Work done by gravity on a ball & the ball on earth I have studied today that  if a ball was to fall a certain height, then the work done by gravity on the ball would equal the work done by the ball's equal and opposite gravitational pull. 
By $W=Fd$, this means that the earth must travel the same distance as the ball.
So if the ball falls 300 meters, the earth would eventually rise 300 meters in the direction of the ball's gravitational pull? 
Can you please give me your input on this particular matter because I don't understand it well ?
 A: The force from the earth on the ball and the force from the ball on the earth are in fact opposite and equal but the amount of work done on each is not the same. The earth is much more massive than the ball so, for an equivalent force, it is going to accelerate much more slowly and move a much shorter distance during the time the ball is falling than the ball itself will move. The forces are the same for the ball and the earth but this distance traveled is much smaller for the earth so the total work done on the earth is much less.
A: Just to expand on vaaaaaal's answer, let's simplify this very slightly by assuming that the ball falls at it's average fall velocity $v$ for the whole height $h$ over a time $t$.  Obviously, $v = \frac{h}{t}$.  Then, we know that total momentum is conserved, so the Earth must fall up with speed $v_{e} = \frac{m}{M_{e}}v$.  Thus, over the whole time of falling, the earth moves a distance $d = v_{e}t = \left(\frac{m}{M_{e}}v\right)\frac{h}{v} = h \left(\frac{m}{M_{e}}\right)$.  Since the forces are the same, the work done on the earth is smaller by a factor of the ratio of the masses of the ball and the Earth.
A: Consider a simpler case, the earth is only twice as massive as the ball, and suppose they are 3 meters apart.
(And suppose they are tiny, so their radius does not matter.)
They fall together and meet at their common center of mass, which is 1/3 of the way from earth to the ball.
So, the work done on the earth is F * 1 meter, and the work done on the ball is F * 2 meters.
So the work is not the same.
