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:

Where, 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 6 times higher, they don't cause nearly as much gravitational pull as those close to you.  With the Moon, all of its particles are close enough that they apply quite a lot of gravitational pull.
 A: 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.
A: 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.
A: 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
$$g_{moon} \approx \frac{13}{82} g_{earth} \approx 0.16 \times g_{earth} \tag{3}$$
A: 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.
