What happens in the center of mass of binary star system ?

[WallaceHS]

Good night!

Please, what happens to an object (eg a spaceship) placed exactly in the center of mass of a binary star system (composed by two similar stars)? Would it be compressed by a huge gravitational mass (something like the mass of both stars)? Would it be stretched and torn? Or would it be in a safe place with no gravitational forces acting?

Note: I did not found the especific answer on this topic:
https://www.physicsforums.com/threads/center-of-mass-binary-stars.56922/

Thank you in advance!

 

[marcus]

The object would fall towards the more massive of the two.

To see this, notice that the center of mass is closer to the more massive star. So the object is placed closer to the more massive star to begin with. It experiences a stronger attraction to the more massive star both because it is closer and because the the star is more massive.

 

[DaveC426913]

Anticipating what I think you're asking, the ship would not experience undue tidal forces.

Gravity acts on a mass as a net force, so two forces acting in opposite directions cancel out.

It is loosely comparable to being at the centre of the Earth. You do not feel pulled in multiple directions, you feel the net effect - which is to not be pulled in any direction.

Now, what is not a net effect is the gravitational potential. The spaceship would expierience time dilation from being at the bottom of a gravitational well.

 

[WallaceHS]

Dear Mr. Dave and Mr. Marcus, thank you very much!

I would like to add a question: when they say that 'the curvature of spacetime is directly related to the energy and momentum of whatever matter and radiation are present. The relation is specified by the Einstein field equations, a system of partial differential equations', does that specific point, right in the center of mass of a binary star system, has any effect in warping spacetime? Would space be warped there?

Thank you again!

 

[marcus]

Hi Wallace, does HS mean "high school"? When I was in high school we had wood shop and metal shop. If you have a long thin piece of wood and you clamp it at various points in different vises, oriented in different directions. You are only applying forces at a few points along the board but you can have curvature all along the board.

With three clamps or three vises say one at each end and one halfway along, you can force the board to take on an S-curve shape.

Spacetime geometry can certainly have curvature at points where there is no matter or any significant amount of radiation.

I'm not sure what you are driving at with your question.

Even if there is just one star (you don't have to imagine two) the surrounding geometry is curved even where it is just empty space. the orbit tracks that a test particle would take are the "straight lines" or geodesics of that geometry.

==============
BTW the board with a few clamps along scattered along its length is NOT meant as a good analogy to General Relativity. I only wanted to make the point that if curvature of something is governed by a differential equation and there are lots of different diff. equations, the einstein GR equation is just one of many many...if something is governed by a differential equation then you only have to hit it at a few spots in order to produce shape all over the place, shape even where you don't hit it.
this may be familiar to you.
I may be saying something obvious. Your questions may be about something else, that I'm not getting. We like questions here. So you are encouraged to keep asking :w

 

[WallaceHS]

Thank you very much, Mr Marcus!

Yes, it is a reference to The Wallace High School :)

When Mr Dave says that 'the gravitational potential is not a net effect" and that 'the spaceship would experience time dilation from being at the bottom of a gravitational well', - Does the gravitational potential and the gravitational well causes spacetime to warp? (Not only time, but space too?)

Best Regards!

 

[phinds]

I think WallaceHS has a good question. I too am confused. If the NET gravitational effect is zero at some point, why would that point act like it was in a deep gravity well and exhibit gravitational time dilation relative to somewhere a million miles away from the center of the binary system? What are he and I missing in this regard?

 

[Ken G]

I think WallaceHS has a good question. I too am confused. If the NET gravitational effect is zero at some point, why would that point act like it was in a deep gravity well and exhibit gravitational time dilation relative to somewhere a million miles away from the center of the binary system? What are he and I missing in this regard?

You are missing that time dilation is not a local effect-- you can't, like the proverbial groundhog, just peak at your own shadow to see what your time dilation should be. Time dilation is a comparison between two points, and everything that goes on between those points contributes to that time dilation effect. So that's why it's the gravitational potential, not the Newtonian local gravity, that matters for time dilation, because to get a gravitational potential difference between two points, you have to integrate over all the gravitational effects along a path that connect the two points.

To see this, imagine you are stationary deep in the center of some well, and some distant clock is stationary far from the well, and you reckon, by exchanging light signals, that your clock is ticking more slowly than theirs. Let's say you want to account for this difference, you want to understand what is responsible for it by figuring out where the difference appeared. So you connect yourself, and that distant point, by a chain of hovering stationary observers, and you notice that you get a smoothly varying answer for where the time dilation is happening. The only observation you can make of your own local environment to tell you this time dilation is happening is you could measure the tidal stresses of gravity in your vicinity, but that will only explain why you get a tiny time dilation between yourself and the nearby observers in that chain you set up. To get the full time dilation between you and that distant point, you have to add up the tidal effects all along that path, so it is an integral over the path that determines the total time dilation. It's just not a local effect, it's a global effect. The only local effect of gravity is a tidal stress that curves inertial paths in spacetime, and that has to be added up over the whole path to get the cumulative effect.

The same is actually true of the speed a particle acquires if you drop it from some distant observer and it falls to the center of the well, where you are-- the local Newtonian gravity around you (which is essentially zero except for the tidal stresses) won't tell you that either, you have to integrate the gravitational acceleration over the whole path. So just think of time dilation as being more analogous to the speed a falling particle acquires, than to the local gravitational acceleration. (And by the way, we should repeat for clarity marcus' point above-- a particle at the center of mass of a binary is not generally in a place of zero net gravity, but we could interpret your question in the case of two stars of equal mass, and then it is.)

 

[phinds]

Thanks Ken. That clears up my confusion.

 

[DaveC426913]

a particle at the center of mass of a binary is not generally in a place of zero net gravity, but we could interpret your question in the case of two stars of equal mass, and then it is.)

Yeah, I was going to clarify this (or beat a dead horse, depending on your viewpoint).

As an analogy, the Earth-Moon's gravitational equilibrium is at Lagrange Point L1, which is about 3/5ths of the way from Earth to Moon, whereas the barycentre of the Earth-Moon system is actually inside the Earth about 1000 miles.

http://citizenscientistsleague.com/wp-content/uploads/2013/03/Feature_1_Fig_5_June_2010.jpg