Physics by Rohit: Analyzing orbit through differential equation (courtesy: wikipedia)

Students,

We discussed in class on how to express equation of conic sections (circle, ellipse, parabola, hyperbola) in polar coordinates. Now we can write differential equation for motion of planet around sun. Solving this differential equation we get equation of general conic section. Depending on value of constants we will end up with equation of circle/ellipse/parabola/hyperbola.

Note: All these derivations are only for reference and are not part of board or IIT-JEE syllabus. But since some of you have frequently raised queries on how we arrive at trajectory so this discussion is just to answer your query.

To analyze the motion of a body moving under the influence of a force which is always directed towards a fixed point, it is convenient to use polar coordinates with the origin coinciding with the center of force. In such coordinates the radial and transverse components of the acceleration are, respectively:

$a_r = \ddot{r}-r\dot{\theta }^2 \,$

and

$a_\theta =\frac{1}{r}\frac{d}{dt}\left( r^2 \dot \theta \right).$

Since the force is entirely radial, and since acceleration is proportional to force, it follows that the transverse acceleration is zero. As a result,

$a_\theta = 0. \,$

After integrating, we have

$r^2 \dot \theta = \text{constant} \,$

which is actually the theoretical proof of Kepler's second law (A line joining a planet and the Sun sweeps out equal areas during equal intervals of time). The constant of integration, h, is the angular momentum per unit mass. It then follows that

$\dot\theta =\frac{h}{r^2} = hu^2 \,$

where we have introduced the auxiliary variable

$u = { 1 \over r }.$

The radial force ƒ(r) per unit mass is the radial acceleration a_r defined above. Solving the above differential equation with respect to time^[9](See also Binet equation) yields:

$\frac{d^2u}{d\theta^2} + u = -\frac{f(1 / u)}{h^2u^2}.$

In the case of gravity, Newton's law of universal gravitation states that the force is proportional to the inverse square of the distance:

$f(1/u) = a_r = { -GM \over r^2 } = -GM u^2$

where G is the constant of universal gravitation, m is the mass of the orbiting body (planet) - note that m is absent from the equation since it cancels out, and M is the mass of the central body (the Sun). Substituting into the prior equation, we have

$\frac{d^2u}{d\theta^2} + u = \frac{ GM }{h^2}.$

So for the gravitational force — or, more generally, for any inverse square force law — the right hand side of the equation becomes a constant and the equation is seen to be the harmonic equation (up to a shift of origin of the dependent variable). The solution is:

$u(\theta) = \frac{ GM }{h^2} + A \cos(\theta-\theta_0)$

where A and θ₀ are arbitrary constants.
The equation of the orbit described by the particle is thus:

$r = \frac{1}{u} = \frac{ h^2 / GM }{1 + e \cos (\theta - \theta_0)},$

where e is:

$e \equiv \frac{h^2A}{G M}.$

In general, this can be recognized as the equation of a conic section in polar coordinates (r, θ). We can make a further connection with the classic description of conic section with:

$\frac{h^2}{GM} = a(1-e^2).$

If parameter e is smaller than one, e is the eccentricity and a the semi-major axis of an ellipse.
(courtesy: wikipedia)

Physics by Rohit

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Thursday, February 28, 2013

Analyzing orbit through differential equation (courtesy: wikipedia)

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