Gravitational potential and partial derivatives app's

[stanley.st]

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 $$u:\mathbb{R}^{n}\to\mathbb{R}^m$$. We defined a gradient too. In my notes is written:

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

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


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 $$F\sim\nabla H\sim\Vert F\Vert=\kappa\frac{mM}{\Vert x\Vert}$$?

Thank you in advance :-)

 

[clamtrox]

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 $$F\sim\nabla H\sim\Vert F\Vert=\kappa\frac{mM}{\Vert x\Vert}$$?

(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 $$\mathbb{R}\to\mathbb{R}$$ 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) $$H \sim 1/r (r = || x ||)$$ 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 $$F = \nabla H \sim \mathbf{r}/r^3$$ (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.

 

[astrokreng]

Hello,

First of all, I commend you for your interest in applied mathematics and economics. Gravitational potential is a concept used in physics to describe the potential energy of an object in a gravitational field. It is a scalar quantity, meaning it only has magnitude and no direction. In simpler terms, it is the amount of work that must be done to move an object from one point to another in a gravitational field.

To understand why gravitational potential must solve Laplace's PDR, we need to first understand what Laplace's PDR is. It is a partial differential equation that describes the distribution of a scalar quantity in a given region. In the case of gravitational potential, Laplace's PDR describes the distribution of potential energy in a gravitational field. Since gravitational potential is a scalar quantity, it must satisfy this equation.

The last part of your question is asking about the relationship between the force of gravity (F), the gradient of gravitational potential (\nabla H), and the distance between two objects (\Vert x\Vert). This is known as the inverse-square law of gravity and it states that the force of gravity between two objects is directly proportional to the product of their masses (m and M) and inversely proportional to the square of the distance between them (\Vert x\Vert). The constant \kappa is known as the gravitational constant and it helps to quantify the strength of the gravitational force.

I hope this helps to answer your questions. Keep exploring and learning about the applications of mathematics in different fields!