Orbital Mechanics: The Mathematics of Elliptical Orbits

Conserve angular momentum: R^2 Omega stays constant as a body moves. Angular momentum L equals M R^2 Omega, linking radius and tangential velocity to orbital motion around a fixed point.

The lecture shows that Earth's orbit conserves angular momentum and velocity scales as one over r squared, and that total energy equals kinetic plus gravitational potential energy, remaining constant.

Derive the equations of motion from the Lagrangian, defined as kinetic energy minus potential energy, using r and theta with velocities r_dot and theta_dot, and show consistency with Newtonian results.

Derive the energy and angular momentum relations and rewrite the motion equation for a gravitational two-body system. Set up the integral relating theta and radius to show an ellipse.

Analyze integrating the equation of motion, part 2, to reveal a closed elliptical trajectory with perihelion and aphelion defined by energy and angular momentum.

Derives the integral for the equation of motion in elliptical orbits, linking perihelion and aphelion through theta and radius, detailing upper and lower trajectory branches via arc cosine.

Derive the trajectory of a planet in polar coordinates and relate the ellipse parameters P and E to aphelion, perihelion, and the sun’s focus.

By placing the origin at the focus, prove the trajectory is an ellipse and, with the center shift Delta, derive the ellipse's semi-axes in terms of p and e.

Proves that the sun lies at one focus of the ellipse by showing the sum of distances to the two foci is constant, linking to Kepler's first law.

Recognize that the radius vector sweeps out equal areas in equal times, so r^2 theta_dot is constant, implying faster motion near perihelion and slower near aphelion.

Animate an elliptical orbit to show the area spanned by the radius vector stays constant. The motion speeds up near perihelion and slows near aphelion with acceleration components toward sun.

Derives the integral in the orbital period formula by differentiating with respect to lambda and applying Euler’s formula to obtain pi lambda divided by sqrt(1 minus lambda squared).