Ok so the total energy of a body following a given trajectory around a much larger body (eg. earth and sun), is described by :(adsbygoogle = window.adsbygoogle || []).push({});

E(total) = (1/2)mv^2 + U (where U = grav. potential energy)

E(total) = (1/2)mv^2 - (GMm)/r

(1/2)mv^2 can then be expanded to give :

E(total) = (1/2)m( (dr/dt)^2 + (r^2)(w^2) ) - (GMm)/r

(where w = omega = angular velocity)

then using angular momentum L = m(r^2)w, we obtain

E(total) = (1/2)m((dr/dt)^2) + L^2/2mr^2 - (GMm)/r

where (L^2)/2mr^2 - (GMm)/r is the 'effective potential' , as referred to by my lecturer.

My main problem is the expansion of v^2 into (dr/dt)^2 + (r^2)(w^2).

My thoughts :

dr/dt represents the rate of change of *distance* between the orbiting body and the large body, or in other words the velocity in the direction of the force, aka the radial velocity.

rw is obviously the tangential velocity.

What I really need some help on is how each term in (dr/dt)^2 + (r^2)(w^2) is derived from v^2.

v^2 = (dr/dt)^2 + (r^2)(w^2) looks to me like a declaration of Pythagoras' law, where v is the resultant vector of the two perpendicular vectors (dr/dt) and (r)(w).

For a circular orbit, dr/dt = 0 so the expression is simply v^2 = r^2w^2.

For an elliptical orbit, r is constantly changing so dr/dt needs to be accounted for. The vectors dr/dt and rw, however, are only perpendicular at the points of closest and farthest approach, and so my idea about a Pythagorean triangle seems to go out of the window for any other points on the trajectory.

Some help with this derivation would be invaluable.

As a final request, can someone give some meaning to the term 'effective potential' ? Even my 600 page physics dictionary does not offer a definition

Many thanks!

Trev

Edit : Here's a link to the pdf file written by my lecturer with the relevant material in it - see section 3.3

http://www.kcl.ac.uk/kis/schools/phys_eng/physics/courses/CourseList/CP1400%20(0506).pdf [Broken]

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# Homework Help: Kepler orbital trajectories and 'effective potential'

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