<p>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 <span class="math-container">$G\frac{M}{R^2}$</span>, where <span class="math-container">$M$</span> is the mass, and <span class="math-container">$R$</span> 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 <span class="math-container">$100/16 \approx 6$</span> 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.<br> On a side note, the moon's density is around 60 per cent that of the Earth.</p>

CC BY-SA 4.0
2019-12-12T16:36:24.753Z
... and that 60% density value should derive directly from knowledge of the radius and gravitational force at that radius  :-)
Carl Witthoft
CC BY-SA 4.0
2019-12-12T22:00:36.357Z
Low density because Theia drops [most of its iron core onto proto-Earth](https://en.wikipedia.org/wiki/Theia_(planet)) at collision time. Still the 2nd densest moon in the solar system; [only Io is denser](https://www.space.com/18135-how-big-is-the-moon.html)
David Tonhofer
CC BY-SA 4.0
2019-12-14T05:26:28.430Z
@DavidTonhofer that must mean Io is worth more, since it has a higher metal content. How do we go about trading?
corsiKa
CC BY-SA 4.0
2019-12-14T12:06:47.463Z
@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
CC BY-SA 4.0
2019-12-12T07:48:11.517Z
2019-12-12T07:48:11.517Z
Tosic
<p>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? </p>

CC BY-SA 4.0
2019-12-12T15:59:33.360Z
The moon being made from other stuff than the Earth probably factors into this. A lot fewer liquids there.
Parrotmaster
CC BY-SA 4.0
2019-12-12T18:08:15.773Z
Do you know the formula for determining the gravitational force between two masses?
Eric Lippert
CC BY-SA 4.0
2019-12-13T13:43:26.497Z
@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
CC BY-SA 4.0
2019-12-14T02:26:36.207Z
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
CC BY-SA 4.0
2019-12-14T04:43:51.053Z
@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
CC BY-SA 4.0
2019-12-14T17:19:42.157Z
@John I find it more helpful to think if it had the same density, it would have the 1/ (the cube root of 80) the surface gravity. That's a bit less than a quarter.
richardb
<the-moon><gravity><mass>
Moon's unusual gravity
John Canon
# Moon's unusual gravity

(Asked by: username_0 on 2019-12-12)

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?

**Tags:** the-moon, gravity, mass

> **Comments:**
> **username_1**: The moon being made from other stuff than the Earth probably factors into this. A lot fewer liquids there.  
> **username_2**: Do you know the formula for determining the gravitational force between two masses?  
> **username_2**: @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.  
> **username_0**: 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.  
> **username_3**: @username_0:  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.  
> **username_4**: @username_0 I find it more helpful to think if it had the same density, it would have the 1/ (the cube root of 80) the surface gravity. That's a bit less than a quarter.  

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## 1 answers

### Answer by username_5 on 2019-12-12. Score: 56 | Accepted answer

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.  
 On a side note, the moon's density is around 60 per cent that of the Earth.

> **Comments:**
> **username_6**: ... and that 60% density value should derive directly from knowledge of the radius and gravitational force at that radius  :-)  
> **username_7**: Low density because Theia drops [most of its iron core onto proto-Earth](https://en.wikipedia.org/wiki/Theia_(planet)) at collision time. Still the 2nd densest moon in the solar system; [only Io is denser](https://www.space.com/18135-how-big-is-the-moon.html)  
> **username_8**: @username_7 that must mean Io is worth more, since it has a higher metal content. How do we go about trading?  
> **username_9**: @username_8 That's easy. I give you \$10 for a bit of the Moon and you give me \$50 for the same amount of Io.  

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