# B Local gravity vs dark matter

#### Viopia

Well, the net effect of gravity can be zeroed out as well, say, by having mass equally distributed about the target mass.
Hi. Could the local effects of gravity between nearby stars at the edge of a galaxy cause these stars to rotate around the centre of the galaxy more quickly than predicted if the prediction is based upon individual stars only. This effect would be similar to the way the Earth "drags" the Moon around the Sun in a different orbit than if the Moon was alone.

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#### sophiecentaur

Science Advisor
Gold Member
Hi. Could the local effects of gravity between nearby stars at the edge of a galaxy cause these stars to rotate around the centre of the galaxy more quickly than predicted if the prediction is based upon individual stars only. This effect would be similar to the way the Earth "drags" the Moon around the Sun in a different orbit than if the Moon was alone.
Of course. It's like the three body problem only much worse. Two stars, if they are far enough away from others, will 'orbit' around each other approximately but they will be affected by the next nearest ones in the details of their mutual orbit and also by the whole galactic mass. Sometimes the galactic orbit of the two will be sped up a bit and sometimes it will slow down. This is what produces the density variations which are seen as spiral arms.
The same sort of effect causes planetary rings, which are kept in fairly stable orbits due to larger masses which 'shepherd' the small particles. I have a feeling that planets without moons are unlikely to develop rings, for this reason.

#### Viopia

It must be extremely difficult to calculate how the outermost stars of a galaxy are affected by "local" gravity in a complex galaxy. Is it possible that the calculations, which led to the conclusion that there is a mysterious "dark matter" were incorrect?

#### DaveC426913

Gold Member
It must be extremely difficult to calculate how the outermost stars of a galaxy are affected by "local" gravity in a complex galaxy. Is it possible that the calculations, which led to the conclusion that there is a mysterious "dark matter" were incorrect?
No. Just because two local stars happen to be gravitationally influencing each other doesn't mean that their combined centre of mass is affected. Whether singly, or as a pair, they will still orbit the galaxy at the same rate.

In other words, your example of the Earth-Moon system is incorrect. Without the Moon, the Earth would still take exactly one year to orbit the sun. Which means we can use Earth's period to figure out how massive the sun is.

This, by the way, is why small service modules can dock with the massive ISS. Neither satellite's orbital velocity is related to its mass.

Same goes for stars at the edge of the galaxy. Their orbital velocity is not related to their mass.

#### Viopia

Thanks. I think I understand what you are saying. I have been searching the internet but I cannot find a definitive answer to the following question. The Earth and Moon are only an insignificant mass compared to the mass of the Sun. You say that "Without the Moon, the Earth would stilltake exactly one year to orbit the sun" even though their combined mass is less. Would it still take one year for the Earth to orbit the Sun if the Earth was the same mass as the Sun and would the orbital speed of the Earth still be the same? If so, my question has been fully answered and I have no more questions on this subject. Thank you.

#### Viopia

Just a clarification. By orbit I mean the orbit around the Earth and Sun's combined centre of gravity.

#### DaveC426913

Gold Member
Would it still take one year for the Earth to orbit the Sun if the Earth was the same mass as the Sun and would the orbital speed of the Earth still be the same?
No.

This calculator indicates that a binary system with two suns the same distance apart as Earth's orbit would have a period of 0.707 years (which coincidentally, equals 1 / sqrt(2) or sin(45) ).

(but that's a system that has twice the mass of ours)

If you keep the system's mass constant- then you have two suns - each .5 of the sun's mass - orbiting each other. And their period will be ... 1 year.

#### Viopia

Thank you for the info. Taking the "same combined mass" scenario the common centre of gravity with the Earth and the Sun with the same mass as they have in reality, the rotation point would be near the Sun's centre and so the Earth would complete a circle with a radius of 93 million miles in one year. In the "each .5 of the sun's mass - orbiting each other" combined mass scenario the rotation point would be half way between the Sun and the Earth and their period will be still be 1 year. Does this mean that the Earth would be orbiting around the centre of gravity at a lower speed than it does at the moment with the mass of the Earth and the Sun the same as it is in reality?

