1. Home
  2. Electronics Design
  3. [OT] Spheres in collision

[OT] Spheres in collision

Like any of you ever thought my posts were on topic ...

(crossposted, trim) I'm sure this will interest the s.e.d regulars and several can solve it, and s.s.science & rec.models.rockets are my best guess for where it actually is on topic. Any other groups likely to be appropriate? This is related to my post about eMachineShop in other groups this week.

Given two spheres of mass \\r and density \\rho, each has a mass of \\rho V = 4/3 \\pi \\rho r^3.

Separate them, center to center, by 3r. That is, surface to surface separation of r, then the gravitational force between them is

F = G m m / (9r^2)

Using F = ma and waiving hands wildly while students sleep ...

a = -G m / (9r^2) a = -4/27 \\pi \\rho G r

Using iridium (for small spheres), \\rho = 22,160 kg/m^3 G=6.673E-11 m^3 / (kg s^2) a = -688E-9 m/s^2 r

Notice, a is reletive to center of mass, not sphere to sphere, explaining a factor of 2 that likes to hide.

Solving s=1/2at^2 and bounding t by the initial value of a:

t = 1,205s or 20 min. Notice, it doesn't matter what r is. ..... Now, call the original position \\s_0 at time t=0 and place the origin at the center of mass. At time t, the position is s, and the separation between centers will be 2s. There is surely more elegant notation.

a = 4/3 \\pi G \\rho r^3 / s^2

s(t) = \\int \\int (4/3 \\pi G r^3) / (2s)^2 dt dt

Maybe I'm slow this morning, but I don't see the solution to this. Can someone with untarnished calculus or mathematica give a solution to this, for arbitrary s_0? Is there some aerospace engineering text where this is treated as an example or homework problem?

Reply to
Aubrey McIntosh, Ph.D.
Loading thread data ...

Hello Aubrey,

You might want to rethink what you are doing. The equation s(t) = ..5at*t is the second equation of motion for a CONSTANT acceleration.

In the case of two bodies falling toward each other, the accelleration is NOT constant but varies over the distanance of seperation between the two objects.

You will discover that you will eventually reach a First Order Differential equation for v(r).

First:

F = ma = m dv/dt and F = GMm / r*r (note I am not using 'r hat' vectoring)

Second: dv dy dv a = dv/dt = -- . -- = v * -- (see you had too many variables) dy dt dy

As these spheres are "falling" towards one another, you have distance and time changing ... that is too many variables changing and not enough equations to solve them.

Third:

By setting ma = GMm / r*r and subbing 'a' from above on the left side of the equation you reach:

dv mGM v -- = - --- (Now I am factoring in 'r hat' vector notation, '-') dy r*r

This is a first order differential equation that is solved to the form:

v*v GM --- = --- + c => v(r) 2 r

Now I am guessing you are interested in r(t). Well I will leave that up to you as 1) I don't care to do the math, 2) only you know what circumstance you are looking for. This will require a Second Ordered Differential Equation and a bit off work. But with a PhD, I am SURE you can manage ... just makes me wonder WHY you are using a constant acceleration equation, to solvle a coupled (r,t) non-linear situation .... drink a double espresso :)

I wish I could be more helpful but it has been many years since I left college and I just can't think as sharply as I once did ... time, time dumbs us all to death eventually. Email me in the end and let me see the entire equation derived and solved please.

Aubrey Mc> Like any of you ever thought my posts were on topic ...

and

best

be

\\rho V

at

Can

where

Reply to
lunarlos

On 14 May 2005 15:11:59 -0700, " snipped-for-privacy@juno.com" wrote

But whereas brute computational abilities may decline with age and disuse, shear sneakiness and guile more than compensate. In other words, you can always hire kids to to the heavy lifting.

John

Reply to
John Larkin

Aubrey McIntosh, Ph.D. wrote in sci.electronics.design:

[...]

[...]

By differentiation wrt. to t (twice) it becomes an ordinary differential equation. In Mathematica notation:

s''[t] == 4/3*Pi*G*r^3/(2s[t])^2

or

deq = s''[t]*s[t]^2 == Pi*G*r^3/3

Mathematica solves it, but the result isn't pretty (see below). You'll note the solution is in terms of two nasty transcendental equations. I haven't followed this further.

Anno

--------------------------------------------------------------------------- In[10]:= DSolve[ deq, s[t], t]

Solve::tdep: The equations appear to involve the variables to be solved for in an essentially non-algebraic way.

Solve::tdep: The equations appear to involve the variables to be solved for in an essentially non-algebraic way.

3 Out[10]= {Solve[C[2] - Sqrt[3] ((G Pi r 3 -2 G Pi r + C[1] s[t]

s[t] 3 2 (-(G Pi r ) + C[1] s[t]) 3/2

Sqrt[C[1]] 3 -2 G Pi r + C[1] s[t] s[t] Sqrt[----------------------] s[t]

C[1] 3

3 -2 G Pi r + C[1] s[t]

s[t] 3 2 (-(G Pi r ) + C[1] s[t]) 3/2

Sqrt[C[1]] 3 -2 G Pi r + C[1] s[t] s[t] Sqrt[----------------------] s[t]

C[1]

Reply to
Anno Siegel

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.