Unveiling the Mystery Behind r^2 in F=Gm1m2/r^2

[Mehrsa]

why r^2??

hi
I have a big problem in physics that nobody could answer it until now.I hope that you can help me.
my question is about this formula:F=Gm1m2/r^2
why r^2??
I searched about it very much but the only thing that i could find was this:the answer is not:because of the experiment!
and i found out that it can be proved with graviton.
but i don't know how.
please help me with this.
thank you

 

[HallsofIvy]

Because of the way things "spread out".

Many years ago, a physics teacher of mine had what he called a "butter gun". It was a little squirt gun with four wires spreading out from the barrel. You put your toast inside the four wires and squirt butter at it!

The point was that you can fit 4 slices of "toast" instead of 1 if you move it twice as far from the barrel. You can use "similar triangles" to prove that for the pyramid formed by the "toast" and 4 wires in this case, but if you have a sphere it's the same thing: the surface area is proportional to the radius- the distance from the center.

If you have the same total force (or area) spread over the greater area, with the area increasing in proportion to r², then the strength in any area must decrease in proportion to 1/r².

 

[Mehrsa]

thank you

thank you for your help
but can you explain it to me more?I got the point but I want to prove it.
can you tell me how should i prove this formula?because i have aproject and i should prove this formula in it.

 

[Crosson]

Mehrsa, this is how Newton proved the inverse square law of gravity.

He looked at the 2nd of keplers observations, "equal areas in equal times", and proved that only a radially directed force could cause such behavior (ruling out the previously held notion that angels pushed the planets around with a force tangential to the path).

Then he used the first of kepler's observations "the planets move in ellipses". He managed to prove that the only radially directed force which produces an elliptical orbit is 1/r^2. He did this secretly with differential equations, but his public proof involved very difficult geometric methods.

So that is how the master did it himself, but if you want to know a deeper reason for 1/r^2, then think about the surface area of a sphere, 4*pi*r^2. Because gravity gets "spread out" over the sphere as you go further away, and the sphere gets larger as r^2, there you go. When you study electromagnetism (which also involves one over r^2 laws) you will see better why it is the way it is.

 

[Tide]

Note that "inverse square" for gravity is only an approximation that is valid when the masses involved are not "too large."

 

[masudr]

If you put a small body inside a hollow sphere, you will find that there is no resultant force on the small body inside the sphere. If you go through the full mathematics, integrating the force contribution from small surface elements, you will find that the only way this will happen is if there is an $$r^{-2}$$ dependence. Then assuming that the force is proportional to the product of the masses, we can determine G using two spheres, and Cavendish's classic experiment.

 

[vinter]

If you can lay your hands on "Feynman Lectures on Physics Volume 2", read the first few chapters. He has given a clear description of this 1/r squared thing. The experiment is similar to keeping a mass inside a spherical shell. It doesn't involve too much of integration or complicated mathematics, though. Only English.

 

[Andrew Mason]

thank you for your help
but can you explain it to me more?I got the point but I want to prove it.
can you tell me how should i prove this formula?because i have aproject and i should prove this formula in it.

In physics, proof is provided by physical evidence. If one accepts Kepler's laws for planetary motion as correct physical evidence, which Newton did, it is relatively straightforward to deduce the 1/r^2 central force:

The centripetal force on any orbiting body is $F_c = m\omega^2r$ so

$$a_c = \omega^2r = \frac{r}{(2\pi T)^2}$$.

If the square of the period varies as the cube of the radius (Kepler's third law): $T^2 \propto r^3$, then:

$$a_c = a_{gravity} \propto \frac{r}{4\pi^2 r^3} \propto \frac{1}{r^2}$$

AM

 

[Mehrsa]

thank you verymuch.I got it compeletly!

 

[Allegro]

Hi,

More about this subject,including exponents of r different from 2, see here:
http://www.binaryresearchinstitute.org/pdf/gravi.pdf

 

[Andrew Mason]

I noticed that I put $\omega = \frac{1}{2\pi T}$ It should be, of course: $\omega = \frac{2\pi}{T}$

So:
$$a_c = \omega^2r = r(\frac{2\pi}{T})^2 = \frac{4\pi^2r}{T^2}$$

If the square of the period varies as the cube of the radius (Kepler's third law): $T^2 \propto r^3$, then:

$$a_c = a_{gravity} \propto \frac{r}{r^3} \propto \frac{1}{r^2}$$

AM

 

[dextercioby]

I have strong reasons to doubt that Newton was aware of the concept (and hence the expression) of the centripetal acceleration.

Nonetheless,your calculations are correct...

Daniel.

 

[rbj]

i know you said you "got it completely" but another useful way to look at this inverse-square law thingie is at wikipedia:
http://en.wikipedia.org/wiki/Inverse-square_law

wikipedia is getting to be about as useful as google.

r b-j

 

[dextercioby]

Yeah,but you've got to admit you can find out about Wikipedia searching on Google and I'm not sure for the reverse,though...

Yes,it's turning into an useful site and will soon be in the same privileged category with PF & Mathworld...

Daniel.

 

[Andrew Mason]

I have strong reasons to doubt that Newton was aware of the concept (and hence the expression) of the centripetal acceleration.

Newton appears to have coined the term: "centripetal force". It appears throughout his Principia. See for example: http://members.tripod.com/~gravitee/booki2.htm

It is true that he did not speak about acceleration, but he obviously understood the term as he speaks of "accelerative force" (Prop. III, Theorem III).

AM

 

[dextercioby]

That figures.It's been ages since reading "Principia" and it was really in latin and at that time i was keen on learning Ancient Greek.

I couldn't do any of them properly... Had Newton known Ancient Greek,i would have been twice as smart...

Daniel.

 

[Mr. Therefore]

my question is about this formula:F=Gm1m2/r^2
why r^2??

Consider just the pyramid that is in the inverse square explanation:
http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html

When you double the height of a pyramid, you quadruple the base. (Don't confuse this with keeping the base the same and increasing the height.)

When you double the height of a pyramid, the base is proportionately squared, because the base becomes four times bigger.

A radius doubling is a squaring of the base of a pyramid. This is where r^2 originates.

Why is it used to divide?

The denominator is the way we choose to package the numerator. If we have 12 eggs as a numerator, we can put them into packets of two by dividing the dozen by two.

In the gravity equation, we are putting the numerator (Gm1m2) in packets of r^2.

If the masses are closer, we have smaller r and the packets are smaller. 2^ is 4 (small packet)

If the masses are further, we have larger r and the packets are larger. 10^ is 100 (big packet)

When you divide by a small number, your quotient increases. (More force of gravity, F = Gm1m2/4)

When you divide by a large number, your quotient decreases. (Less force of gravity, F = Gm1m2/100)


So, a smaller radius means a larger quotient. A larger quotient is more force.

The area of force in question prevails in same space we would say the area of base of the triangle exists.