Gravitational acceleration for a rotating torus

In summary, the equations for gravitational acceleration towards a rotating torus are-F=-\frac{Gmx}{\left(x^2+a^2\right)^{3/2}}g=\frac{Gm}{\left(r-r_t \right)^2}g_t=\left|\frac{Gmr}{\left(r^2+r_t^2\right)^{3/2}}sin^2\theta\right|+\left|\frac{Gm}{\left(r^2-r_t^2 \right)}cos^2\theta\right|
  • #1
stevebd1
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gravity of a rotating torus

While Gm/r^2 seems to apply only for gravitational acceleration towards spheroids, what would be the equation for gravitational acceleration towards a rotating torus? I'm sure the equation would be the same for spheroids and toroids at large distances but what would be the equation in close proximity to the torus (i.e. the outer surface)?
 
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  • #3
Thanks for link Shalaya, while the mathpage is informative and backs up the idea that gravity cancels out at the centre of the torus and that anything that falls inside the torus is pulled away from the centre, is there a more 'rule of thumb' equation on the lines of Gm/r^2?
 
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  • #4
While it doesn't relate to a rotating torus, I did find an equation that looks at the gravitational field on the polar axis of a torus-

[tex]F=-\frac{Gmx}{\left(x^2+a^2\right)^{3/2}}[/tex]

where x is the distance from the centre of the torus (the torus being on the y,z plane) and a is the radius of the torus.
source- http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/MoreGravity.htm

There seems to be a discrepancy with Gm/r^2 over very large distances but it is at least within the same order and it could be argued that over a certain distance, Gm/r^2 is used. Within about 50a, the results are very similar to Gm/r^2 until within close proximity of the torus (~5a) where the gravitational acceleration begins to reduce until it reaches zero at the centre.

I had previously assumed that gravity on the same plane as the torus would be something like-

[tex]g=\frac{Gm}{\left(r-r_t \right)^2}[/tex]

where rt is the radius of the torus. The resulting negative figure for the gravitational field once inside the torus doesn't imply 'negative gravity' as such, just that now as you move to the centre of the torus, gravity is pulling the other way (i.e. to the inside face of the torus). Based on the above, this could be rewritten as-

[tex]g=\frac{Gm}{\left(r^2-r_t^2 \right)}[/tex]

providing a very rough rule of thumb equation-

[tex]g_t=\left|\frac{Gmr}{\left(r^2+r_t^2\right)^{3/2}}sin^2\theta\right|+\left|\frac{Gm}{\left(r^2-r_t^2 \right)}cos^2\theta\right|[/tex]

so that the first equation dominates at the poles and the second equation dominates at the equator. This produces fairly similar results to Gm/r^2 (up to about 50r from the torus) until about 5r where the results for the pole and the equator begin to diverge. While not perfect, it does produce a fairly simple alternative to Gm/r^2.

Steve
 
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  • #5
Once inside the torus, the gravitational field in the equatorial plane needs to take into account multiple sources and gradually reduce to zero so r is introduced to the top half of the fraction. This also appears to take care of any geometric singularities that might have cropped up at the centre on the equatorial plane.

Gravity for a torus where r<rt-

[tex]
g_t=\left|\frac{Gmr}{\left(r^2+r_t^2\right)^{3/2}}sin^2\theta\right|+\left|\frac{Gmr}{\left(r^2-r_t^2 \right)}cos^2\theta\right|[/tex]

The way I see it is that this gives an approximation of the gravitational field at any point around the torus, relative to the part of the torus that is most local to the point being looked at.
____________

Re post #4, the pole equation is actually good to infinity (as is the equatorial equation).
 
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  • #6
I've looked at this in a bit more detail in the blog linked below, establishing points of reference on the torus which makes it appear as a ring of beads and then calculating the proper distance from each bead-

https://www.physicsforums.com/blog.php?b=334

Attached below are diagrams that demonstrate the trigonometry involved and an excel spreadsheet based on 8 and 16 points of reference. The only bits of info that need to be punched in are mass, r (radius), radius of torus and plane angle-
 

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What is gravitational acceleration for a rotating torus?

Gravitational acceleration for a rotating torus is the acceleration experienced by an object due to the force of gravity in a rotating torus-shaped object. This is also known as the "centrifugal acceleration."

How is gravitational acceleration for a rotating torus calculated?

The formula for calculating gravitational acceleration in a rotating torus is: g = (4π²R/T²) * (1 - cosθ), where g is the gravitational acceleration, R is the radius of the torus, T is the time period of rotation, and θ is the angle of rotation.

What factors affect the gravitational acceleration in a rotating torus?

The gravitational acceleration in a rotating torus is affected by the radius of the torus, the time period of rotation, and the angle of rotation. It is also affected by the mass and density of the objects within the torus.

How does the gravitational acceleration in a rotating torus differ from that of a non-rotating torus?

In a rotating torus, the gravitational acceleration is affected by the centrifugal force, which is caused by the rotation of the object. This force is absent in a non-rotating torus, resulting in a lower gravitational acceleration.

What is the significance of studying gravitational acceleration in a rotating torus?

Studying gravitational acceleration in a rotating torus can help us understand the effects of gravity in different environments, such as in space stations or habitats on other planets. It can also aid in the design and engineering of structures that rotate, such as centrifuges or amusement park rides.

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