GREPhysics.NET: GR0177 #3

GR0177 #3

Problem

GREPhysics.NET Official Solution

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Mechanics $\Rightarrow$ }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 $T \propto R^{3/2}$ , as in choice (D).

Alternate Solutions

rifraff 2016-08-06 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. Reply to this comment
rifraff 2016-08-06 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. Reply to this comment
nathan12343 2008-08-13 13:44:56	One can always remember that $U = \frac{GMm}{R}$ And, from the Virial Theorem, $U = \frac{3}{2}T = \frac{3}{4}mv^2$ From this we get $v \propto \sqrt{\frac{GM}{R}}$ Now, the orbital Period, $\tau$ , is nothing but $\frac{2 \pi R}{v}$ So, we have, $\tau \propto R^{\frac{3}{2}}$ Which is the answer Reply to this comment

Comments

rifraff 2016-08-06 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. Reply to this comment
rifraff 2016-08-06 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. Reply to this comment
secretempire1 2012-08-28 06:29:51	The general velocity equation is $V=\frac{d}{t}$ . Orbital velocity $V_{orbit} = \sqrt{\frac{GM}{R}}$ . so if you consider your distance in the equation to be the orbital path, then $d = 2\pi R$ . Rearranging the general equation to solve for time leaves $t=\frac{d}{V}$ . Plugging in yields $t=\frac{2\pi R}{\sqrt{\frac{GM}{R}}} = \frac{2\pi R \sqrt{R}}{\sqrt{GM}} = \frac{2\pi R^{\frac{3}{2}}}{\sqrt{GM}}$ . Therefore $t \propto R^{\frac{3}{2}}$ . Reply to this comment
nathan12343 2008-08-13 13:44:56	One can always remember that $U = \frac{GMm}{R}$ And, from the Virial Theorem, $U = \frac{3}{2}T = \frac{3}{4}mv^2$ From this we get $v \propto \sqrt{\frac{GM}{R}}$ Now, the orbital Period, $\tau$ , is nothing but $\frac{2 \pi R}{v}$ So, we have, $\tau \propto R^{\frac{3}{2}}$ Which is the answer Reply to this comment
michealmas 2007-01-01 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: $\frac{GMm}{r^{2}} = \frac{m v^{2}}{r}$ using $v = \omega r$ and $T = \frac{2 \pi}{\omega}$ . Reply to this comment

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The general velocity equation is $V=\frac{d}{t}$ . Orbital velocity $V_{orbit} = \sqrt{\frac{GM}{R}}$ . so if you consider your distance in the equation to be the orbital path, then $d = 2\pi R$ . Rearranging the general equation to solve for time leaves $t=\frac{d}{V}$ . Plugging in yields $t=\frac{2\pi R}{\sqrt{\frac{GM}{R}}} = \frac{2\pi R \sqrt{R}}{\sqrt{GM}} = \frac{2\pi R^{\frac{3}{2}}}{\sqrt{GM}}$ . Therefore $t \propto R^{\frac{3}{2}}$ .

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