When a stone falls from a certain height above the Earth's surface, it accelerates towards the center of Earth under the influence of Earth's gravity. According to Newton's 3rd law, the stone also exerts an equal force on the Earth, but towards itself. So the Earth accelerates towards the stone, even though it is very insignificant.
What type of force does the stone exert on the Earth?
I know it is a reaction force, but what type of force is it?
Is it also gravity? Or am I misunderstanding this?
Best Answer
So you know that the gravitational force between two objects can be described as $$\vec{F_{12}}=G*\frac{m_1*m_2}{{r}_{12}^2}\hat{r_{12}}$$
and according to Newton's 3rd law the force that the rock exerts on Earth is the same and opposite as the Earth's on rock.
They attract one another (not in a weird way mind you). However, because the mass of the rock is negligible compared to the Earth's, the Earth doesn't really budge ($\vec{F}=m\vec{a}$). If it did budge, well then anyone dropping a rock anywhere would cause some major catastrophes.
Furthermore, you are attracted to the furthest star known to man, but because the $\vec{r}$ term is so large (even more so squared), you barely feel the connection (which makes the star sad).
This is in a nutshell answer. If you want to read up, I recommend University Physics or Berkeley's Mechanics (http://www.astrosen.unam.mx/~posgrado/libros/1_Mechanics_Kittel_BPC.pdf)