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So lets say if there is a hypothetical space craft capable of surviving the gravity of the neutron star and the heat, and say they are on the surface of a neutron star that is about 2.2 times the mass of the sun. How perfect would their escape have to be to actually escape, assuming they are on full thrust. Would any slight disruptions cause them to crash land into the surface? Is this true for light as well?

MiltonTheMeme
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  • That's very hypothetical. You need very good unobtainium to avoid being crushed into neutronium just by the gravity, even before you take off and subject your ship and crew to the astronomically high thrust of the rocket engine. ;) – PM 2Ring Jun 07 '23 at 22:44
  • Not all matter on a neutron star turns into neutronium though right? The surface is made of iron so what if it is an iron spacecraft. The spacecraft's mass if considerably smaller maybe about 100 tons should not affect the neutron star too much. – MiltonTheMeme Jun 08 '23 at 15:00
  • Ok, the material on the neutron star surface tends to be iron-peak elements. ProfRob has several posts on neutron star composition, structure, etc, eg https://physics.stackexchange.com/a/225530/123208 & https://physics.stackexchange.com/a/762359/123208 But my point is that the gravity is orders of magnitude stronger than the mere chemical forces that maintain the shape of the spaceship. – PM 2Ring Jun 08 '23 at 15:31

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When you try to escape the influence of any gravitational body (not just neutron stars), you can have three results:

  1. Parabolic trajectory - crashing to the surface
  2. Elliptical trajectory - an orbit
  3. Escape trajectory

If they have sufficient thrust to escape, but their engine cuts out for just a moment and then recovers, they can still escape so long as their thrust is restored in time to deliver enough acceleration to their spacecraft that is doesn't crash back down first.

Those are your only options. If you lack the ability to make it into an escape trajectory, then you can either recover into an elliptical trajectory or crash down to the surface

As for light, it always moves at $c\approx3\times10^{8} m/s$. Gravity does, however, influence its direction of travel, and in extreme cases can even cause it to orbit or fail to escape.

Logan J. Fisher
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  • But in the case of a neutron star, even being millidegrees off can result in crashing back because neutron stars if dense enough can have the point of no orbit. – MiltonTheMeme Jun 07 '23 at 22:18
  • Yes, that's true. Neutron stars are dense enough to have an innermost stable circular orbit. Aside from that, it's worth clarifying that there's no real difference in this problem between a neutron star and Earth. It's the exact same laws of rocketry. – Logan J. Fisher Jun 07 '23 at 22:21
  • Does an object trying to orbit in the ISCO crash onto the surface? – MiltonTheMeme Jun 07 '23 at 22:24
  • At the ISCO? No. It's stable. Ever so slightly inside? Yes, unless they provide a thrust to main their distance. – Logan J. Fisher Jun 07 '23 at 22:27
  • Ok thanks for the answer – MiltonTheMeme Jun 07 '23 at 22:29
  • Actually, your case 1 is also an elliptical arc. We just normally approximate it as parabolic because we assume that g is constant. But that's not a good approximation on a neutron star. And really, we should be using GR, but fortunately the ballistic trajectories of a test particle "look" Newtonian, if we use Schwarzschild coords, but bear in mind that the proper time on the orbiting body is rather different to the Schwarzschild time, which is the proper time of the observer at infinity. – PM 2Ring Jun 07 '23 at 22:40
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    PM 2Ring. Fair point about it being inappropriate to consider g as being approximately constant for a neutron star. – Logan J. Fisher Jun 07 '23 at 22:48
  • Points 1 and 2 are not true in the (applicable) Schwarzschild metric. – ProfRob Aug 06 '23 at 20:54