# 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. – Parrotmaster Dec 12 '19 at 15:59
• Do you know the formula for determining the gravitational force between two masses? – Eric Lippert 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. – Eric Lippert 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. – John Canon 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. – Ross Millikan 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. – Andrew Leach Dec 14 '19 at 12:06