# Would our orbit really remain the same if the sun were a black hole of equal mass?

There seems to be an idea floating around that the sun could be replaced by anything of equal mass with no consequence to our orbit.

It seems to me that if the mass of the sun were confined to a single point that the local geometry of space would be different, therefore our orbit and those of the other planets would also be different.

1. Am I mistaken to think of this as analogous to distribution of weight on an elastic surface? - (A more confined distribution of mass would create a deeper gravity well, so to speak).

2. I'm thinking that angular momentum would need to be conserved and that might result in frame dragging which may need to be accounted for, but I'm not sure and I don't know how to calculate it (yet).

3. Anything I'm overlooking?

P.S. I know black holes don't come in our size.

Thanks!

## 3 Answers

It depends what you mean by 'Would remain the same'. If you are asking about the exact precise orbit, the answer is yes, it would change, since the Schwarzschild (and also Kerr solution of a rotating black hole) solution is a vacuum solution of the Einstein equations, so it describes strictly speaking only Vacuum, with a curvature singularity in the centre. Since the Sun is made out of matter, has a finite size and you need many, many variables to fully describe it (for example it is non-spherical) the metric around it is not Schwarzschild (remember that a Schwarzschild black hole is only described by one parameter).

However, the Schwarzschild metric is a good approximation around the Sun, and for current observational purposes it describes nature well. What one should do, is to calculate the perturbations around the Schwarzschild metric in some form of power series in the characteristic length scale of the Sun over the Schwarzschild radius of the Sun. You will then find that these corrections are negligible at our current measurement capabilities. This last statement is with the exception of the advancement of the perihelion of Mercury, where we can now measure the quadrupole moment $$J_2$$ of the sun to be non-zero. This means that one should replace the Sun not with a Schwarzschild black hole, but with a Kerr black hole (to achieve a non-vanishing quadrupole moment).

As an aside, it is also important to note that the parameter that describes a Schwarzschild black hole $$M$$ is precisely constructed such that the physics around that central object reduces to the orbits of a body with mass $$M$$ in Newtonian gravity. It is therefore not surprising that the orbits do not change that much far from the event horizon.

• Thanks konstle. That's very insightful. Apr 10, 2022 at 0:45
• “ You will then find that these corrections are negligible at our current measurement capabilities.” Citation needed. These corrections are certainly significant for Mercury’s orbit. Apr 10, 2022 at 6:52
• @mmeent are you sure about that? Note that we're not talking about corrections due to general relativity versus Newtonian gravity, but rather (if I understand correctly) corrections due to the sun's mass distribution not being perfectly spherical. Apr 10, 2022 at 7:08
• I misremembered the effect of the quadrupole moment by about an order of magnitude. It is currently on the cusp of being significant for Mercury, which makes it certainly insignificant for the Earth’s orbit. Nonetheless the assertion that it is insignificant in this answer needs to be backed up with something more factual. Apr 10, 2022 at 7:29
• @mmeent I added a comment and a reference to the answer. For example from measurements with MESSENGER spacecraft one finds a a non-vanishing quadrupole moment of the Sun. This means that a Schwarzschild black hole is insufficient to explain the metric surrounding it. Apr 10, 2022 at 15:25

There seems to be an idea floating around that the sun could be replaced by anything of equal mass with no consequence to our orbit.

I have to correct my former answer. That idea is correct!

Let's assume the earth is orbiting in the static spherically symmetric spacetime shaped by the sun (we neglect the earth contribution to spacetime curvature). The corresponding metric is the famous Schwarzschild solution of Einstein field equations (https://arxiv.org/abs/physics/9905030) : $$$$ds^2=\Big(1-\frac{2GM}{c^2 r}\Big)~c^2 dt^2-\Big(1-\frac{2GM}{c^2 r}\Big)^{-1} dr^2-r^2 d\Omega^2.$$$$ That metric depends only on the parameter $$M$$, thus, any other spherical gravitational object with the same mass, including a black hole, will have the same exterior spacetime geometry. Orbits, as geodesics of that spacetime, will be identical, too.

• Thanks Jan, this is a great answer, specifically in that address is point #1 of my question quite succinctly. This may be a lot to ask, but if you happen to have a link to the compactness parameter so that I might read more about it, that would be fantastic. In any case, thanks for the insight. Apr 10, 2022 at 20:48
• @StuartB, if you like take a look into physics.stackexchange.com/a/679431/281096. However, today I have become doubts whether my answer is correct. I am checking it now using mathematics from arxiv.org/abs/0812.1806v3. Apr 11, 2022 at 11:20
• @StuartB, I have had to correct me, see the answer again! Apr 11, 2022 at 14:48
• Thanks for setting this straight Jan! Apr 12, 2022 at 1:51

Yes, the number of gravitational degrees of freedom remains the same if the Sun were magically turned into a black hole. Now apply that to disk-like interaction, and you have focused, super strong gravitation. Count.