- #1

JDoolin

Gold Member

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- 9

[tex]F=-G \frac{m_1 m_2}{r^2}[/tex]

I'd like to set up the problem so the particle begins at t=0 at radius r=r

_{0}and radial velocity v

_{r}=v

_{0}. And there is only a component of velocity, in the radial direction. (The particle is going straight toward the gravitating body)

The above equation becomes

[tex]F=m a_r= m_2 \frac{\mathrm{d}^2 r}{\mathrm{d} t^2}= G \frac{m_1 m_2}{r^2}[/tex]

And so the most simply I can give the differential equation is:

[tex]\left (\frac{\mathrm{d}^2 }{\mathrm{d} t^2} \right )r = -G \frac{ M}{r^2}[/tex]

Does this equation have a closed-form solution?

If not, is there a well-known technique of solving it numerically? Here is what I did numerically.

(1) determine r0 and v0.

(2) determine the acceleration at your current radius a=-GM/r^2

(3) choose a small Δt during which the acceleration is approximately constant.

(4) Find a final velocity by multiplying the acceleration times the Δt.

(5) Find the average velocity from (initial velocity + final velocity)/2

(6) Multiply average velocity by Δt to find Δr.

(7) Add Δt and Δr and Δv to your previous values of t, r, and v, and start over at step (2)

One particular set-up is to determine the initial velocity at r

_{0}to be the negative of the escape velocity at that same r

_{0}. It stands to reason that if a particle is traveling at a speed equal and opposite to its escape velocity as it falls in, that conservation of energy will maintain this, so that the particle will fall in at its escape velocity the entire way.

I tried out this numerical method and the velocities calculated numerically did match the escape velocities pretty well. But this gives us a function v(r) when I am wanting a function r(t), and it also is a solution that requires a specific pre-defined initial radial velocity, where I would like a function where the initial velocity can be decided arbitrarily.

I appreciate any help anyone might think of. Thanks.