#### DaveC426913

Gold Member
Thank you for the info. Taking the "same combined mass" scenario the common centre of gravity with the Earth and the Sun with the same mass as they have in reality, the rotation point would be near the Sun's centre and so the Earth would complete a circle with a radius of 93 million miles in one year. In the "each .5 of the sun's mass - orbiting each other" combined mass scenario the rotation point would be half way between the Sun and the Earth and their period will be still be 1 year. Does this mean that the Earth would be orbiting around the centre of gravity at a lower speed than it does at the moment with the mass of the Earth and the Sun the same as it is in reality?
Er.. what?

Scenario 1: Earth and Sun: Earth completes an orbit in one year
Scenario 2: Two suns, each with .5 Sol mass: they complete an orbit in one year

I cannot parse this:
"Does this mean that the Earth would be orbiting around the centre of gravity at a lower speed than it does at the moment with the mass of the Earth and the Sun the same as it is in reality?"

What two scenarios are you comparing?

Are you creating third scenario, with 2 Sol-mass stars and the Earth?

If so, essentially, the Earth will orbit the heliocentre of the two stars (all three will form an equilateral triangle ). And the whole system will complete an orbit in one year.

#### Viopia

Sorry. My question was not very clear. Basically I am trying to find out if the the rotation curve (velocity curve) of a disc galaxy can be different because the mass within the galaxy is distributed differently than in the Solar System. The solar system has most of it's mass concentrated in the central region (within the Sun itself) but only a 200th of the mass of a galaxy is centrally placed within the supermassive black hole. I was trying to ascertain if the speed of rotation of a two body system can be altered by re-distributing the mass while keeping the total mass of the system the same. The best representation if a galaxy that I can think of is a 3 body system. This would be a central "sun" with two planets, each half the mass of the "sun", on opposite sides of the "sun" rotating in the same orbit. This would mean that the planet on the other side of the "sun" would never be seen by it's sister planet because the "sun" would always be in the line of site of this planet. This would also mean that there would be no gravitational "wobble" of the "sun" because both planet's gravities cancel each other out at the "sun". If this configuration would produce the same rotation (or velocity) curve as a real galaxy I believe this would negate the need for "dark matter".

#### DaveC426913

Gold Member
Yes. The fact that galaxies are gravitationally extended objects is taken into account.

#### Viopia

Thanks. In my last 3 body example did you calculate the orbital speeds of the two large sister planets at varying equal distances from the "sun" to see how the velocity curve is affected? At Earth distance from the "sun" would the planets not complete an orbit in less than 1 year because of the added gravitational "pull"of it's sister planet? I'm pleased that the gravitational effects of the opposite side of galaxies have been taken into consideration when calculating the velocity curves.

#### Viopia

If you think it is worth doing the calculations I have suggested please adjust the masses so that all the 3 bodies each have a mass which is half the mass of our own Sun so that the results can be compared to the Earth's orbit.

#### sophiecentaur

Science Advisor
Gold Member
The best representation if a galaxy that I can think of is a 3 body system.
Your suggested three body system would not be stable, though, whereas all the existing galaxies demonstrate that the distributed arrangement is, in fact, dynamically stable (i.e. there is a lot of shuffling around inside, over time but the overall shape is maintained. I think the whole analysis should b e very different from what you propose. I'm sure Google would throw up something to help you.

#### Ibix

Science Advisor
I think the whole analysis should b e very different from what you propose.
I think as long as you idealise it to two identical bodies diametrically opposite one another and in identical circular orbits it's possible. There's no net force on the central body and both forces on the orbiting bodies are colinear. So it ought to be straightforward. As you say, it's unlikely to be stable and the instant it's not symmetric you need to go for numerical simulation. And the whole thing is irrelevant to galaxies, where (unless you're going for a star-by-star simulation) I expect you'd just model it as more or less continuous and apply Gauss' theorem.

#### Viopia

I agree it would not be stable in the real world. I was only using it as an example of how the mass on the one side of a galaxy may be significant enough to exert sufficient gravitational "pull" to stars on the edge of the opposite side of the galaxy causing the velocity curve to change shape because of the distribution of mass within the galaxy itself. This effect would not be apparent in the Solar system because most of the total mass is concentrated at the centre of rotation (the Sun itself) and so the gravitational "pull" from planets on the opposite side of the Solar system would be insignificant. Thanks for your suggestion, I will do some further research on Google.

