# Newton's 3rd law of motion, related to Earth's gravity

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?

• Yes, it's also gravitational force. Action and reaction forces must be of the same type. – PeaBrane Mar 18 '18 at 9:20

## 4 Answers

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)

• Should add a vector to the right side of your equation, e.g. a unit vector $\hat{r}_{12}$. – user1583209 Mar 18 '18 at 9:32

Yes, the force is also gravity. All action reaction pairs are of the same type.

The stone, by virtue of its mass, has a gravitational force that it exerts on every other massive object in the universe, including, of course, the earth!

• Please accept the answer if it satisfactorily answers your question. – user1936752 Mar 18 '18 at 9:10

From a fundamental point of view there is nothing special about earth (or the heavier object) compared to the stone (or lighter object) here. Gravity is a force that acts between masses. You can also see this from the law of gravity:

$$F=G\frac{mM}{r^2}$$

which is completely symmetrical between the stone ($m$) and earth ($M$). Or said differently, you could have formulated your question as follows without any problem.

When earth falls from a certain height above a stone's surface, it accelerates towards the center of the stone under the influence of the stone's gravity. According to Newton's 3rd law, the earth also exerts an equal force on the stone but towards itself. So the stone accelerates towards the earth even though it is very insignificant. What type of force does the earth exert on the stone?

The reason that we usually think of the stone being affected by earth's gravity and not the other way around, even though both forces are equal in magnitude, is basically Newton's second law: $F=ma$. The force acting on the light stone (small mass) will cause a large acceleration, while the same force acting on the heavy earth (large mass) will cause a tiny acceleration.

Consider the Earth and the falling stone as one system with no external forces acting on the system.

There are two internal forces which are the force on the Earth due to the gravitational attraction of the stone and the force on the stone due to the gravitational attraction of the Earth.
These two forces are a Newton third law action and reaction pair of the same type (gravitational) and must be equal in magnitude but opposite in direction.
Because there are no external forces the centre of mass of the Earth & stone system does not accelerate and so if the stone is accelerating towards the Earth the Earth must be accelerating towards the stone.

If the mass of the stone is $m$ then the force on the stone due to the Earth is $mg$ where $g$ is the acceleration of free fall of the stone towards the Earth.
If we assume $g \approx 10\, \rm m\,s^{-2}$ and $m \approx 0.1 \,\rm kg$ then the force on the stone due to the Earth is $10 \times 0.1 = 1 \, \rm N$.

This is the force on the Earth due to the stone and if the mass of the Earth is $6\times 10^{24} \approx 10^{25} \,\rm kg$ then the acceleration of the Earth is $\frac {1}{10^{25}} = 10 ^ {-25} \,\rm m \, s^{-2}$ which is such an extremely small value that it is normally neglected.