# Math question (in context of Gravity)

Registered User regular
edited December 2008
I've been looking into the physics of gravity and the like, out of both preperation for next sem and curiosity. I'm trying desperately to understand the theory of gravitation, specifically through the "Feynman lectures on physics volume 1,2 and 3". I have a relatively good understanding of Algebra and Calculus, but I'm having difficulty understanding some of the concepts.

Specificically I'm here to ask what this means: "It is that every object in the universe attracts every other object with a force which for any two bodies is proportional to the mass of each and varies inversely as the square of the distance between them."

I can't visualize how something acts inversely as a square. Can someone show me mathematically what this means and maybe direct me towards a diagram or something? I've looked it up but I can't manage to get my mind around it, especially in the context of gravitational pull. Thanks in advance!

Also: this is from the chapter "the theory of gravitation" on page 69 of the PDF file.

blue powder on

## Posts

• Columbia, SCRegistered User regular
edited December 2008
An inverse square relationship means that if you double the distance between the objects, the force is a fourth of what it was (1 / 2^2 = 1/4).

Diagram (forget the labels, just look at the curve):

FunkyWaltDogg on
• Registered User regular
edited December 2008
it means that the the equation for gravity is

g (force) = (G (constant) * M1 * M2 (masses of the objects in question)) / (x^2) (where x is the distance between them)

basically it means that the force of gravity gets weaker quadradically with respect to distance and linearly with respect to the mass of one object.

you can also think of it as

g = G * (M1/x) * (M2/x)

• Registered User regular
edited December 2008
If A varies proportionally to B, then A = kB, where k is some constant. If C varies inversely proportional to D, then C = k/D, where k is some constant. Qualifiers such as "the square" or "the cube" etc, just refer to the exponent, so if E varies proportionally to the square of F, then E = kF^2, where k is some constant.

HalberdBlue on
• Registered User regular
edited December 2008
It means that the driver of the force is this part of the equation 1/(d^2), where d is your distance (inverse because it lowers it as your distance increases. To be directly proportional to a squared distance would be where d^2 is the driver. The rest is a bunch of constants and mass variables.

Zombie Nirvana on
• Registered User regular
edited December 2008
Diagrams like the one in this article helped me visualize how forces governed by the inverse square law worked.

Djeet on
• Registered User regular
edited December 2008
Dammit, someone posted the article that has the good explanation! I'm going to say it anyway, you can't stop me!

Inverse square works on a variety of waveforms. We don't know if gravity is a waveform, but it does work on inverse square to a good approximation. The basic idea is that there's only so much power (using that term loosely) in the wave, and that the wave expands as it travels.

So the total effect is the same but the area of the wave is going up. And, as you learn late in algebra (or at least I did), the area increases as the square of the distance it has traveled.

So

effect*area = constant = effect*(something*distance^2)

effect = constant/(something*distance^2)

and there's your inverse square law.

ProPatriaMori on
• Registered User regular
edited December 2008
If A varies proportionally to B, then A = kB, where k is some constant. If C varies inversely proportional to D, then C = k/D, where k is some constant. Qualifiers such as "the square" or "the cube" etc, just refer to the exponent, so if E varies proportionally to the square of F, then E = kF^2, where k is some constant.

Okay, I think I can understand this, and reporduce it mathematically! So if E varies inversely proportional to the square of F, then E = k/F^2?

Thanks a lot, guys!

blue powder on
• Registered User regular
edited December 2008
If A varies proportionally to B, then A = kB, where k is some constant. If C varies inversely proportional to D, then C = k/D, where k is some constant. Qualifiers such as "the square" or "the cube" etc, just refer to the exponent, so if E varies proportionally to the square of F, then E = kF^2, where k is some constant.

Okay, I think I can understand this, and reporduce it mathematically! So if E varies inversely proportional to the square of F, then E = k/F^2?

Thanks a lot, guys!

That is correct, yes.

grungebox on
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