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 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum Jul 10th 2019, 06:15 PM #1 Junior Member   Join Date: Jul 2019 Posts: 1 What is the period of the binary orbit? Okay, here is the problem I'm trying to solve: There are 2 masses, one with 90% the mass of earth and one with 50% the mass of earth. Let's call the bigger one m1 and the smaller one m2. m1 has a radius of 95% that of earth and m2 has a radius of 40% that of earth. These 2 masses are in a binary orbit. The minimum distance in this binary orbit is the earth-moon distance. The maximum distance in the binary orbit is 2.25 * the earth-moon distance. At minimum distance, the 2 masses are at their maximum velocity and at maximum distance, they are at their minimum velocity. Calculate the period of the binary orbit using Newton's laws of motion and the Law of Universal Gravitation. Use calculus if necessary. Okay, so let's simplify things and look at the ideal case, where the 2 masses are the same. Well, this gives us a sine wave pattern to the orbital velocity. Consequently, the acceleration would follow a cosine wave since the derivative of velocity is acceleration and the derivative of sine is cosine. Since this acceleration is from the gravitational force of the 2 masses, the gravitational force would have to change in proportion to cos(t). Now, getting to the real problem. Figuring out gravitational force: $F=G∗\frac{m1∗m2}{r^2}$ $G=6.67∗10^{−11}\frac{Nm^2}{kg^2}$ $m1=5.97∗10^{24}kg∗.90$ $m2=5.97∗10^{24}kg∗.50$ $r_{min}=3.84∗10^6m$ $r_{max}=1.158∗10^7m$ Plugging these values in I get $F_{min}=7.25∗10^{25}N$ and $F_{max}=1.81∗10^{25}N$ $F_{min}$ is at minimum distance and $F_{max}$ is at maximum distance That is a pretty strong gravitational force between the 2 masses. Gravitational acceleration of each mass at minimum distance: $a1_{min}=13.49\frac{m}{s^2}$ $a2_{min}=24.29\frac{m}{s^2}$ Gravitational acceleration of each mass at maximum distance: $a1_{max}=3.37\frac{m}{s^2}$ $a2_{max}=6.06\frac{m}{s^2}$ Where $a1$ is the acceleration of $m1$ and $a2$ is the acceleration of $m2$. If I equate the gravitational force, which is what I have calculated, with the centripetal force, I can get the velocity at minimum and maximum distance just doing some algebra. Here is what I get after doing the algebra: $F_c=F_g$ $F_c=\frac{m∗v^2}{r}$ $F_c∗r=m∗v^2$ $\frac{F_c∗r}{m}=v^2$ $sqrt{\frac{F_c∗r}{m}}=v$ And then if I plug in the masses, gravitational force, and distance at aphelion(maximum distance) and perihelion(minimum distance), I will get these velocites for $m1$: $v_{peri}=7198.23\frac{m}{s}$ $v_{aph}=5394.95\frac{m}{s}$ And for $m2$: $v_{peri}=9657.45\frac{m}{s}$ $v_{aph}=7238.09\frac{m}{s}$ But now, what do I do? I know the velocities of both masses at perihelion and aphelion as well as their accelerations at those points and the gravitational force at those points. But how am I supposed to go from knowing the velocities, accelerations, and the gravitational force at 2 points to figuring out the period of the binary orbit?   Jul 10th 2019, 09:24 PM #2 Forum Admin   Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 2,818 I haven't checked the numbers but they look decent. Typically when you have a binary system it can be simpler to get information about periods and the like by converting it from a two body problem to a single body problem. See here. See what you can do with this and check back if you still have difficulties. -Dan __________________ Do not meddle in the affairs of dragons for you are crunchy and taste good with ketchup. See the forum rules here.  Tags binary, orbit, period Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post ChemBlitz Advanced Thermodynamics 0 Oct 19th 2016 02:32 PM critten92 Thermodynamics and Fluid Mechanics 0 Mar 12th 2016 12:39 PM JamesKeene General Physics 0 Jul 28th 2010 08:43 AM Apprentice123 Kinematics and Dynamics 0 Apr 20th 2009 05:14 PM 