Suppose you had a small body in a circular orbit. At some point, you turn on a thruster pointed directly away from the primary, ostensibly pushing yourself down toward the planet.

Now the force between the two bodies is described as:

$$ F = \frac{k_1 } {d^2} + k_2 $$

What is the resulting motion? Obviously it does not change the angular momentum, so it will remain bound and will enter a new steady state, not spiral in progressivly and crash like bad Sy-Fy movies would have.

But what happend? An ellipse, a spirograph drawing? Or will the state indeed progress and get a lower and lower perige over time?

Update: the Wikipedia page indicates that this case, $ F(r) = 0r^{-3} + Br^{-2} + Cr + 0 $, “has solutions in terms of circular and elliptic functions” but doesn’t state what that solution is.

  • $\begingroup$ P.S. my TeX skills have atropated — can someone fix my \frac please? I got a + in the denominator and the desired contents missing! $\endgroup$ – JDługosz Jun 1 '17 at 0:18
  • $\begingroup$ $\sigma = r \times p$ can be conserved while $\|p\| \to +\infty$ and $\|r\| \to 0$, no? $\endgroup$ – user154997 Jun 1 '17 at 13:40
  • $\begingroup$ @HritikNarayan ah! I tried to wtite it infix! Thanks. $\endgroup$ – JDługosz Jun 1 '17 at 14:09

Wikipedia's page on the central force problem has quite a lot of detailed information on this, including which potentials have exact solutions.

I'd expect your orbit (path) to be like a spiral away from the central body as its radial velocity will continually increase, past escape velocity if you can continue thrusting that long.

Once thrust stops, the path will again become one of the conic sections.

  • $\begingroup$ Thanks for the pointer to the Wikipedia page. But, your description doesn't match. Why would it spiral awaywhen I'm thrusting toward the center? And the beginning of the page confirms that this situation follows the rules for an effective potential energy, so it can’t escape! $\endgroup$ – JDługosz Jun 1 '17 at 13:24
  • $\begingroup$ @JDługosz Ah!! I initially made the same mistake that StephenG made. You have pointed the thruster away from the primary. That makes the thrust (aka, force) itself to be toward the primary. $\endgroup$ – Bill N Jun 1 '17 at 13:41

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