# Is the net work done on earth-object system equal to the change in potential energy of the system?

My book says:

Let us derive an expression for the potential energy associated with an object at a given location above the surface of earth. Consider an external agent lifting an object of mass $m$ from an initial height $yi$ above the ground to a final height $yf$. We assume the lifting is done slowly with no acceleration so the applied force from the agent can be modelled as being equal to in magnitude to the gravitational force on the object. The work done by the external agent on the earth-object system as the object undergoes this upward displacement is given by : $Wnet = \vec{Fapp}\cdot \Delta \vec{r} = mg\hat{j}\cdot[(yf - yi)\hat{j}] = mgyf - mgyi$

Where this result is the net work done on the system because the applied force is only force by the environment. The equation represents a transfer of energy into the system and the energy appears in a different form called potential energy. Therefore we call the quantity $mgy$ as the gravitational potential energy $Ug$. $Ug \equiv mgy$.

Therefore $Wnet = \Delta Ug$

What I don't understand is , why is the net work done only by the external force? Why didn't the author include the work done due to gravity? Isn't the net work done on a system the sum of all the works done by both external and internal forces?

## 1 Answer

If the system is the object and the Earth the work is done on the system by two external forces equal in magnitude to $mg$ and opposite in direction acting on the Earth and the object.
As the separation between the object and the Earth the potential energy of the object-Earth system increases.

The centre of mass of the system does not change and because the mass of the Earth is so much greater than the mass of the object the distance moved by the force acting on the Earth can be neglected compared with the distance moved by the force acting on the object $h$.
So the work done by the external forces is $mgh$ and this is the increase in the potential energy of the object-Earth system.

The two external forces have two equal and opposite internal forces (gravitational attraction) which are also acting on the object and the Earth.
The displacement of each of those two internal forces is opposite to the direction of each of those two forces.
So the work done by those two internal forces is negative and essentially equal to $mgh$ since the Earth does not move very much compared the object.
We interpret negative work done by the system as an input of energy to the system and call it potential energy.

• Yeah I read that answer on some other related post. The work is done by one external force, gravitational force is internal to the system. – Noah Cygnus Sep 17 '17 at 14:10
• @NoahCygnus if you think about it, to separate the object and the Earth and not move the centre of mass of the system requires two external forces. One on the object and the other on the Earth. – Farcher Sep 17 '17 at 14:13
• If we are considering earth and the object as a system, the gravitational force is internal to that system, as earth exerts $$\vec{F_{g}}$$ on the object and the object exerts $$\vec{-F_{g}}$$ on the earth. We are considering earth and the object as a system, the forces they exert on each other are deemed internal. If I am wrong, please correct me. – Noah Cygnus Sep 17 '17 at 14:25
• @NoahCygnus Yes, the gravitational forces exerted on the object and the Earth in the object-Earth system are internal forces. – Farcher Sep 17 '17 at 14:28
• Then why are you stating that there are two external forces on the system? – Noah Cygnus Sep 17 '17 at 14:40