Osculating Circles and Trajectories Just Kidding

Osculating Circles and Trajectories r  v M m Mechanical Energy

Circular Orbit M m v F How do we get the mass, M, of the gravitating object? Apply Newton’s Second Law to the mass, m  M C r

General Orbits, Conic Sections circle,  =0 ellipse, 0<  <1 parabola,  =1 hyperbola,  1  r In polar coordinates

General Orbits, Conic Sections, polar unit vector notation In polar coordinates In terms of unit vector notation we have: Differentiating the position vector with respect to time we have the velocity: or, and, so,

General Orbits, Conic Sections, polar unit vector notation In polar coordinates Differentiating the velocity vector with respect to time we have the acceleration: or, so,

General Orbits, Conic Sections, polar unit vector notation In polar coordinates Recall that And note that Then we get for the acceleration, Combining terms in unit vectors,

General Orbits, Conic Sections, normal and tangential unit vector notation In polar coordinates In terms of unit vector notation we have: Differentiating the velocity vector with respect to time we have the acceleration: or, and, so, dd  ds Note that and so, C

General Orbits, Conic Sections In polar coordinates We now have the acceleration in normal and tangential unit vector notation Compare this with the acceleration in polar coordinate unit vector notation

General Orbits, Conic Sections, continued When  = 0  we have v ellipse, 0<  <1 circle,  =0 parabola,  =1 hyperbola,  1

General Orbits, Conic Sections, continued part 2 When  = 0  we have v At the minimum distance from the massive body the acceleration is purely radial or normal depending upon your point of view. Let’s write the acceleration at  for each basis. becomes and becomes

General Orbits, Conic Sections, continued part 3 When  = 0  we have v In the r,  basis we have: In the t,n  basis we have: Because we get

The Osculating Circle at closest approach v C We have and because we get Let’s obtain

Polar coordinates Let’s re-write this as And take two successive derivatives with respect to time to obtain Now let  = 0 to get which now gives or

The Osculating Circle at closest approach v C We now know so and or

The Osculating Circle at closest approach continued v C Compare this with the polar form of a conic section Clearly The semi-latus rectum for conic sections is nothing more than the radius of the osculating circle at the distance of closest approach.

Polar Coordinates The equation for a conic section could now be written as v C m M Apply Newton’s second law to m Solving for M we get or

General Orbits, Conic Sections When  = 0  we have v ellipse, 0<  <1 circle,  =0 parabola,  =1 hyperbola,  1 M CcCc CeCe CpCp ChCh