submitted 9 months ago by GardenVarietyAnxiety@lemmy.world to c/askphysics@lemmy.world

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Is it possible to determine the percentage of the gravitational force at a specified distance using only the geometry of the planet?

Example: The ISS at ~420km altitude "weighs" about 90% of what it would on the Earth's surface.

Is there an equation using only geometrical values that would give you this info?

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[-] ArbitraryValue@sh.itjust.works 9 points 9 months ago* (last edited 9 months ago)

What do you mean by "geometry of the planet"?

According to Newton's shell theorem,

A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center.

Thus, if you treat the Earth as a sphere with a radius of 6,371 km, then the force of gravity at an altitude of 420 km will be (6,371 / (6,371 + 420))^2 = ~0.88 times as strong as the force of gravity on the surface. The general equation is (force_at_radius_r1 / force_at_radius_r2) = (r2/r1)^2.

Note that the relative strength of the force depends on the ratio of the distances rather than their absolute values. The space station is 1.066 times further from the center of mass than the surface is so the force on the space station is (1/1.066)^2 times the force on an object at the surface.

[-] GardenVarietyAnxiety@lemmy.world 2 points 9 months ago

That was a lot simpler than I thought. Thanks!

What do you mean by "geometry of the planet"?

I just meant measurements such as radius.

this post was submitted on 23 Jan 2024

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