# Gravitational potential energy of an object on Earth compared to Moon

The gravitational potential energy of a $$200\,\text{kg}$$ satellite $$3000\,\text{km}$$ above the surface of the Earth is $$U=-\frac{GMm}{r}=-8.5 \times 10^9\,\text{J}$$.

Would the gravitational potential energy of this satellite be greater or less if it was at an altitude of $$3000 \,\text{km}$$ above the Moon? Explain.

The answers say that the gravitational force of attraction on the Moon is less than that of on Earth, so the potential energy is less.

I am struggling to understand this: $$g=\frac{GM}{r^2}$$ is a smaller number on the Moon.

If we write $$U=-\frac{GMm}{r}=-\frac{GM}{r^2}\times mr$$

From this we see the quantity $$\frac{GM}{r^2}$$ is a smaller number on the Moon, and $$r$$ is a smaller number on the Moon (radius of Moon is less than that of Earth). So, $$U$$ should be a smaller negative number, meaning $$U$$ is greater on Moon than that on Earth.

This goes against the answers and against the intuition that indeed if force of gravity is less than you have less gravitational potential energy.

How can we mathematically explain this situation?

• Yeah, the question is just poorly written. Notice also that "3000 km above the surface of the Earth" does not correspond to the same $r$ as "3000 km above the Moon", since $r$ is measured from the center of the object, not the surface. Nov 25, 2022 at 14:22
• @MichaelSeifert the quoted value for U (in the 3000km above Earth condition) does assume $r=r_\oplus+3000km$
There is an ambiguity here in what is meant by "greater potential energy." This is because the point where potential energy is zero can be set anywhere without changing any physics. When close to the Earth's surface, the ground is a convenient place to set $$U=0,$$ such as $$U = mgh.$$ For orbits, setting $$U = 0$$ at an infinite distance away is convenient because the form of the equation is simple: $$U = GMm/r$$ instead of $$U = GMm(1/r - 1/r_0)$$ where $$r_0$$ is the radius of the relevant star or planet.