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GR9677 #23
Problem
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Mechanics$\Rightarrow$}Stability of Orbits

The gravitational force suspect to a bit of perturbation is given as $\vec{F}_{12} = \hat{r}_{12} Gm_1 m_2/r_{12}^{2+\epsilon$.

One can narrow down most choices by recalling some basic facts from central force theory:

(A) No mention is made of frictional effects, and thus energy should be conserved.

(B) Angular momentum is always conserved since the net torque is 0 (to wit: the force and moment arm are parallel).

(C) This is just Kepler's Third Law applied to this force. (Recall the following bromide: The square of the period is equal to the cube of the radius---for the inverse square law force. For a perturbed force, the bromide becomes: The square of the period is equal to the $3+\epsilon$ power of the radius.)

(D) Recall Bertrand's Theorem from Goldstein. Stable non-circular orbits can only occur for the simple harmonic potential and the inverse-square law force. This is of neither form, and thus this choice is FALSE.

(E) Circular orbits exist for basically all potentials. A stationary orbit exists if and only if the following conditions are satisfied: $V'=0 \; \;V''>0$. Recall that the potential is related to the force by $-V'=F\Rightarrow V=-\int F dx$. Use $V\propto 1/r^n$, and recalling the extra term added to the effective potential to be $L^2/(2mr^2)$, one chunks out the derivatives to get the condition that $n<2$, as a potential exponent, ($n<3$, as a force exponent) for stable orbit. One can remember this result or re-derive it whenever necessary. For $n < 3$, (the power exponent of the force equation), a stable circular orbit exists. Since $\epsilon$ is presumably less than 1, the planet does, indeed, move in a stationary circular orbit about the sun.

Alternate Solutions
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tweetiebird
2014-11-24 21:09:21
Is there a typo in the original answer for (C)? Should it be (2+e)/2 instead of (3+e)/2
?
 tweetiebird2014-11-24 21:10:50 Never mind. I forgot kepler's law. T^2 = kd^3
cczako
2013-10-17 18:09:04
The way that I solved this problem is that if only one could be false so it had to be D or E. And if C is true then (circular orbits) D must be false.
kaic
2013-10-15 11:17:44
the simplest way to answer this question is to know that no elliptical orbit can be stationary: elliptic orbits always precess, as demonstrated in this nice picture
http://en.wikipedia.org/wiki/File:Precessing_Kepler_orbit_280frames_e0.6_smaller.gif
r10101
2007-10-30 17:03:27
So... (D) or (E)? And why?
 r101012007-10-30 20:35:40 Oops, scratch that.
mhas035
2007-03-21 21:44:37
Has anyone else got a paper that says that D is the right answer?
 k932008-03-29 23:25:05 then why does the answer have E as the right not D like in the solutions to the test posted
 dogsandfrogs2009-10-07 13:54:34 Remember, the question asks which statement is FALSE. D is false, as Yosun says, so D is the answer.

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