User:Karl Dickman/Threads/06/12/11a

Gravitational flux density of Earth
All the formulas you need to calculate it:

http://www.imath.kiev.ua/~symmetry/Symmetry2001/Bedrij589-601.pdf

≈ jossi ≈ (talk) 18:40, 8 December 2006 (UTC)


 * I already know it, and you can actually calculate it extremely simply, without any of the formulas listed there. Have you calculated it?  If you haven't, you'll be pleasantly surprised by the result.  Karl Dickman talk 23:10, 11 December 2006 (UTC)

I am all ears... ≈ jossi ≈ (talk) 23:13, 11 December 2006 (UTC)

$$G=\frac{F_G}{m}=\frac{m g}{m}=g$$, where $$G$$ is the strength of the field, $$F_G$$ is the force caused by the field, $$m$$ is the mass of an object in the field, $$g$$ is the acceleration due to gravity. Now, lete me explain what this means.

Gravitational and electric fields
The gravitational flux density is the same quantity as the gravitational field strength, just like electric flux density is the same quantity as electric field strength. What do I mean by this? Allow me to explain using a common technique for visualising electric and magnetic fields (it can also be applied to gravitational fields) is the concept of "lines of force", which was developed by Michael Faraday.

In the case of electric fields, lines of force start at positive charges and terminate at negative charges. Consider the case of a single positively charged body: it should have lines of force radiating from it in all directions like the spokes of a wheel. The larger the charge of the body, the more lines of force it has. For a negative charge, the lines of force should enter it from all directions; the field looks the same as that generated by a positive charge, but the arrows point in the opposite direction. Now consider two charges of opposite sign but the same magnitude (illustrated here). Some of the lines of force generated by the positive charge radiate into space, while others curve around and end at the negative charge.

Now consider a small positively charged body with a charge $$q_0$$, such that $$q_0$$ is negligible relative to the charge $$\pm q$$ of either charge in the dipole ($$q_0 \ll |q|$$). In Physics, we call this a test charge. If the test charge is acted upon by no other force than that generated by the electric field, it will follow a line of force to come to rest on the negative charge.

A similar visualisation can be used for gravitational fields: lines of force start and end at a mass. But with gravitational fields, the concept of direction gets a little confusing, so I'll try not to go any further.

One final thing: if I define a region $$S$$ with an area $$A_S$$, then the number of field lines that flow through $$S$$ is equal to the electric or gravitational flux (flux=flow) through that region. The number of field lines per unit area is the flux density. Remember that we said that the density of the field lines is proportional to the strength of the field? Flux density is the same thing as field strength.

Vector fields
These days, we express electric, magnetic, and gravitational fields as vector fields. I'm sure your at least acquainted with vectors. They are mathematical constructs with magnitude and direction. One way of visualising vectors is to draw an arrow from the origin (at point $$(0,0,0)$$) to a point $$P(x_0,y_0,z_0)$$. That vector can be expressed in coordinate form as $$\langle x_0,y_0,z_0 \rangle$$. In a vector field, each point is assigned a vector, and each component of the vector is a function of its coordinates. So a vector at any point $$(x,y,z)$$ will have coordinates $$\langle P(x,y,z),Q(x,y,z),R(x,y,z)\rangle$$ where $$P$$, $$Q$$, and $$R$$ are functions of the variables $$x$$, $$y$$, and $$z$$. I'm sorry if this is to mathy for your; but having some idea of what vector fields are is extremely useful in understanding some of the math. Go to the articles on electric field, vector field, and dipole; you can see some nice pictures of vector fields there.

Here's a great example of a vector field equation that defines the force due to gravity ($$\mathbf{F_G}$$) under the influence of a single mass $$M$$: $$\mathbf{F_G}=\left \langle\frac{-GMx}{(x^2+y^2+z^2)^\frac{3}{2}},\frac{-GMy}{(x^2+y^2+z^2)^\frac{3}{2}},\frac{-GMz}{(x^2+y^2+z^2)^\frac{3}{2}}\right\rangle$$. Using the very useful substitution $$\mathbf{r}=\langle x,y,z \rangle$$, we can simplify this to $$\mathbf{F_G}=-\frac{GM\mathbf{r}}{|\mathbf{r}|^3}$$.

