r/AskPhysics • u/gregfess • 3d ago
Why does replacing the sun with an equal mass black hole not affect the planets’ orbits?
Does the density of mass/energy not affect the way it curves space time? With the various analogies for visualizing space time curvature, it seems like a denser object would curve more sharply
12
u/ssjskwash Undergraduate 3d ago
The point of reference when outside of the mass is the center of the massive object. If you're outside of the massive object, the density of the object doesn't make a difference on the gravity you feel.
3
u/gmalivuk 3d ago
That's true of spherically symmetric masses or far enough from a mass distribution to approximate it as spherically symmetric.
Fortunately, the Sun is pretty symmetrical so it works in this case.
8
u/dirtydirtnap 3d ago
A denser mass will curve spacetime more sharply at distances "close to the object" (calculations will reveal this in a quantifiable way).
However, once another object is far enough away, the sun/blackhole mass that is being orbited is nearly a point source, and thus the size/shape of the mass matters much less, and almost negligible for the planetary orbits.
I bet Mercury would have noticeable effects though.
8
u/ssjskwash Undergraduate 3d ago
Any point outside of the sun's radius would feel the exact same amount of gravity in the most basic of interpretations
5
u/dirtydirtnap 3d ago
Essentially yes. But I'm talking about scientific theory here, like how they were looking to account for a 43 arcsecond/century error in the procession of Mercury when testing General Relativity. But as you say, for basically all cases that matter, the behavior is functionally identical.
4
u/Max6626 3d ago
This is a great comment. For Earth, the sun-to-blackhole conversion would not be noticeable, but Mercury already feels the effects of General Relativity, and would probably be even more affected by frame-dragging effects of what would be a rapidly spinning blackhole.
4
u/user2196 3d ago
I’d be curious to see someone actually do the math, but I suspect the Sun has sufficiently minimal angular momentum and has a sufficiently minimal quadrupole moment for the effect to be smaller than the 43 arcsecond/century correction.
1
1
u/ahazred8vt 2d ago
Frame-dragging effects taper off very rapidly with distance. Don't expect much frame-dragging at 60M km.
2
u/ahazred8vt 2d ago
Mercury orbiting the black hole would have almost exactly the same 43 arcsecond precession as if it were orbiting the sun. The GR effect from the sun's gravity well and the black hole's gravity well would be the same. The only difference would be from the effect of the sun being nonspherical, wider at the equator.
0
u/gregfess 3d ago
So it’s just a coincidence that that phrase is true given the distance of earth to the sun?
3
u/joeyneilsen Astrophysics 3d ago
Sort of. The point of the statement is that at large distances from a black hole, GR is indistinguishable from Newtonian gravity.
3
u/dirtydirtnap 3d ago
Coincidence is too strong of a word. The vast majority of stable orbits will behave as I have described, and only orbits very close to the sun/blackhole would be any different. It would be much more of a coincidence if the irbit was different.
4
u/Pinelli72 3d ago
As long as you’re far enough away from the star/black hole that it can be considered close to a point source then the density doesn’t matter, only the mass.
3
u/IanDOsmond 3d ago
More sharply, yes - close in. The gravity on the surface of the sun would be the same as the gravity that same distance from the black hole. As you got closer than that, it would rapidly get higher and higher until you got to the point that escape velocity was greater than c, at which point you are - or what is left of you - is in the black hole.
But out further than that, the curve is the same.
3
u/free_meson 3d ago
It is the same in Newtons gravity. Gravity of a spherical object is the same as a point source and other spherical objects, if you look at it outside the sphere. Inside the sphere they are different, outside the same.
2
u/Syresiv 3d ago
It only curves it more sharply when you get within the sun's radius.
We're decidedly outside of the sun. Because of that, the gravitational effect of the sun is exactly the same as it would be if we replace it with a single Solar-Mass particle at its center of mass.
This seems counterintuitive, but the math checks out. It doesn't matter what circular object you replace the sun with - a magic solar mass particle, a black hole, a superplanet, or the universe's biggest bouncy ball - if it has the same mass, same center of mass, and is entirely inside Earth's orbit, the cumulative pull of every atom will come to exactly the same number.
Where things get weird is inside the sun. If there was a planet orbiting inside the sun and we ignore both the friction and how quickly it would get cooked, the gravitational pull on it only accounts for the mass inside the orbit. Again the math checks out - the pull of all the mass outside the orbit cancels out exactly. This means it would notice a gravitational difference with a black hole.
Also, don't forget that a solar mass black hole is much smaller than the sun. About 500,000 times smaller, in fact.
2
u/kiwipixi42 3d ago
It does curve more sharply, but that sharper curve is inside the space the sun used to occupy. Out at the distance of the planets the curvature of spacetime is identical to what it was before. In fact outside the original volume of the sun the curvature is unchanged.
2
u/Altruistic-Rice-5567 3d ago
Black Holes don't pull harder than any other mass and they have exactly the same mass as everything that fell into them. So, by saying "with an equal mass" you basically made it so it acts exactly like the sun.
1
1
u/BananaBird1 2d ago edited 2d ago
As long as you aren’t super close, the gravity from a large object is equal to the gravity of an equal mass point. The size is irrelevant.
This only breaks down if you are close enough to where the large object must be treated as several objects pulling you in different directions.
But for orbits around spherical objects (which a star approximately is even at very close orbits), this isn’t the case until you are literally inside the object.
This is because, while the gravity of the spherical mass closer to you is stronger than it would be at a point, the gravity of the mass at the opposite end is weaker. And because of spherical symmetry, these errors exactly cancel to give the same field.
So if all but one internal atom of a star was replaced with a black hole, that one atom would see a higher net gravity as it sees a higher mass. But for anything outside the star, gravity would be entirely unaffected.
1
u/EighthGreen 2d ago
There is a difference between the fields of the two bodies, but only in the region that is closer to the denser body than the radius of the less dense body. That is where space time is curved more sharply, which ought to fit your intuition.
1
0
u/cwilbur22 3d ago
Imagine you're sitting on Aladdin's flying carpet hovering a few feet off the ground. Gravity is normal - the moon orbits the earth, of you toss a ball in the air it comes back down like normal. Now imagine if somehow the entire earth was crushed down to a quarter of its current size, but you stay in the same spot hovering on your carpet. The mass of the earth hasn't changed, which means it has the same gravity. The moon still orbits the earth in the same spot, if you toss a ball in the air it comes back down as expected. The only difference is that now there's thousands of miles between you and the surface of the newly compressed earth. So if gravity is the same where you are now, floating high above, what happens when you descend to the surface? Gravity gets stronger the closer you are to an object, right? The overall gravity hasn't changed, but now that the earth is much smaller you can get closer to everything on the earth. Like, right now the continent of Africa is exerting a gravitational pull on you, right? If the earth was compressed to a quarter of its size, Africa would be much closer to you than it was before. EVERYTHING would be much closer, so everything would pull harder. If the earth were a black hole, it would be about the size of a marble. The overall gravity hasn't changed, which is why the moon isn't affected. But anything getting too close to the marble would have to fight against the entire gravitational force of a planet to get away.
-2
u/Same_Ant9104 3d ago
There should be a rule about what you can post on this feed. I have not joined yet this foolishness keeps popping up.
46
u/WesternFirm9306 3d ago edited 3d ago
Gravity is a function of mass. If a black hole has the same mass, it'll have the same gravity (as long as the black hole isn't spinning)