What is an example Kepler's Second Law practice problem?

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 =]