Gravitational plot

1. Nov 8, 2015

Invutil

Is this the correct plot of gravity and motion?

x(t) = x0 + v0 t + 1/2 a t^2

F = G m1 m2 / r^2

x(t) = x0 + v0 t + t^2 / (x2 - x(t))^2

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2. Nov 8, 2015

Chandra Prayaga

No. The first equation assumes that the acceleration is a constant. The second equation does not give a constant acceleration, so you cannot use the acceleration from the second in the first equation I don't know where you got equ 3 from. Perhaps if you stated the problem?

3. Nov 8, 2015

PeroK

Can you explain what you're trying to do? I don't understand the third equation, nor the graphs.

4. Nov 8, 2015

Invutil

I'm just plotting gravity free-fall.

r = (x2 - x) for x2 > x where x2 is the coordinate of mass 2

Is this correct?

x(t) = x0 + int from {0} to {t} ( v0 t + 1/2 a0 t^2 + int from {0} to {t} ( 1/2 t^2 da/dt ) dt ) dt

a0 = F/m1
a0 = G m2 / r0^2
r0 = (x2 - x0)

x(t) = x0 + int from {0} to {t} ( v0 t + 1/2 G m2 t^2 / (x2-x0)^2 + int from {0} to {t} ( d( 1/2 t^2 G m2 / (x2 - x(t))^2 )/dt ) dt ) dt

I can't get this to plot, but should there be anything special at negative radius values?

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5. Nov 8, 2015

PeroK

You seem to be mixing up two different gravitational equations: one for a constant force (valid, for example, near the surface of the Earth); and one for a variable force (valid, for example, for planetary motion).

The first of these is mathematically simple; the second is mathematically much more complicated.

6. Nov 8, 2015

Invutil

I guess I'm dealing with planetary scales so radius comes into effect. Is there a formula I can use? Is there anything special that happens at negative radius, like, a Kerr black hole, using a Newtonian equation? Would the constant-acceleration formula generally be correct, as it is a Taylor expansion? If so, how do you interpret the trajectory after r < 0?

7. Nov 8, 2015

PeroK

It depends how much maths you know. The planetary motion equation leads to a second-order differential equation, from which Kepler's laws can be deduced. For free fall, say of an asteroid towards the Earth, the equation can be solved with some difficulty.

8. Nov 8, 2015

Chandra Prayaga

I think you are trying to over-simplify the problem. Your equations are still valid only for constant acceleration. Planetary motion, including that of asteroids, is described using universal gravitation for the force, and the acceleration is not constant. If you want to use the constant acceleration formula, than you have to numerically integrate the equations of motion, using short intervals of time, during which you assume that the acceleration is constant. That becomes a job to be done by a computer, and not a simple formula into which you can substitute numbers.