GREPhysics.NET
GR | # Login | Register
   
  GR9677 #23
Problem
GREPhysics.NET Official Solution    Alternate Solutions
This problem is still being typed.
Mechanics}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+\epsilo<i>n</i>.

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.



See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
There are no Alternate Solutions for this problem. Be the first to post one!
Comments
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
?
tweetiebird
2014-11-24 21:10:50
Never mind. I forgot kepler's law. T^2 = kd^3
NEC
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.NEC
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
NEC
r10101
2007-10-30 17:03:27
So... (D) or (E)? And why?
r10101
2007-10-30 20:35:40
Oops, scratch that.
NEC
mhas035
2007-03-21 21:44:37
Has anyone else got a paper that says that D is the right answer?
hamood
2007-04-09 11:59:45
yes..you can download the original exams from http://www.physics.ohio-state.edu/undergrad/ugs_gre.php
k93
2008-03-29 23:25:05
then why does the answer have E as the right not D like in the solutions to the test posted
dogsandfrogs
2009-10-07 13:54:34
Remember, the question asks which statement is FALSE. D is false, as Yosun says, so D is the answer.
NEC

Post A Comment!
You are replying to:
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.

Username:
Password:
Click here to register.
This comment is best classified as a: (mouseover)
 
Mouseover the respective type above for an explanation of each type.

Bare Basic LaTeX Rosetta Stone

LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces .
type this... to get...
$\int_0^\infty$
$\partial$
$\Rightarrow$
$\ddot{x},\dot{x}$
$\sqrt{z}$
$\langle my \rangle$
$\left( abacadabra \right)_{me}$
$\vec{E}$
$\frac{a}{b}$
 
The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...