GR0177 #3


Problem


This problem is still being typed. 
Mechanics}Kepler's Third Law
Recall Kepler's Third Law stated in its most popular form, The square of the period is proportional to the cube of the orbital radius. (Technically, the orbital radius is the semimajor axis of the ellipse.)
Recast that commonsense fact above into equations to get , as in choice (D).


Alternate Solutions 
rifraff 20160806 22:59:29  (I apologize if this was a double post. The first post was acting funny.)\r\n\r\nIf anyone wants a solution that\'s a bit easier by \"feel,\" maybe:\r\n\r\nThe radial acceleration, a, is proportional to v^2/R.\r\n\r\nThat same radial acceleration is proportional to 1/R^2, since the force is similarly proportioned.\r\n\r\nv^2/R ~ 1/R^2,\r\nso v ~ 1/√R.\r\n\r\nFinally, the time it takes to complete an orbit is t = d/v, where the \"circumference\" d is proportional to R.\r\n\r\nso t ~ R/v, or t ~ R^(3/2).\r\n\r\nThis was easier for me after the first 2 answers were ruled out with some thinking.\r\n\r\nMan, I wish I knew LaTeX.   rifraff 20160806 22:58:18  If anyone wants a solution that\'s a bit easier by \"feel,\" maybe:\r\n\r\nThe radial acceleration, a, is proportional to v^2/R.\r\n\r\nThat same radial acceleration is proportional to 1/R^2, since the force is similarly proportioned.\r\n\r\nv^2/R ~ 1/R^2,\r\nso v ~ 1/√R.\r\n\r\nFinally, the time it takes to complete an orbit is t = d/v, where the \"circumference\" d is proportional to R.\r\n\r\nso t ~ R/v, or t ~ R^(3/2).\r\n\r\nThis was easier for me after the first 2 answers were ruled out with some thinking.\r\n\r\nMan, I wish I knew LaTeX.   nathan12343 20080813 13:44:56  One can always remember that
And, from the Virial Theorem,
From this we get
Now, the orbital Period, , is nothing but
So, we have,
Which is the answer  

Comments 
rifraff 20160806 22:59:29  (I apologize if this was a double post. The first post was acting funny.)\r\n\r\nIf anyone wants a solution that\'s a bit easier by \"feel,\" maybe:\r\n\r\nThe radial acceleration, a, is proportional to v^2/R.\r\n\r\nThat same radial acceleration is proportional to 1/R^2, since the force is similarly proportioned.\r\n\r\nv^2/R ~ 1/R^2,\r\nso v ~ 1/√R.\r\n\r\nFinally, the time it takes to complete an orbit is t = d/v, where the \"circumference\" d is proportional to R.\r\n\r\nso t ~ R/v, or t ~ R^(3/2).\r\n\r\nThis was easier for me after the first 2 answers were ruled out with some thinking.\r\n\r\nMan, I wish I knew LaTeX.   rifraff 20160806 22:58:18  If anyone wants a solution that\'s a bit easier by \"feel,\" maybe:\r\n\r\nThe radial acceleration, a, is proportional to v^2/R.\r\n\r\nThat same radial acceleration is proportional to 1/R^2, since the force is similarly proportioned.\r\n\r\nv^2/R ~ 1/R^2,\r\nso v ~ 1/√R.\r\n\r\nFinally, the time it takes to complete an orbit is t = d/v, where the \"circumference\" d is proportional to R.\r\n\r\nso t ~ R/v, or t ~ R^(3/2).\r\n\r\nThis was easier for me after the first 2 answers were ruled out with some thinking.\r\n\r\nMan, I wish I knew LaTeX.   secretempire1 20120828 06:29:51  The general velocity equation is .
Orbital velocity .
so if you consider your distance in the equation to be the orbital path, then .
Rearranging the general equation to solve for time leaves .
Plugging in yields .
Therefore .   nathan12343 20080813 13:44:56  One can always remember that
And, from the Virial Theorem,
From this we get
Now, the orbital Period, , is nothing but
So, we have,
Which is the answer   michealmas 20070101 16:00:24  That's an elegant way to solve it, but if you didn't remember it, you could equate the forces, Newtons Law of Gravitation with the centripetal force for uniform circular motion:
using
and
.  




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