Timeline for $1/r^2$ gravitational force of triangles in 2D
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 3, 2012 at 19:44 | comment | added | Helder Velez | Sorry, Yesterday I was at the funeral of a friend and I did not read the question properly. | |
| Apr 3, 2012 at 19:44 | history | edited | Helder Velez | CC BY-SA 3.0 |
the objections are valid
|
| Apr 3, 2012 at 7:48 | comment | added | Manishearth |
@annav: That's only in a uniform gravitational field. Yeah, you can approximate, but only if the bodies are pretty far away. The question states relatively close to each other
|
|
| Apr 3, 2012 at 6:57 | comment | added | anna v | @Manishearth I do not understand your comment. en.wikipedia.org/wiki/Center_of_mass . The center of mass of any shape body is also the center of gravity. Depending on the distance between the two triangles, to first order it will be a good approximation to use the center of mass. When the distances are the same order of magnitude as the size of the triangle then higher order terms will enter because of the distribution of mass of the triangles.( including tides) | |
| Apr 3, 2012 at 5:05 | comment | added | Helder Velez | i did not said that they will not rotate but that they will converge to the colective centre of mass. If a triangle is long then it can be further subdivided into smaller triangles. if the bodies are near the tidal force is relevant and elasticity shoud be considered in the deformation of the materials. In the OP I did not noted that the triangles are near. Sorry. | |
| Apr 3, 2012 at 2:57 | comment | added | Manishearth | @JayLemmon: You can't. Center of mass in gravitation works only for bodies with angular symmetry(basically spheres) in 3D, circles in 2D(with $1/r$ law)--as you learned in the comments above. Center of mass is only really useful for rotational calculations and the like (in which case the moment of inertia is also required). | |
| Apr 3, 2012 at 1:11 | comment | added | Jay Lemmon | Are you sure you can just use the triangle centroids and treat them as point masses? I know you can for circles. But it's not obvious to me that you can treat all objects as point masses if you have arbitrary shapes like triangles. Consider the case of a long thin triangle near a very massive point. Wouldn't the gravity force cause a torque on the triangle? If so, we can't treat the triangle as a point mass, certainly. | |
| Apr 2, 2012 at 22:48 | history | edited | Helder Velez | CC BY-SA 3.0 |
completeness
|
| Apr 2, 2012 at 22:34 | history | answered | Helder Velez | CC BY-SA 3.0 |