Scenario help orbiting a neutron star

skypickle

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Certainly, tidal forces would rip most things apart as they approach the neutron star. But there is a solution!

Consider making a space ship which has 6 asteroids orbiting it in a 'polar' orbit (relative to its approach to the neutron star). Make these asteroids super massive. Because the space ship is at the center of these orbiting bodies, you would still be weightless AND not be pulled off center due to your excellent balancing skills. Now get the whole parade to gingerly approach the neutron star and enter an orbit around it. The tidal forces of the neutron star on your ship should be cancelled out by the orbiting super heavy asteroids.

yes?

Can I make this in Orbiter?
 

dbeachy1

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An interesting idea! But wouldn't the neutron star's gravity destabilize the orbits of the six super-massive asteroids orbiting about their barycenter, where your ship is? By definition, the neutron star would pull more on the asteroid(s) closest to it, and over time, at least, the barycenter would shift and thus the orbits of the six asteroids would destabilize, with the asteroids being pulled away from each other. But I'm not a physicist. :)
 

Urwumpe

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The simplest thing to prevent this would be staying out of the Roche limit.... You would likely still get fried by radiation there, but gravity would not cause problems.

For spacecraft and human bodies, which are not kept together by gravity alone, you could even get much closer, without having to fear tidal forces. Theoretically, if you are inside a spacecraft orbiting a neutron star, you could orbit the star so close, that you could get 1G apparent gravity alone by standing on a anti-radial wall inside a spacecraft orbiting the star and some distance away from the CoG of the spacecraft (including you). Yes, that could also apply to space stations.

Of course, its still better to avoid neutron stars, if possible, for various other reasons than just gravity...
 

n72.75

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I'm not sure of the logic that's leading you to the conclusion that appearing weightless would mean that you're not subject to tidal forces.
 

Linguofreak

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An interesting idea! But wouldn't the neutron star's gravity destabilize the orbits of the six super-massive asteroids orbiting about their barycenter, where your ship is? By definition, the neutron star would pull more on the asteroid(s) closest to it, and over time, at least, the barycenter would shift and thus the orbits of the six asteroids would destabilize, with the asteroids being pulled away from each other. But I'm not a physicist. :)

Not only that, but the asteroids themselves would be pulled apart by the tidal forces. A neutron star is roughly the radius of this asteroid, and weighs twice as much as the sun. This corresponds to densities on the order of 10^18. A cubic meter of the neutron star weighs at least ten times as much as an asteroid the size of the whole neutron star. The neutron star is as much denser than rock as rock is denser than the vacuum in the beam pipes at LIGO. The orbital period at the surface corresponds to frequencies in the audible range (1500 Hz or so).

You'd have more success figuring out a way to survive standing at ground zero when the Chicxulub impactor hit than you would trying to figure out how not to be ripped apart in low orbit around a neutron star.

---------- Post added at 17:52 ---------- Previous post was at 17:25 ----------

The simplest thing to prevent this would be staying out of the Roche limit.... You would likely still get fried by radiation there, but gravity would not cause problems.

For spacecraft and human bodies, which are not kept together by gravity alone, you could even get much closer, without having to fear tidal forces.

Well, at 2500 km, you'd get a gradient of about 3.5 gee / meter, which would be pretty near the redout limit (only half the blood in your body would be rushing to your head, instead of all of it, as in a proper negative-g situation, so you might not be *totally* incapacitated), and at 2000 km (still about 100 times the radius of the neutron star), that becomes 7 gee /meter, which would be incapacitating if not fatal (about 12 gee across your entire body, 6 gee from your midpoint to your head or feet).

Theoretically, if you are inside a spacecraft orbiting a neutron star, you could orbit the star so close, that you could get 1G apparent gravity alone by standing on a anti-radial wall inside a spacecraft orbiting the star and some distance away from the CoG of the spacecraft (including you). Yes, that could also apply to space stations.

Anti-radial walls on the anti-radial side of the spacecraft, radial walls on the radial side, and interior walls on the center plane.

EDIT: Just for perspective on what you're dealing with at low altitudes, the surface tidal gradient on a neutron star is on the order of 6 *million* gee / meter.

In other words "friends don't let friends orbit neutron stars".
 
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Urwumpe

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Well, at 2500 km, you'd get a gradient of about 3.5 gee / meter, which would be pretty near the redout limit (only half the blood in your body would be rushing to your head, instead of all of it, as in a proper negative-g situation, so you might not be *totally* incapacitated), and at 2000 km (still about 100 times the radius of the neutron star), that becomes 7 gee /meter, which would be incapacitating if not fatal (about 12 gee across your entire body, 6 gee from your midpoint to your head or feet).


Well, considering that we have 140% the mass of the sun concentrated into a tiny sphere, that is merely 40 km or so in diameter, I would say proper orbital distances are should also follow solar scales - and be measured in milli-AU (One sun radius = 4 mAU... can we maybe coin the unit for such small distances "meow"?)
 

Keatah

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Your asteroids would have to be as massive as a neutron star in order to cancel anything out. And then there'd be tidal forces from two objects.
 

Linguofreak

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Well, considering that we have 140% the mass of the sun concentrated into a tiny sphere, that is merely 40 km or so in diameter, I would say proper orbital distances are should also follow solar scales - and be measured in milli-AU (One sun radius = 4 mAU... can we maybe coin the unit for such small distances "meow"?)

I was actually assuming a 2 Msun neutron star, not a Chandrasekhar mass neutron star.

Anyways, the size of the star is on the scale of 100s of nano-au, and the tidal limit for humans is in the 10-20 micro-au range.

---------- Post added at 05:40 ---------- Previous post was at 05:35 ----------

Your asteroids would have to be as massive as a neutron star in order to cancel anything out. And then there'd be tidal forces from two objects.

Well, they'd have to be as dense as a neutron star to avoid tidal disruption, but as long as they were closer, they could still possibly provide similar tidal gradients. But even so, gravity just can't do what the OP is proposing. And even if it could, they'd be dragged out of formation by tidal forces, as dbeachy mentioned.
 
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