View Full Version : Distance and Derivatives for Gravity

gillardrs

2007-Sep-26, 08:04 PM

I am trying to devise an equation to describe distance, velocity, and acceleration with non-constant acceleration due to gravity changing with respect to distance. For instance, at 100,000 meters the gravitational acceleration is let's say X, yet at 1,000 meters the gravitational acceleration is 10,000X. This gives different distance, velocity, and acceleration equations. Can anyone help me? Thanks in advance!

tusenfem

2007-Sep-27, 02:07 PM

Just a little math you gotta do. Your acceleration will either be a function of x (which is a bit tricky, but hey, who cares)

so you just say that g = g(x) m/s2, let's keep it one dimensional for the moment.

now you have to solve the equotion of motion of a particle.

F(x) = dt p(x,t) = m g(x,t)

now keeping it real simple (non-relativistic) you will find that the time derivative the impuls (m v(x,t)) will just be the time derivative of the velocity, the mass falls out at both sides (unless your mass also changes with x but that is unnecessarily complicated)

so we have dt v(x,t) = g(x,t) and than you have to rewrite that a little to get the equation:

(dx(t) / dt) (d / dx) v(x,t) = g(x,t)

where, dx(t) / dt is equal to v(t) and g(x,t) = g(x) to make it only dependent on x and not on t. and then it is just a lot of math, which I cannot do in this window, sitting at a conference :-)

Have fun with the intro.

tusenfem

2007-Sep-28, 06:47 AM

So you go on with substituting v(x,t) for dx(t)/dt and then do some strange stuff, which is okay here, you move the v into the derivative to x and "multiply both sides by dx" which will leave you with

d 0.5 v2(x,t) = g(x) dx

and this you can integrate from x = a to x = b and you have to choose some "nice" starting points (e.g. a = 0 etc.)

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