Dealing with variable force

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Firstly, I'd like to announce that this is not a homework question, it is an example of a problem I thought up to illustrate what I have trouble with:

Two spheres, each with a mass of 1 kg and a radius of 1 meter, lie in space. Their centers are 10 meters apart. When will they make contact?

Now obviously the initial force acting on each sphere is easily calculable from the gravity equation. But an infinitesmall period of time after they have moved towards each other, their distance apart will have changed, and the force acting on each sphere will increase. Not only are they accelerating towards each other because of gravity, but they are also increasing their acceleration as they get closer to each other.

I am sure calculus will be involved and that's fine. What's the simplest method that you would use to solve a problem where force changes with distance/time?

Thanks!
 

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  • #2
arildno
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Well, you might do it like this:
Call the spheres A and B, and let the origin of the coordinate system lie at the midpoint of the line segment defined by their centres.
Calling the sphere centres' positions as functions of time [itex]x_{A}(t),x_{B}(t), x_{A}(0)=-5,x_{B}(0)=5[/tex], respectively, we define the distance function between them as:
[tex]D(t)=x_{B}(t)-x_{A(t), D(0)=D_{0}=10[/tex]

Setting up Newton's 2.law for both, we get, with unit masses:
[tex]\frac{d^{2}x_{A}}{dt^{2}}=\frac{G}{D^{2}}, \frac{d^{2}x_{B}}{dt^{2}}=-\frac{G}{D^{2}}[/tex]
whereby the equation for d(t) is readily derived:
[tex]\frac{d^{2}D}{dt^{2}=-\frac{2G}{D^{2}} (*)[/tex]

We assume that the initial velocities are 0, i.e, [tex]\frac{dD}{dt}\mid_{t=0}=0[/tex]

Let us multiply (*) with dD/dt:
[tex]\frac{d^{2}D}{dt^{2}}\frac{dD}{dt}=-\frac{2G}{D^{2}}\frac{dD}{dt}[/tex]

Integrating both sides from t=0 to some arbitrary t-value, taking due notice of the initial conditions, yields:
[tex]\frac{1}{2}(\frac{dD}{dt})^{2}=\frac{2G}{D}-\frac{2G}{D_{0}}[/tex]

multiplying with two, taking the square root and remembering that D(t) will be decreasing, we get the diff. eq:
[tex]\frac{dD}{dt}=-\sqrt{\frac{4G}{D_{0}}\sqrt{\frac{D_{0}-D}{D}}[/tex]
This is a separable diff.eq; we write:
[tex]\sqrt{\frac{D}{D_{0}-D}}dD=-\sqrt{\frac{4G}{D_{0}}dt[/tex]

We now remember that when they spheres make contact, D(T)=2, where T is the time we're looking for!

Thus, we get the equation for T, integrating both sides:
[tex]\int_{10}^{2}\sqrt{\frac{D}{D_{0}-D}}dD=-\sqrt{\frac{4G}{D_{0}}T[/tex], or, equivalently:
[tex]T=\sqrt{\frac{D_{0}}{4G}}\int_{2}^{10}\sqrt{\frac{D}{D_{0}-D}}dD (**)[/tex]

In order to crack that integral, let us set:
[tex]u=\sqrt{\frac{D}{D_{0}-D}}\to{D}=D_{0}-\frac{D_{0}}{1+u^{2}}[/tex]
Thus, [tex]dD=\frac{D_{0}2u}{(1+u^{2})^{2}}du[/tex]
The limits are [tex]D=10\to{u}=\infty,D=2\to{u}=\frac{1}{2}[/tex]
We thereby get the expression for T in u:
[tex]T=D_{0}\sqrt{\frac{D_{0}}{G}}\int_{\frac{1}{2}}^{\infty}\frac{u^{2}du}{(1+u^{2})^{2}}[/tex]

This can quite readily be solved by standard techniques.
 
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  • #3
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I will guess 92,249.8 seconds from numerical integration.

edit: I left out a constant. I get 2,917,196 seconds.
 
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  • #4
arildno
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Well, we can make it a bit more explicit than that, Bob S!

We have:
[tex]\int\frac{1}{1+u^{2}}du=\frac{u}{1+u^{2}}+\int\frac{2u^{2}}{(1+u^{2})^{2}}du[/tex], by doing integration by parts.
Therefore, we get:
[tex]\int\frac{u^{2}}{(1+u^{2})^{2}}du=\frac{1}{2}(\arctan(u)-\frac{u}{1+u^{2}})+C[/tex]

whereby we get:
[tex]T=\frac{10^{\frac{3}{2}}}{2\sqrt{G}}(\frac{\pi}{2}-\arctan(\frac{1}{2})+\frac{2}{5})[/tex]
 

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