As far as I know *g* is based on the Newtonian equation:

Fg = GmSIZE=1]1[/SIZE] m[SIZE=1]2/SIZE] /r^2

If this is the case, larger objects should descend towards the earth at *slightly* larger accelerations. My physics teacher says not.

As far as I know *g* is based on the Newtonian equation:

Fg = GmSIZE=1]1[/SIZE] m[SIZE=1]2/SIZE] /r^2

If this is the case, larger objects should descend towards the earth at *slightly* larger accelerations. My physics teacher says not.

Well, yeah.

The more massive object is pulling the Earth up a little bit more than a less massive one.

A very little bit more.

If you dropped the two things at the same time, though, as Galileo did, they’re both pulling on the Earth together, so the rate of falling should be identical.

You forgot about F=MA.

The larger object has more force pulling it towards the Earth because it has more mass, but the force is acting on more mass, so the acceleration is the same in both cases. If you double the mass you double the force, but you have to divide the mass by the force to get the resulting acceleration.

The math:

**F** = G(m[sub]1[/sub]m[sub]2[/sub] / r[sup]2[/sup]) = m[sub]1[/sub]**a**

m[sub]1[/sub], the mass of the falling body, cancels out of the equation. Thus, the acceleration, **a**, of any falling body is constant for a given m[sub]2[/sub].

The explanation:

The force of gravity on the more massive object is, indeed, greater. However, the more massive the object, the more resistent to acceleration the object is - it’s being pulled harder, but it has more mass to resist the pull of gravity. The two effects cancel each other.

Damn coding, that should be Fg = G*m1*m2/r^2.

Thanks **bup**. So the only thing that will effect gravity on earth is distance from the center of the earth?

The above neglects, of course, air resitance or the existence of any other outside forces aside from gravity.

I think I get where QWERTY is coming from… Let’s try this thought experiment:

I have a cannonball 10Kg, suspended 9.8m above the surface of the Earth. I let it go, it accelerates straight down at 9.8m/s[sup]2[/sup], thusly impacting on the surface 2 seconds later. [sub]I hope that’s right[/sub]

Next, I have a slightly heavier cannonball, with a mass equal to 1EarthMass, suspended 9.8m above the surface of the Earth. I let it go, it accelerates straight down 9.8m/s[sup]2[/sup]. However, in this scenario, the Earth accelerates straight up at 9.8m/s[sup]2[/sup] and meets the cannonball at the middle. Note that the acceleration of the cannonball does not change with mass, only the acceleration of the earth…

Does the second scenario take less time for impact than the first? I believe that it does, and thus we may say that mass does affect the rate that bodies fall, if we base that rate on the time it takes before impact. Since my Physics degree is about 8 years old and totally unused, I’d appreciate some comments. Obviously, this effect is negligible under normal circumstances.

No discussion of this matter is complete without mention of the Staff Report by Ian, and the subsequent classic thread by **Joe_Cool** commenting on it. See also the sequel, Galileo, Hammer, Feather II: The Electric Boogaloo.

To sum up, if the Earth is the same in both cases (not a trivial assumption; see the above threads) then the heavier mass will hit earlier, but the effect is extremely negligible for something as lightweight as a bowling ball or a hammer.

Given that both start at the same height and start at zero velocity relative to earth - at any given height:

A baseball will take longer to hit the ground then a neutron star.