# Why is the Moon's gravity so high compared to its mass?

According to Wikipedia, the Moon's mass is about 1.23% of Earth's, but its gravity is 0.1654g or 16.5%. If gravity is proportional to mass, why isn't the Moon's gravity 1.23% of Earth's?

EDIT: Presumably, something like this applies:

If the top circle is the Moon and the bottom is the Earth, the particles closer to you apply more gravitational pull. Therefore, whilst the further-away particles on Earth cause the gravity to be six times higher, they don't cause nearly as much gravitational pull as those close to you. With thMoonon, all of its particles are close enough that they apply quite a lot of gravitational pull.

• "but its gravity is 0.1654g" This doesn't actually mean anything. Note that the Wikipedia article you quote says "Its surface gravity is about one-sixth of Earth's". Why did you omit the "surface", and can you tell why the omission matters? Commented Jul 25, 2022 at 16:15
• if you recast the problem in terms of density then you get $g=\phi r\rho$ where $\phi$ is a constant, $r$ is the distance from the center and $\rho$ is the density of the object. In other words for what you are saying to be true, the density of the earth and moon should be same, which is not the case. Commented Jul 25, 2022 at 18:59

Because what it lacks in mass it gains in radius.

Let $$g$$ be the acceleration due to gravity at the surface of a spherical object with radius $$R$$, mass $$M$$, and surface area $$S=4\pi R^2$$. Combining Newton’s second law and gravitational force we get

$$mg=G\frac{Mm}{R^2} \tag{1}$$ Meaning that surface gravitational acceleration ratio of object 1 by object 2 is $$\frac{g_{1}}{g_{2}}=\frac{M_{1}}{M_{2}} \times \frac{R_{2}^2}{ R_{1}^2} =\frac{M_{1}}{M_{2}} \times \frac{4\pi R_{2}^2}{4 \pi R_{1}^2}=\frac{M_{1}}{M_{2}} \times \frac{S_{2}}{S_{1}} \tag{2}$$

The acceleration ratio is proportional to the ratio of masses and inversely proportional to the surface area’s ratio. In case object 1 being moon and object 2 earth, the ratio is

$$g_{moon} \approx \frac{13}{82} g_{earth} \approx 0.16 \times g_{earth} \tag{3}$$

• Sphere surface area is $4 \pi R^2$ instead, it can be seen by differentiating sphere volume $\frac {\partial}{\partial R} 4/3 \pi R^3$. Your second equation doesn't stand to dimensional analysis,- $g_1/g_2$ ratio must be unit-less, but it have units $m^3 kg^{−1} s^{−2}$ due to $G$ constant. I think $G$ simply factors-out in division. Lastly if we could define mass compression factor $\Omega = m/S$, then acceleration due to gravity would be $g = 4 \pi G~\Omega$, which makes ratio of accelerations simply expressed as ratio of mass compression factors, with no "added noodles". Btw, great post Commented Jul 25, 2022 at 19:46
• Btw, as an added bonus, Earth mass compression is $\approx 1~\text{billion grams}/\text{ cm}^2$. Commented Jul 25, 2022 at 20:10
• @AgniusVasiliauskas. My mistake. Corrected the equations. Sphere surface area obviously isn’t $2\pi R^2$ and a ration of $(1)$ clears the $G$ factor. This is why it didn't appear in $(3)$. Thanks for your observations. Commented Jul 26, 2022 at 9:14

The acceleration due to gravity at the surface of a spherical object is given by $$g=\frac{GM}{R^2},$$ $$M$$ being the mass of the planet, $$R$$ being its radius. Now, all you have to do is now plug in the numbers.

• I think I get it - so the further away from you the particles of the earth are, the less gravitational pull they give, meaning that the higher radius of the earth causes a significantly lower gravitational pull for particles on the other side of the earth than those right next to you.
– Jez
Commented Jul 25, 2022 at 16:25
• @Jez yes. Adding up the gravitational pull from all particles comprising a spherical body, taking their respective distances into account (in other words: integrating over the volume of the sphere) result in a net force as if all the mass was concentrated in a point at the sphere‘s center. When you stand on the planet, the distance to the center happens to be the radius of the planet. The formula also holds for greater distance, i.e. above the surface. BUT: for distances smaller than the radius (i.e. standing in a hole) it is not correct anymore because there is attracting mass above you. Commented Jul 25, 2022 at 20:15

Provided $$r>R_{m}$$ the gravitational pull at a given radius is proportional to $$M_{m}$$

If the Moons mass is 1.23% the mass of the earth, then the strength of the gravitational force is 1.23% that of earth, PROVIDED you are evaluating the gravitational field strength at the same radius.

The radius of the earth is larger than the radius of the moon, hence the surface gravity doesn't have to be 1.23% of the surface gravity of the earth.

Because gravity is proportional to more than mass. The acceleration at the surface due to gravity is given by:

$$g = GM/r^2$$

where $$G$$ is the gravitational constant, $$M$$ is the mass of the planet/moon, and $$r$$ is the radius. The radius also matters, and the different radii of the Moon and the Earth is where the difference is coming from.