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Newtonian Gravitation & a differential equation
By Michael Doré (P904) on Sunday, June 6, 1999 - 12:08 pm:

Hi, my name is Michael Doré, and I have recently discovered nrich and become a member. I wonder if you can help me with something I have been puzzling over.


Is it possible to solve the differential equation


d²y = A/y² (A is constant)
dx²


analytically? (i.e. without using iterative techniques). I understand that a solution in terms of all the normal trigonometric and exponential functions is probably impossible but even a solution as an infinite polynomial would be most helpful.


Here is the context in which the problem appears:


Suppose you have 2 particles with masses of m1 and m2. These particles interact through gravitation, with the usual inverse square law. Their initial velocities are u1 and u2 and initial position vectors from a fixed origin are r1 and r2. (r2>r1) The vectors u1, u2, r1 and r2 are all parallel so the particles only travel in a straight line. The question I wanted to answer was what are the particles' displacement functions x1 and x2 (from the origin) at a time t? (While x2>x1, after which they collide).

Let v1 and v2 be the velocity of the particles after time t and let a1 and a2 be the particles' acceleration after time t.

By applying Newton's 2nd law of motion and Newton's law of gravitation:

m1a1 = Gm1m2/(x1-x2)²

m2a2 = -Gm1m2/(x1-x2)²

So

m1a1 + m2a2 = 0


Integrating:

m1v1 + m2v2 = C


But C = m1u1 + m2u2, so:


m1v1 + m2v2 = m1u1 + m2u2


Integrating once more, and again applying initial conditions:


m1x1 + m2x2 = m1u1t + m2u2t + m1r1 + m2r2


This can be solved for x1, then substituted into the equation for x2 above:

m2a2 = -Gm1m2/(x1-x2)²

to get:

a2 = -Gm1/(u1t + m2u2t/m1 + r1 + m2r2/m1 - m2x2/m1 - x2)²


By writing a2 as the 2nd derivative of x2 (with respect to t) and then making a linear substitution (of the part of the formula to be squared) the equation can be made into the form given at the start.


I would appreciate any comments,


Michael

By Alex Barnard (Agb21) on Tuesday, June 8, 1999 - 03:15 pm:

Actually this has a solution in terms of what you would call normal functions (square root and log). However to get it you have to solve for the other variable (ie. for x instead of y).

From the fact that (dx/dy) = 1/(dy/dx) we can change the differential equation to be:

y2 × (d2x/dy2) = -A×(dx/dy)3

You should now be able to solve this for (dx/dy) as it is just a separable differential equation.

To solve for x you'll need to be able to integrate things like Sqrt(C+1/y). To do this you use a substitution like y=cosh(z)^2. And you'll eventually get:

Integral[Sqrt(C+1/y)] == y×Sqrt(C+1/y) + Sqrt(1+Cy)×Log[Sqrt(Cy) + Sqrt(1+Cy)] / Sqrt[C+yC2]

Using this you'll be able to find a formula for x in terms of y. You won't be able to invert this to give y in terms of x without having to use power series or much more complicated functions.

Hope this sketch helps... Write back if you want anything in more detail.

AlexB.