So, each kind of force field can be expressed as a vector function, magnetic fields are usually denoted by $$\mathbf{B}$$ (magnetic flux density/magnetic field strength) and electric fields by $$\mathbf{E}$$ (electric flux density/electric field strength). I will use $$\mathbf{G}$$ to denote gravitational field strength. I use bold to denote vector quantities here; that is the difference between $$\mathbf{G}$$ (gravitational field) and $$G$$ (Newton's constant): the former is a vector, the latter is just a number (or scalar).

One final properties to remember about vectors: their magnitude. Remember how we said that a vector can be represented by an arrow drawn from the origin to a point $$P(x_0,y_0,z_0)$$? The magnitude is equal to the length of that arrow, which, by the Pythagorean theorem, is $$\sqrt{{x_0}^2+{y_0}^2+{z_0}^2}$$. The magnitude of a vector is denoted by $$|\mathbf{a}|$$. Magnitude is a scalar quantity.

How fields act on charges and masses
At any given point in an electric field, the force that is caused by the field on a particle of charge $$q$$ is given by the equation $$\mathbf{E}=\frac{\mathbf{F_E}}{q}$$. Likewise, a graviational field acts on a particle of mass $$m$$ with the relation $$\mathbf{G}=\frac{\mathbf{F_G}}{m}$$.

This last equation is the one that I introduced you to way up at the top of this rant. By Newton's second law, every force is equivalent to a mass times the acceleration caused by the force. Thus, we can say that $$\mathbf{F_G}=m\mathbf{a_G}$$, where $$m$$ is the mass of the object being acted upon by the field and $$\mathbf{a_G}$$ is the acceleration caused by the force of gravity. Plugging the expression $$m\mathbf{a_G}$$ into the equation for $$\mathbf{F_G}$$, we arrive with the relationship $$\mathbf{G}=\frac{m\mathbf{a_G}}{m}=\mathbf{a_G}$$.

Remember that at the earth's surface acceleration is approximately g  (9.80665 m/s²). By definition, the magnitude of the acceleration due to gravity must always be g ($$g=|\mathbf{a_G}|$$). So we are left with the conclusion that the local gravitational flux density $$\mathbf{G}$$ is the same thing as the local gravitational acceleration.

Conclusion
At this point, you're probably wondering why I said that the solution was so simple. Well, the solution really is very simple: I could have just told you that $$\mathbf{G}=\frac{\mathbf{F_G}}{m}=\frac{m\mathbf{g}}{m}=\mathbf{g}$$. Partly, I felt that if I just tossed the equation at you you wouldn't appreciate the beauty of the result; partly, I just love lecturing on my favourite physics topics. I did my best to make it easy to understand, but you have to realise that I've packed about two weeks of math and three weeks of Physics into a few paragraphs.

Now it's time for me to tell a story. At the end of my AP Physics class my senior year of high school, we had spent a lot of time talking about electric and magnetic fields, field lines, fluxes, flux densities/field strength, electric and magnetic forces, etc. I had known for a while that gravitation can be portrayed with the same kind of fields that electric and magnetic forces can. So I sat down and thought about the implications of this. I decided to calculate the field strength at the earth's surface, and realised, with a shock that it was g. All that crazy stuff about vectors, lines of force, fluxes and flux densities boild down to simple acceleration. It was like some brilliant piece of poetry or an essay.

There are stories about how Einstein, upon learning Maxwell's equations, was so excited by them that he would lurk in coffehouses and corner passerby to lecture them on Maxwell's equations. If you can get to the point where you understand Maxwell's equations, they have a kind of indescribable beauty, like your favourite song or favourite sonnet.

Anyway, sorry about the rant. Hope you enjoyed it. Karl Dickman talk 04:16, 12 December 2006 (UTC)


 * Beware of what you ask for... I should tell myself :) But I really enjoyed it. You really have a knack for explaining things. Thanks a lot. ≈ jossi ≈ (talk) 20:41, 12 December 2006 (UTC)