When is an orbital body going it’s slowest? When it is closest to the object it orbits, or farthest?
Kepler’s Second Law states that the line a satellite makes with what it’s orbits sweeps out equal area within equal intervals of time.
Now we know from assumptions that ##(r)/(r’) < 1## as ##r’## (the radius farthest away) is just that: the farthest away.
If we consider small, equal intervals of time ( ##Delta t < < T##, where ##T## is the period of orbit) then we can approximate the areas as triangles such that:
##A = (r Delta theta)/2## and ##A’ = (r’ Delta theta’)/2##
Where ##Delta theta## is the small angular distance that will be swept through during ##Delta t##. We also assume that ##r## and ##r’## do not change much during these intervals.
Since we are considering equal intervals of time ##Delta t## it follows from Kepler’s Law that
##Delta t = A = A’## or ## A/A’ = (r Delta theta)/(r’ Delta theta’) = 1 ##
Since ##r/(r’) < 1## we know that ## (Delta theta)/(Delta theta’) > 1##. As well, angular velocity is defined as ##omega = (Delta theta)/(Delta t)##.
Hence, it must follow that ##omega/omega’ > 1## or ##omega > omega’## !
Without even using any numbers we have shown that the angular velocity of a body is greater at the point during it’s orbit where it is closest to the planet/star it is orbiting!
Hope this helps =]
