field acceleration

yourdadonapogostick

the magnitude of the gravitational field vector, [tex]\vec{G}[/tex], is equal to g at that point. does that mean that magnitudes of [tex]\vec{B}[/tex] and [tex]\vec{E}[/tex] are equal to the acceleration due to magnetism and electric fields, respectively, at a point?

sniffer

no.
gravity is different, since the force acting on a mass is proportional to the mass, thus you get a "constant".

yourdadonapogostick

with electricity, the force is proportional to the charge.

marlon

When you write down the equation for the gravitational force you get :

[tex] \frac {Amm'}{r^2}[/tex]

A is the universal gravitational constant, m amd m' the two masses and r is the distance between those two masses. The above formula is ofcourse the component of the interaction along the axis that connects the two masses.

Now write this force as [tex]mG[/tex] then [tex]G = \frac {Am'}{r^2}[/tex]

Suppose you look at an object with mass m on this earth. You describe the gravitational interaction between this object and the earth by setting m' equal to the earth's mass, A is a universal constant, and r is the earth's radius. Now, if this object is 100 above the earth's surface, you should have written for r the value of the earth's radius PLUS 100m. But since the earth's radius is much bigger, just forget about the 100m

If you fill in these values for G, you will get the 9.81 m/s^2 that we all know.
The expression for G which depends on the mass m' and the distance between m and m' also suggest why the gravitational constant is not everywhere the same value on this earth. Well, the earth is not a perfect sphere right


hope that helps

marlon

Meir Achuz

E and B produce force by the Lorentz force equation:
F=q[E+vXB].
This equals the rate of change of momentum: dp/dt.
Non-relativistically, dp/dt=ma, but in SR the acceleration is much more complicated.
dp/dt is still relatively simple in SR.