# Moon's unusual gravity

We know that the Moon's gravity is about one-sixth that of Earth. Then I recently read that the mass of the Moon is about one-eightieth the mass of Earth. Since gravity depends on the mass of the 2 objects, shouldn't the Moon's gravity be one-eightieth the gravity of Earth? Is the Moon extra dense?

• The moon being made from other stuff than the Earth probably factors into this. A lot fewer liquids there. Dec 12 '19 at 15:59
• Do you know the formula for determining the gravitational force between two masses? Dec 12 '19 at 18:08
• @Alchemista: If the Moon had its current mass but was the size of the Earth -- and was therefore enormously less dense -- then it would have 1/80th of the surface gravity. But the Moon is not the size of the Earth, and that matters. I asked if the original poster knew the formula as a nudge towards thinking about the fact that the distance between centers of the masses is just as important a factor as the masses. Dec 13 '19 at 13:43
• I get it. If the Moon had its same mass, but with a radius the same as Earth, then the surface gravity would be 1/80th. The distances would correspond, and mass would be the only variable. The Moon's smaller radius from center to surface gives us the 1/6 instead of 1/80th. No wonder it's so hard to split an atom. Dec 14 '19 at 2:26
• @JohnCanon: The difficulty of splitting an atom is not due to the gravitational force holding it together, but that belongs on physics.se Gravity is truly negligible at that scale. Dec 14 '19 at 4:43

As you said, the mass of the Moon is 1.2 percent that of the Earth. Now, if you mean the gravitational acceleration at the surface, it is calculated like this $$G\frac{M}{R^2}$$, where $$M$$ is the mass, and $$R$$ is the radius of the celestial body. The moon's mass is a hundred times smaller, but the radius is four times smaller, meaning its surface gravity will be $$100/16 \approx 6$$ times smaller. Considering the factor is mass over radius to the power of two here, density alone does not help you determine the ratio of surface accelerations.
• @corsiKa That's easy. I give you \$10 for a bit of the Moon and you give me \$50 for the same amount of Io. Dec 14 '19 at 12:06