#### Viopia

I think as long as you idealise it to two identical bodies diametrically opposite one another and in identical circular orbits it's possible. There's no net force on the central body and both forces on the orbiting bodies are colinear. So it ought to be straightforward. As you say, it's unlikely to be stable and the instant it's not symmetric you need to go for numerical simulation. And the whole thing is irrelevant to galaxies, where (unless you're going for a star-by-star simulation) I expect you'd just model it as more or less continuous and apply Gauss' theorem.
Exactly.

#### Viopia

But I don't think it is irrelevant to the way galaxies behave because it could model how the velocity curve can be affected by significant mass on the opposite side of the rotation point.

#### sophiecentaur

Science Advisor
Gold Member
But I don't think it is irrelevant to the way galaxies behave because it could model how the velocity curve can be affected by significant mass on the opposite side of the rotation point.
If you want an approximate way to the behaviour then assume a uniform disc of mass and calculate how the Potential varies with position. Forget about using just two / three masses and treat the galaxy as a distributed mass. For a spherical galaxy, there is no contribution from the shells outside the orbit.

#### Viopia

If you want an approximate way to the behaviour then assume a uniform disc of mass and calculate how the Potential varies with position. Forget about using just two / three masses and treat the galaxy as a distributed mass. For a spherical galaxy, there is no contribution from the shells outside the orbit.
Yes, I can see it would be a better representation but more difficult to calculate. By using the 3 bodies I was merely trying to ascertain if the velocity curve can be changed by a significant mass on the other side of the rotation point. I am not interested in the values but only if my assumption is correct. Using the 3 bodies the velocities can be plotted against the distance from the "sun" which can be increased incrementally to give specific values. These can then be plotted on a graph to show the shape of the velocity curve. If my assumption is correct, and the shape of the curve can be altered by different mass distributions, it could provide evidence for an alternative explanation to "dark matter" being present. I am not a physicist and so my assumption may be incorrect, but if there is no obvious flaw in my reasoning it would be very easy to produce such a graph.

#### Ibix

Science Advisor
But I don't think it is irrelevant to the way galaxies behave because it could model how the velocity curve can be affected by significant mass on the opposite side of the rotation point.
Depends what you mean. Modelling any three bodies is pretty pointless because you are neglecting the other hundred billion bodies that make up the galaxy. The obvious approach to it is just to model the thing as a rotating body of fluid, and work out the gravitational effect of all the fluid. Gauss' Theorem cuts out the hard work of that if you assume the galaxy is rotationally symmetric, in which case you can say that the velocity at some radius $r$ depends on the radius $r$ and the total mass inside $r$ and everything else drops out. If you want to handle asymmetric situations you'll typically have to go to numerical simulation, and you'll need a good integrator.

You may well choose to simply add up the force from all the stars, but you need to consider all the stars. Triples in highly specific orbits won't help because those orbits won't typically exist in the more complex situation.

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#### Viopia

Good idea, but I am not trying to model a whole galaxy. I am trying to model something which confirms my assumption.

#### DaveC426913

Gold Member
Viopia, astrophysicists know how the gravity in extended bodies such as galaxies works.

No matter how you cut it, the gravitational force on a given star in a galaxy is the sum of the forces of each individual star in the galaxy - as per the gravitational formula: F=Gm/r^2.

There are ways of simplifying this for the sake of calculations - but good lord, do you think the people who whose career it is to know this stuff don't know how gravity works?

#### Viopia

You make a good point, although I can think of instances where physicists have been wrong. It was only around 100 years ago that most physicists believed that an ether permeated the whole of space. I believe that gravity is still not fully understood at the quantum level. Also, physics needs a healthy degree of scepticism to function properly. The formula you have stated indicates that gravity follows the inverse square law and so, in my 3 body model, I believe there may be a mismatch in the comparative relationship between the gravity from the "sun" and from the opposite sister planet at varying distances. Could this affect the shape of the velocity curve? The gravitation formula is simple but it's amazing how complicated things can become when there are hundreds of billions of gravitational bodies in a system. Of course, I am not a physicist and so I may be wrong about all this.

#### Viopia

Clarification: I am referring to the comparative effect that the gravity from the "sun" and the planet have on the sister planet at varying distances.

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