation regime, can be approximated very well as an extremely high-eccentricity orbit: actually, maximal eccentricity, $e = 1$. Thus the fall time is half an orbital period, which is given by Kepler's third law, which gives

$$T_\mathrm{orbit} = \sqrt{\frac{4\pi^2 a^3}{GM}}$$

where $a$ is the semi-major axis, so

$$T_\mathrm{fall} = \frac{1}{2} \sqrt{\frac{4\pi^2 a^3}{GM}}$$

Taking $a = 10\ \mathrm{Ym}$ just same gives a fall time of about 530 Ys, or $5.3 \times 10^{26}\ \mathrm{s}$. For comparison, the age of the Universe is only about 435 Ps (13.8 billion years), and this is $530 \times 10^{9}$ Ps, so over a billion times longer. The star would die long before the particle even got underway.

And moreover, this is only possible in an empty(!) Universe with no dark energy. In our Universe, the cosmic expansion alone would tear the ball away from the star.

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