Maths in gravitation law equation?

Hey, are you referring to Kepler's third law?

 

distance "in there"...

Hey buddy...

I know your feeling, I had the same question a long time ago...I think that I've found the answer on electromagnetig fields :tongue2:

Ok, ok, let me explain...it will be almost a philosophic answer but...I can live with that

When I saw that electric forces depends also from the square of the distance (F = K *|Q|*|q|*distance^-2) , and also electric field even from a single point of charge (E = K*Q*distance^-2). It means that it necessarily HAS to interact to something.

The field must "travel" from somewhere to somewhere.

Just like gravity! :-)

You have to have two points of references, one mass and the another...F = G * M * m * distance^-2, right? From one to other and vice versa.

So, the field will "travel" from one to another and vice verse...then you have, like..."two distances" instead of one...

Did I help or now its worse? :yuck:

 

This actually is a common misconception.

The force need not always follow an inverse square law! For example, let's say you have an "infinite" line of uniform mass distribution. If you calculate the gravitational force due to that line mass, you will NOT get an inverse square law as you vary your distance from that line mass. The same with an infinite plane of mass (which is why, sufficiently close to the surface of the earth, we assume "g" to be constant and not vary inversely).

What this means is that the inverse square law is a function of the geometry of the problem. The inverse square law appears when one solves for spherical sources, or, in general, finite bound sources. Since in celestial mechanics, we normally deal with "spherical" or bound mass from quite a distance, it means that the inverse square law appears ubiquitously all over the place, leading to many misconception that this is the only description of gravity. It isn't.

The mathematical form to arrive at the description of the field as a function of distance can be presented in the most general form as the Gauss's Law equivalent for gravitational field, such as

[tex]\nabla \cdot g = -4\pi G\rho[/tex]

where g is the gravitational acceleration, G is the gravitational constant, and [itex]\rho[/itex] is the mass density of the source. One solves this for the particular geometry of the problem.

Zz.

 

I think it is fair to say "yes, the universe has a mathematical scaffolding." This may be somewhat easier to accept if you think of math as "the way in which we understand the universe". It is believed that all phisical events can be desdcribed mathematically, even though some are so complex that we haven't yet figured out the equations for them. This is no more (and no less) surprising than the fact that you can take a bucket with two apples in it, and put in one more apple, and you can know without even conting that there are three apples in the bucket. The universe is mathematically consistent.

As for the inverse square law, it is not restricted to gravity and EM fields. It applies to virtually everything that has a radius. Throw a rock into a pond, and waves will travel out from the point of impact. As the waves get further from that central point, they get weeker by the sqaure of the distance. Set up a spherical sprinkler-head (like a fountain or something) that sprays out droplets fo water; the number of droplets in a cubic meter of air reduces by the square of the distance.

In all these cases, we are talking about taking a set amount of something (water droplets, wave energy, gravitational or electromagnetic potential), and spreading it out over an increasing area with a circular cross-section. Therefore, the area being covered is the area of a circle, and the area of a circle is a function of the square of its radius. So, as we get further from the source (and further from the center of the circle), the amount of energy/droplets/potential remains the same, but the area it covers increases, as the "stuff" gets sperad out over a larger and larger circular area. The same amount of "stuff" spread out over a larger area = less density. Since the area increases by the square of the distance, the density decreases by (or "is inversely protional to") the square of the distance.

But why is it that the area of a circle is a function of the sqaure of its radius? For that matter, why is it that "two apples" + "one apple" = "three apples"? These things we do not yet know.

 

Hi Lurch,

Don’t mind, but let us keep the facts straight. The energy in the ripples in a pond do not decrease in an inverse square fashion, neither does EM or gravitational potential. In the former case, by your own argument the energy in the waves, being constrained to the surface, should grow weaker inversely as the distance. (I don’t want to get into how much energy is transmitted into the water at every stage – I am just focussing on the spirit of the argument.)

In one dimension, as in a string, there is no weakening of the waves if we neglect friction.

In fact, your argument is absolutely correct for things that we can measure or count when they spread out from a central source in a 3-d space. But when the Law of Gravitation or Coulomb’s law was first propounded, it was the force which was found to decrease as inverse square. There was no a priori reason to believe that it would be the force which would follow this law. Why not something else? On the other hand, the inverse square decrease of light intensity was not surprising, because light, whether thought to be corpuscular in nature or a form of energy, was presumed to be a measurable quantity.

This led to the very fruitful concept of the lines of force, the flux of which does decrease as inverse square due to geometry.

Now I have a question: When the droplets from a lot of sprinklers pass through a point, do we get in effect,

[itex]
\nabla \cdot g = -4\pi G\rho
[/itex] ? That's only to be expected beacuse of the [itex]1/r^2[/itex] !

 

The answer lies here :http://en.wikipedia.org/wiki/Area_of_a_disk