# Gravitational potential energy defined as the work done on a mass

Our physics sir made us write that gravitational potential energy is the work done in bringing a mass from infinity to a point without acceleration, but I am confused because if acceleration is $$0$$ it means that the external force is 0, and hence net work done should always be zero. Then how can potential energy be anything other than zero?

...if acceleration is $$0$$ it means that the external force is $$0$$...

No. If acceleration is $$0$$ then net force is $$0$$.

...and hence net work done should always be zero.

Yes, in this scenario the net work is in fact $$0$$, since the net force is $$0$$. However, this means that there are (at least) two forces acting on the object in question: gravity $$F_g$$ and an external force $$F_e$$. These two forces must be equal and opposite.

This is a standard treatment/explanation of potential energy. We move the body with a constant velocity, as $$F_g=-F_e$$, and so the work done by the external force $$W_e$$ is equal to the negative of the work done by gravity $$W_g$$. By definition, the work done by gravity is also equal to the negative change in potential energy $$\Delta U$$. Finally, if we start "at infinity" where $$U(\infty)=0$$ and end at position $$x$$, then $$\Delta U=U(x)-U(\infty)=U(x)$$,

Therefore, we have $$W_e=-W_g=-(-\Delta U)=\Delta U=U(x)$$ So then we have what you stated at the beginning:

gravitational potential energy is the work done in bringing a unit mass from infinity to a point without acceleration.

• Regarding your edit on the question : gravitational potential is the gravitational potential energy per unit mass.
– user249968
Jan 2, 2020 at 13:42
• @JohanLiebert Yes, it is. The OP seemed to be mixing up both, so it could have gone either way. I chose to go the potential energy route. Jan 2, 2020 at 13:53