# Orbit with extra center-directed force?

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.

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