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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?

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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”.

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  • $\begingroup$ That's a very interesting article! My previous answer was indeed incorrect, gravity behaves according to the inverse of the square of the distance and therefore does not always point towards the center of gravity. $\endgroup$ – Karel Peetermans Apr 28 at 17:54
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    $\begingroup$ @ÁrpádSzendrei I suggest asking this in a question, not in a comment. $\endgroup$ – G. Smith Apr 28 at 20:14
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    $\begingroup$ The approximation that the force of gravity is directed towards the center of mass can still be applied when the distance between the objects attracted toward each other is much greater than the greater internal length of both objects. $\endgroup$ – Martigan Apr 29 at 8:10
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    $\begingroup$ @R: I like this answer, though it would be better IMHO for the crux of the explanation linked, to be included here. That said, it seems to me that the core issue here is that the whole idea of "center of gravity" is a construct made up for simplifying certain calculations, in which we arbitrarily "connect" two or more regions of mass, rather than a manifestation of actual physical laws. The actual effect of gravity is the net of all mass distributed throughout the bodies of interest. Gravity can't be directed toward CG, because there's not actually any such thing as CG. :) $\endgroup$ – Peter Duniho Apr 29 at 23:22
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    $\begingroup$ It's worth noting that while total gravitational attraction is proportional to 1/r², the error in the center-of-mass approximation falls off with 1/r³ or faster, which is why the approximation is fine when you're far enough away. $\endgroup$ – Matt Timmermans Apr 30 at 13:18
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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.

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  • $\begingroup$ But it is still true that the net force on A is equal to its mass times the acceleration of its center of mass. Likewise, the net force on BC is equal to its total mass time the acceleration of its center of mass. But as you say the net force on A is not in the direction of BC's center of mass. $\endgroup$ – Not_Einstein Apr 28 at 19:49
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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.

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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.

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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.

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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

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