B Effect of Density and Geometry on Gravity and Luminosity

1. Mar 13, 2016

townfool

Imagine a planet similar to Earth, but exposed to a completely different star. The star has the same mass and emits the same amount of photons as the sun, but it is a huge, extremely slender torus made of 1 mm diameter neon tubing. The planet is on the axis of the torus and at the same distance from the plane of the torus as Earth is from the sun.
A 1mm section of the tubing has a very small mass, so it hardly distorts space locally and it emits very few electrons and gravitrons and extremely few in the direction of the planet and no neutrinos.
An astronomer on the planet is not likely to see the star, unless she happens to focus her telescope exactly in the direction of the torus and even then luminosity would appear low. A double parabolic trough lense would be much better to gather photons arriving from the distant torus, but the parabolic trough has to be alined exactly with the torus.

2. Mar 13, 2016

A.T.

What?

3. Mar 13, 2016

townfool

All photons from the sun arrive from a relatively small area and close to Earth, so we see and feel the sun (unless there is an eclipse, once in a long while). In contrast, photons from the torus arrive from myriad directions and from an extremely long distance (any particle of dust, etc, between the torus, billions of km away and the telescope will block the view. If Jupiter' is between the torus and the planet, it will block its photons from a good size arc for quite a while). Do you think it is easy to see a 1 mm diameter neon tube billions of km from earth, regardless of how long it is?

4. Mar 13, 2016

A.T.

And how big is your torus compared to that?

5. Mar 13, 2016

townfool

assuming that 30 m of 1 mm tubing makes a kg and that the sun's mass is 2 x 10 to 30th kg, the perimeter of the torus is 6 x 10 to 28th km

Last edited: Mar 13, 2016
6. Mar 13, 2016

townfool

Regarding star visibility, the planet itself blocks a huge arc of the torus from the astronomer.

7. Mar 13, 2016

Staff: Mentor

Perimeter only applies to a 2d shape. Do you mean the surface area of the torus?

Also, what is the point of this thread? Did you have a question or something?

8. Mar 13, 2016

townfool

Perimeter is the length of the tubing making up the torus, measured along the center of the tube's section. One can think of a doughnut of 1 mm diameter tubing and 10 to 28 km long as a line, instead of a 3d body.

9. Mar 13, 2016

Staff: Mentor

Alright. So, the circumference of a circle is: C=2πR
Plugging in your numbers and solving for R: R=6x1031/2π (C is in meters)
R = 9.5x1030 meters.
One light-year is about 9.5 x 1015 meters.
That means that the distance from this planet to the ring is 1015 light-years, or 1 petalight-year.
That's 20,000 times the radius of the observable universe.

10. Mar 14, 2016

townfool

Thanks, I chose too small a diameter for the neon tube. However, the point is that density and geometry clearly influence luminosity and gravity. In this case the star would never be seen nor its gravity felt by the astronomer (the planet does not orbit the star) and the star hardly distorts space.
Even with a 100 m diameter, much thicker wall neon tube (a meter long section has a mass of hundreds of kg), the distance between the astronomer and the torus is so long that the astronomer will never see its light or feel its pull.. In contrast, a massive planet moving close to a section of the torus would see its light and feel a little gravity, but it exerts more attraction on the section of the torus close to it than the massive torus exerts on the planet (most of the torus' large mass being very far from the planet, while all the small mass of the planet is close to the section of the torus).

11. Mar 14, 2016

Khashishi

12. Mar 14, 2016

townfool

No it's not for a science fiction story. Why should anyone care about young Einstein riding on a light beam? Thought experiments allow us to look from a different perspective

13. Mar 14, 2016

Khashishi

Brightness doesn't depend on distance, but the apparent size does. Farther things are harder to see because they are smaller. Under normal circumstances, the visual size drops as 1/r^2, but since your torus is so long and thin, and you are looking from near the center of the torus, you should treat it as a special case to calculate the visual size of the object. Compare the thickness of the band you see in the sky (in angular units) vs the angular diameter of typical stars.

14. Mar 14, 2016

Staff: Mentor

Sure. But so what?

Not true. The force on the torus from the planet is equal and opposite to the force exerted on the planet by the torus.

15. Mar 14, 2016

townfool

According to all formulae, density and geometry play no role on gravity, only mass.

Do you think that the force Earth exerts on you is the same as the force you exert on Earth? although the resultant force is the same, the components of the force are clearly different.

16. Mar 14, 2016

Khashishi

Totally false.

17. Mar 14, 2016

A.T.

Yeah, they have the opposite sign.

18. Mar 14, 2016

Staff: Mentor

That's not true. If you want to find the gravitational force from a non-spherical, non-uniform object then you can't just plug a few numbers into the GMm/r2 formula. You actually have to start using more advanced math, like calculus, to figure out the gravitational force from the object. The density and geometry of the object absolutely play a role.

To quote wiki: https://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation

In modern language, the law states: Every point mass attracts every single other point mass by a force pointing along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them.

For a perfectly uniform spherical object, if we add up the gravitational force vectors from every single point in the sphere we get a value that is identical to assuming that the mass of the sphere is packed into a single point at the very center, even though it is not.

For an object that is not a perfect sphere, we would find that the gravitational force vectors all add up to point towards the object's center of mass.

Yes, that is exactly what I said, which was in response to your claim that the planet exerts a larger force on the ring than the ring exerts on the planet, which is false.

19. Mar 15, 2016

townfool

The center of mass of the slender torus is at the same distance from the planet as earth is from the sun and the masses are the same, yet the force is not the same, just like the luminosity is not the same.

20. Mar 15, 2016

Staff: Mentor

Of course not. The Earth essentially lies inside the torus, so the gravity from the different parts of the torus mostly cancel out. The same thing would happen if you were inside a hollow shell.