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View Full Version : Calculating gravity on planet surface?


lbrltom
Dec 16, 2009, 06:50 PM
I know that we normally treat the planet as a point source (take mass of planet, take distance from center of planet to surface, etc.) to determine the gravitational force an object on the surface will experience.

It is also clear that this is simply a rough estimate that is more effective over long distances. The way to get a truer number would be by treating the problem as the integral of two overlapping spheres. The object (let's say a person) is standing on the surface of the Earth, so one metric ton of the planet within one kilometer of that individual would exert a far greater attraction than 1 metric ton on the opposite side of the planet, several thousand times further away.

Furthermore, much of the mass is also working against itself. If our person is standing at the north pole, the band of soil at the equator is pulling at an angle we can break into two vectors, one tangent to the person and the other directly towards the center of the Earth. The tangent vectors would all cancel one another out and leave a far less significant net gravitational force relative to the planet than a point source model would give for the given mass.

Keeping this in mind (and assuming a body of uniform density), I was wondering how significant of a difference in gravitational force answers you would find between the two models (point source and integral of overlapping spheres with vectors considered.) The math for the second model would be fairly bad, but I'm mainly interested in whether it would be negligible, or we are looking at a 5-10% or more difference in calculated gravitational force.

Perito
Dec 17, 2009, 05:22 AM
Basically, if you're concerned about the gravitational attraction on, for example, the side of the moon nearest the earth and the side of the moon distant from the earth, then you can make the calculation and see a small difference. However, if you're concerned about the gravitational force on the moon (or other body) as a whole, then the concept of the centroid comes into play. In this case, there is absolutely no difference. The differences drop out of the equation.

ebaines
Dec 18, 2009, 12:08 PM
However, if you're concerned about the gravitational force on the moon (or other body) as a whole, then the concept of the centroid comes into play. In this case, there is absolutely no difference. The differences drop out of the equation.

Right - at least for spherical objects ** EDIT (and asuming that you are not positioned inside the boundary of the sphere). ** There's a good explanation of this effect here:

Gravitational Field (http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/GravField.htm)

For non-spherical objects, or if the density of matter in the object is not uniformly distributed, the gravitational force may indeed be different from what you would calculate if you assumed all the mass was concentrated at the center of mass.