Just curious, does anyone know what are the equations for the displacement-time graph of a Keplerian orbit? The horizontal displacement-time graph should be somewhat sinusoidal, the vertical one cosinusoidal, or vice versa.
A Keplerian orbit is an ellipse, the equation for which is \(r = \frac{r_0}{1- \epsilon \cos(\theta)}\), where \(r_0\) is the radius of the equivalent circular orbit and \(\epsilon\) is the eccentricity. This is the equation for a single planet orbiting a star in an otherwise empty universe, and clearly this is not what we have in reality. Deviations from this orbit will be caused by gravitational interactions between the planet and other massive bodies close by. These interactions make the equations of motion impossible to solve analytically, but the deviations are small for solar systems like ours.
sorry dude, i wanted a cartesian form, not just the shape of the orbit but also the displacement-time graph.
Despite yet another attempt to troll, Tach is right. If you do a google search for "equation of an ellipse in cartesian coordinates" you will get results where the constants are the lengths of the semimajor and semiminor axes, rather than the ones I defined above. They are certainly related to each other in a simple way (that I don't know off the top of my head.)
Oh lol. i know what the equation for an ellipse is. Apparently Tach has needs to get new glasses, because I said I was looking for the displacement-TIME graph. For instance, in uniform circular motion the x-displacement is identical to SHM, and its a sine wave. obviously for an eccentric orbit its similar, but more complicated.
being more specific, if i plot the x-displacement against the y-displacement I should get an ellipse with one foci at the origin. for an arbitrary ellipse with signicant eccentricity where the planet obeys the law of equal areas, what is the equation for its displacement-time graph?
Although he did not yet understand the physics, Kepler certainly opened the door for Newton's discovery of the law of gravitation by his remarkable finding that the laws of planetary motion had underlying mathematical explanations. He was actually pursuing an orderly explanation for the apparent fluctuations in the orbit of Mars which, under the geocentric theory, were not only complicated, but they were irregular. He recalls the day he made his breakthrough: And if you want the exact moment in time, it was conceived mentally on 8th March in this year one thousand six hundred and eighteen, but submitted to calculation in an unlucky way, and therefore rejected as false, and finally returning on the 15th of May and adopting a new line of attack, stormed the darkness of my mind. I love that last phrase. It would have probably be sacrilege in 1618 to call it an epiphany, but here it is: So strong was the support from the combination of my labour of seventeen years on the observations of Brahe and the present study, which conspired together, that at first I believed I was dreaming, and assuming my conclusion among my basic premises. But it is absolutely certain and exact that the proportion between the periodic times of any two planets is precisely the sesquialterate proportion of their mean distances. (sesquialterate: Once and a half times as great as another; having the ratio of one and a half to one.) Of course Latin and Greek were second and third languages to scholars back then. Kepler's second law is best remembered as The square of the periodic times are to each other as the cubes of the mean distances. Yes, if it were a circle you would have a sine and cosine. Scale one and you get an ellipse.
cool backstory on kepler's life. If only it were that simple. But a planet speeds up as it approaches the sun, so unfortunately it is more complex.
\(0 \le e \lt 1 \\ x = -ae + a \; \cos \, \theta y = a \sqrt{1 - e^2} \; \sin \, \theta t = \sqrt{\frac{a ^3}{ G (M + m) }} ( \theta - e \; \sin \, \theta )\) http://en.wikipedia.org/wiki/Kepler_orbit#math_27 \( \frac{d x}{d \theta} = - a \; \sin \, \theta \\ \frac{d y}{d \theta} = a \sqrt{1 - e^2} \; \cos \, \theta \\ \frac{d t}{d \theta} = \sqrt{\frac{a ^3}{ G (M + m) }} ( 1 - e \; \cos \, \theta ) \\ \frac{d x}{d t} = - \sqrt{\frac{ G (M + m) }{a}} \frac{\sin \, \theta}{1 - e \; \cos \, \theta} \\ \frac{d y}{d t} = \sqrt{\frac{ G (M + m) }{a}} \sqrt{1 - e^2} \frac{\cos \, \theta}{1 - e \; \cos \, \theta} \\ x \frac{d y}{d t} - y \frac{d x}{d t} = \sqrt{a G (M + m)(1 - e^2)} \)
Also, \( x^2 + y^2 = a^2 ( 1 - e \, \cos \theta )^2 \left( \frac{dx}{dt} \right) ^2 + \left( \frac{dy}{dt} \right) ^2 = \frac{G (M+m)}{a} \times \frac{1 + e \, \cos \theta }{ 1 - e \, \cos \theta } x \frac{dx}{dt} + y \frac{dy}{dt} = \sqrt{a G (M+m)} \, e \, \sin\theta \frac{1}{G(M+m)} \left( \left( \left( \frac{dx}{dt} \right) ^2 + \left( \frac{dy}{dt} \right) ^2 \right) x - \left( x \frac{dx}{dt} + y \frac{dy}{dt} \right) \frac{dx}{dt} \right) - \frac{1}{\sqrt{x^2 + y^2}} x = e \frac{1}{G(M+m)} \left( \left( \left( \frac{dx}{dt} \right) ^2 + \left( \frac{dy}{dt} \right) ^2 \right) y - \left( x \frac{dx}{dt} + y \frac{dy}{dt} \right) \frac{dy}{dt} \right) - \frac{1}{\sqrt{x^2 + y^2}} y = 0 \) http://en.wikipedia.org/wiki/Eccentricity_vector
Using geometric units w = 2pi/T w^2_phi = (dphi/dt)^2 = M/r^3 [rate of angular velocity for Newton and Kepler] Kepler's Third Law M = w^2_phi * r^3 Or w^2_phi = M/r^3 This is the rate of angular velocity and rate of radial oscillation for Newton. For Einstein orbits these rates are slightly different leading to the natural precession of all Einstein orbits. You should be able build a graph from (dphi/dt)^2 = M/r^3 I think.
now we can "easily" solve this D.E. did newton derive these from Kepler's 1st & 2nd laws, and used this to prove his inverse square law?
It's not only for a circular orbit when the definition of r includes parameters to identify a specific orbit. I'm sure what I wrote down falls short of the best way to do this. rpenner provided the correction for r_elliptical orbit.
I always appreciate what you write down for us. If I wanted could I rename the left side of the last line r_elliptical orbit?
Apparently I'm confused. Not surprising. In Schwarzschild coordinates L/m = r^2 dphi/dTau = r(dt_shell/dTau)(r dphi/dt_shell) = r(gamma_shell)(dx_shell/dt_shell) = r(gamma_shell)v_shell That's what I'm familiar with. I'll see if I can learn something.
