Twitching & Birdwatching1 min ago
Acceleration vs. Time in inverse-square force field
Suppose a fixed particle A attracts movable particle B by an inverse-square force field ( as in the case of gravitation or electrostatic attraction) so that the acceleration of B toward A is
a=K/(r^2)
where K is a constant and r is the distance of particle B from A.
What is a as a function of time t if particle B is released from rest at initial radius R at t=0?
a=K/(r^2)
where K is a constant and r is the distance of particle B from A.
What is a as a function of time t if particle B is released from rest at initial radius R at t=0?
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Hi Ignoblius
Yes but that's no mean task, because the only way you can solve for r is by using setting a value for t and then using the Newton-Raphson
numerical method to iterate to the corresponding value of r.
By the way just one small correction to your last post:
theta=pi/2 at r=0
theta=0 at r=R
You could use a spreadsheet to do the Newton-Raphson or just produce a spreadsheet and do the calculation as I mentioned in my pdf.
Here's a link to a spreadsheet which does it:
https:/
Vasco
Yes but that's no mean task, because the only way you can solve for r is by using setting a value for t and then using the Newton-Raphson
numerical method to iterate to the corresponding value of r.
By the way just one small correction to your last post:
theta=pi/2 at r=0
theta=0 at r=R
You could use a spreadsheet to do the Newton-Raphson or just produce a spreadsheet and do the calculation as I mentioned in my pdf.
Here's a link to a spreadsheet which does it:
https:/
Vasco
I graphed out
t= √(R³/2K) * [acos (√r / √R) + √(r(R-r)) / R]
then flipped it about to get an image of the inverse function.
I can't put a graphic into this post but what I get looks like a half-cycle of a periodic function. It is what you would get if particle B passed through center point A without collision, reaching apogee at -R and falling back to A again. By linking the curve with its mirror image to get the complete cycle, the result looks like a rectangular waveform with the flats & corners rounded. The movement of particle B seems like an elliptical orbit with a minor axis of zero.
If it's a periodic waveform a Fourier transform is possible. The result can be differentiated to get v(t) and a(t).
t= √(R³/2K) * [acos (√r / √R) + √(r(R-r)) / R]
then flipped it about to get an image of the inverse function.
I can't put a graphic into this post but what I get looks like a half-cycle of a periodic function. It is what you would get if particle B passed through center point A without collision, reaching apogee at -R and falling back to A again. By linking the curve with its mirror image to get the complete cycle, the result looks like a rectangular waveform with the flats & corners rounded. The movement of particle B seems like an elliptical orbit with a minor axis of zero.
If it's a periodic waveform a Fourier transform is possible. The result can be differentiated to get v(t) and a(t).
Its a cardioid if you adjust the scales of r and t as follows:
x can be written as:
x=R/2*(1+cos2 theta) and
t=sqrt(R^3/(2K))/2*(2 theta+sin2 theta)
=R/2*sqrt(R/(2K))*(2 theta+sin2 theta)
So if we define T as t/sqrt(R/(2K)) which is just a scaling factor then
if we let a=R/2 then
X=a(1+cos2 theta) and T=a(2 theta+sin2 theta) and this is the equation of a cardioid.
x can be written as:
x=R/2*(1+cos2 theta) and
t=sqrt(R^3/(2K))/2*(2 theta+sin2 theta)
=R/2*sqrt(R/(2K))*(2 theta+sin2 theta)
So if we define T as t/sqrt(R/(2K)) which is just a scaling factor then
if we let a=R/2 then
X=a(1+cos2 theta) and T=a(2 theta+sin2 theta) and this is the equation of a cardioid.
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