I was reading the dictionary and came across the entry for Earth’s Moon wherein it said:
Emphasis mine.
Now I already know that the moon has about 1/4 of the Earth’s gravity, but shouldn’t that mean that it has 1/4 of it’s mass too? I must be missing something obvious here but I thought that gravity was directly proportional to mass. I thought something with 1/80 of Earth’s mass would have 1/80 of Earth’s gravity.
What is the formula for mass:gravity?
The acceleration of gravity at a planet’s surface is g=GM/R^2, where G is the universal gravitational constant, M is the mass of the planet, and R is the radius of the planet. So if the Moon is 1/80 the mass of the Earth, and about 1/4 of its radius, it’s gravity is (1/80)/(1/4)^2=2/10 the gravity on Earth. (The actual figure is that the Moon’s surface acceleration is 0.17 Earth’s.)
Lets see…
F = G ( M * m) / ( r^2)
F is the force, G is the universal gravitational constant, M is the first mass, m is the second mass and r is the distance between them.
Oh yeah. Looks like someone already posted. And Podkayne’s answer is more correct/ relevant. Cool 
http://www.madsci.org/posts/archives/jan99/915808059.Ph.r.html
here’s a link for 'ya…
Maybe if the moon was made of styrofoam it could be the size of the Earth yet mass the same as it does now. That would get you the 1/80G “weight” at its surface.
In a few hundred years maybe we can use orbital solar mirrors to melt the moon, then inflate it with gas and let it cool again. This would give it a much larger surface for habitation, plus lower the escape velocity, and the hollow core would have all sorts of uses.
Or just do it as an artistic statement. By then nobody will be wanting to live on Earth/Moon anyway. It’ll be like going back to live in the African rift valley, or along the Tigris/Euphrates.
In other words, Cisco–
It’s not just total mass, but also the radius (ie, “size”) of the object that counts.
If the mass of the moon were compressed to the size of a marble, the gravitational pull at its surface would be…very very great.
Cisco wrote, in the OP:
Podkayne answered:
Gravitational attraction, for a sphere, acts .AS IF all the mass is located at a point in the center of the sphere. So the mass of an object is not necessarily a direct predictor of gravational attraction.
Yeah, what Scott said.
An object with the mass of the moon and a diameter of 1 meter would have a surface gravity of just shy of 5 trillion times that of earth. A 150 pound person on earth would be a scale busting 369 billion tons, but on the bright side you’d still fit into those new 28 inch waist jeans you got for Christmas.
That’s true when the object being attracted is outside the sphere containing the mass, but not quite true when the object being attracted is inside the sphere containing the mass (the net attraction still acts through the centre, but is only governed by the mass of new sphere with a radius of the distance betwwen the two objects)
Yeah, but you’d probably need a belt AND suspenders to hold them up.
If I might clarify your statement, MC. The law of gravity (and the equation given above) specifies how two objects attract each other. But if you really wanted to figure the attraction to something like the Earth, you need to figure out the attraction to each individual piece of the Earth, and add them all up. This is a job for calculus. But after you work through it once, there’s a brilliant shortcut.
Visualize the Earth as a collection of different sized spheres, all hollow; like those painted, wooden dolls with the smaller ones nested inside the larger ones. For anything outside of a hollow sphere, add up the attraction to every point, and it’s exactly the same as if the sphere was compressed to a single point at the exact center. For anything inside a hollow sphere (and it doesn’t matter where inside it is) the attraction to all points perfectly cancels out.
The result is what SandyHook said, the attractions adds up to a single point in the center. But if you start tunnelling deep into a planet, then you can imagine the collection of nested spheres again. The ones above you cancel out, it’s as if they weren’t even there; the ones below you add up and you feel the gravity from those.
Reading this back, I’m not sure if I clarified anything, but it was worth a shot.
If I remember some distant, drug clouded physic classes correctly, an object at the exact center of a sphere would have 0 weight. You’d be crushed by the pressure, but you’d finally be at that super model weight you’ve desired all these years.
Yes, an interesting afect is, if you imagine a massive but hollow (like every right-minded nutcase believes the earth is) sphere, once you were inside it you would be weightless, even though the people on the surface would be governed by the laws of gravity.
*Ok, you still would be governed by the law of gravity, but you would be weightless
I love seeing a thread that answers everything but the initial error in the OP. 
The Moon has about 1/4 of the Earth’s diameter. However, the volume of a sphere goes as the radius cubed, r^3.
Assuming it has the same density as the Earth, that means the Moon has (1/4)^3 = 1/64 the Earth’s mass. In reality, the Moon is somewhat less dense than the Earth, which is why it has only about 1/81 the Earth’s mass.
hmmm…
… I thought my link answered the question quite well.
I guess it’s not the same as writing in the equations myself, huh?
That’s pretty good, but since we’re cubing the figure, a small error in the radius becomes a larger error in the volume. The moon’s diameter is about 1/3.667 that of the Earth, and so the Moon has about 1/49.3 the Earth’s volume.
I dunno, what does that have to do with the OP anyway?
Re: what it’s like on the inside if the earth is a hollow globe:
You will still feel the effects of centifugal force, varying with the thickness of the global “crust” and your position with respect to the equator. (Anyone care to give the formula?)
But please note:
The inhabitants of Pellucidar apparently have a central sun that (contrary to known physics) acts as a shield against the gravitational forces coming from the distant opposite side of the globe; what remains is the downward pull they feel (“downward” toward the enclosing globe), to which must be added the centrifugal forces noted above.
Ah, I see. The OP was off by saying the gravity was 1/4 the Earth’s, when it’s 1/6th. I read into it that the mass was 1/4, and so I went on about how that depends on volume and density. I misinterpreted the OP a bit.