Gravitational potential and partial derivatives app's

  • Thread starter stanley.st
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Hello,

I'm a student of applied mathematics to economics. Basic course consists of all pure math subjects. We were talking about app's of differentiating the functions [tex]u:\mathbb{R}^{n}\to\mathbb{R}^m[/tex]. We defined a gradient too. In my notes is written:

Gravitational potential is a function [tex]H:\mathbb{R}^{2}\to\mathbb{R}[/tex] and it has to solve Laplace's PDR [tex]\Delta H=0[/tex].

Then we found out, that for example a function [tex]H=\frac{\mathrm{const.}}{\Vert x\Vert}[/tex] solves that equation. Then there's a mystery in the form [tex]F\sim\nabla H\sim\Vert F\Vert=\kappa\frac{mM}{\Vert x\Vert}[/tex]


QUESTIONS:
(1) Can you explain me, what is gravitational potential. Or how to imagine it.
(2) Why it must solve Laplace's PDR.
(3) Why [tex]F\sim\nabla H\sim\Vert F\Vert=\kappa\frac{mM}{\Vert x\Vert}[/tex]???

Thank you in advance :-)
 

Answers and Replies

  • #2
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QUESTIONS:
(1) Can you explain me, what is gravitational potential. Or how to imagine it.
(2) Why it must solve Laplace's PDR.
(3) Why [tex]F\sim\nabla H\sim\Vert F\Vert=\kappa\frac{mM}{\Vert x\Vert}[/tex]???

(1) As you said, gravitational potential is a function of the spatial coordinates. The potential isn't any physical thing as such, but it's simply introduced to make the calculations easier. An easy way to visualize the potential is to think about an one dimensional case, where you just have a simple [tex]\mathbb{R}\to\mathbb{R}[/tex] function. Now, as you move through space (x-axis) you feel different kind of forces depending on the gradient of the curve; your acceleration is the greatest where the slope of potential is largest etc. You can imagine yourself sitting on a sled, riding down the potential function if you like.

(2) That is just the way the world works. Laplace equation is the solution in empty space - usually you have a source term at the right hand side.

(3) [tex] H \sim 1/r (r = || x ||) [/tex] is a solution for the laplacian, as you can easily show by solving the differential equation. Now, the force that a potential H induces is just [tex] F = \nabla H \sim \mathbf{r}/r^3[/tex] (you are missing one r). The constant in front just describes the strength of the gravitational interaction; the mass terms come from the right hand side of the equation, gravitational coupling constant tells what the strength of gravity is in the unit system we're using.
 

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