0
$\begingroup$

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?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

You are correct. Smaller-magnitude negative numbers are greater than larger magnitude negative numbers. There's nothing else to explain, except to suggest that you triple check your textbook to make sure that your representation of the question in this post is an accurate representation of the text before concluding that the text is incorrect.

See both answers here: What is the significance of negative potential energy for an explanation of why we use the convention of negative-signed potential energy and what it means.

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Michael Seifert
    Commented Nov 25, 2022 at 14:22
  • $\begingroup$ @MichaelSeifert the quoted value for U (in the 3000km above Earth condition) does assume $r=r_\oplus+3000km$ $\endgroup$
    – g s
    Commented Nov 25, 2022 at 22:12
0
$\begingroup$

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.

For this question, I would interpret "greater potential energy" in the following way: If you were to drop one 200-kg object from a height of 3000 km above the surfaces of the Earth and do the same on the Moon, which object would have more kinetic energy on impact?

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.