Let's do a practical example. I drop a stone to the earth...

Let's say the stone is a sizable one. 1 KG (2 pounds) in weight. From a height of 1 meter.
Relevant formulae:
F(g)=m*g
F=m*a -> a=F/m : 2nd law of Newton

Now, since earth attracts the stone, earth puts a "pull", a force onto the stone. The same pull the stone puts onto the earth. This means the gravitational constant of the earth can be used for F:
F(g)=m*g
g is a constant, depends a bit where on earth you are. Varies from 9.8 to 9.91 if I'm not misstaken... For me it is 9.81 m/(s*s)
m=1KG... This would result in:
F(g)=9.81

So back to the earlier formula.
a=F/m
9.81=m(earth)*a

Earth's mass is 5,976*10E24 KG

Earth would accellerate with 1,641*10E-24 m/s*s towards the stone. That is 0.0000000000000000000000001641 m/s*s, a small figure indeed.