25

This is a pretty basic question, but I haven't had to think about orbital mechanics since high school. So just to check - suppose a [classical] system of two massive objects in a vacuum.

If the density of either object is the same at a given distance from the center, and both objects are spherical, then both objects can be treated as point-masses whose position is the [geometric] center of the original sphere. In the case that either object is not spherical or has an irregular distribution of mass (I'm looking at you, Phobos!), both objects can still be treated as point-masses but the center of mass rather than the geometric center must be used.

Is this correct?

R. Burton
  • 677

6 Answers6

44

No. For example, the gravity of a cubical planet of uniform density, which can be computed analytically, is not directed towards its center (or any other single point).

You can also imagine a dumbbell-shaped mass distribution where the two heavy ends are very far apart. If you drop an apple near one end it is going to fall toward that end, not toward the middle of the “neck”.

G. Smith
  • 52,489
9

No. It is not correct.

Consider this ridiculously contrived counter-example... Three spherically symmetric bodies (or point masses if you can tolerate this) are at the three vertices of a 45°, 90°, 45° triangle, ABC. The masses of the bodies are: $m_\text{A}=m,\ \ m_\text{B}=M,\ \ m_\text{C}=2M$. Regard the bodies at B and C as a single body, BC; join them, if you like, by a light rod.

The centre of mass of body BC is at point P, $\tfrac23$ of the way between B and C.

But the pull due to BC experienced by $m$, at A, is not directed towards P, as one can easily show by vector addition of the forces due to B and C. [In this case the forces are of equal magnitude, so the resultant bisects angle BAC and clearly doesn't pass through P!] The reason for the discrepancy is the inverse square law of gravitation.

Philip Wood
  • 37,325
1

In order to have gravity to always point to the center of mass, your mass must have a spherical symmetry (be homogenous or at least made of homogenous concentric layers). The approximation can be used (to a certain extent) for bodies that are not symmetrical, but are pretty much apart from each other.

The more the body deviates from the symmetry, the more its gravity deviates from the "point mass" approximation.

Most celestial bodies are in or near a hydrostatic equilibrium that imposes more or less symmetric distribution of mass.

Then again, certain phenomena like tides or sun-synchronous orbits imply non-center-of-mass gravity even for pretty round objects like the Earth, the Sun and likes.

fraxinus
  • 9,585
0

Perhaps I am wrong, given the other answers, but it was my understanding that gravity would indeed always be directed towards the centre of an object's mass. I would argue this by proposing a 2D plane rather than a 3D space. in this example, we would like to see the direction of gravity between a point and, say, a rectangle. The centre of mass, here, is incredibly useful. Because of the definition of gravity, point-mass, and centre of mass, the centre of mass will always be the point at which gravitational forces of the surrounding mass on any opposite sides are exactly equal. If the point for which we are testing gravity is directly above our rectangle's centre of mass, then the gravitational pull from both the right and left sides of the rectangle are perfectly balanced, and the point will be pulled straight down (and, similarly, the rectangle will be pulled straight up, assuming that the point in question has some mass). I hope this helps, and I look forward to hearing from other responders regarding this.

Sky
  • 1
0

Well the gravity can be treated as a point mass if the object is a sphere. However even this approximation would not work if you are going inside the object. Also this is if the object is a sphere. This is because really the mass and gravity technically comes from all parts of the planet or spherical body. It just can be assumed it comes from the center but only to a limit.

Roghan Arun
  • 1,554
0

In this case the center of gravity won't coincide with center of mass.cog would be towards the more denser or heavier part to neutralize moment of torque.But center of mass would be the average center (require integration). consider a very large rod (≈Radius of earth) of uniform density the gravity force would vary greatly over height but center of mass would be at geometric center.the gravity force acts at cog which will be lower than the com

Guji2203
  • 53