Why do masses cancel out in free fall equations?

[misogynisticfeminist]

I was wondering something about gravity and I think someone has posted this before here but i think the thread is gone. Galileo's experiment showed that everything falls onto the Earth at equal acceleration.

So, i messed around and, say if I equate $$F=ma$$ with $$F=\frac {Gm_1m_2}{D^2}$$, the masses actually cancel out. Is this the right way to do it? and to explain why the mass of an object doesn't matter when free falling on the Earth's gravitational field?

but also if we take $$m_1$$ as the mass of the particular object, would $$m_2$$ be the mass of the Earth? If so, i would get, $$m_2 = \frac {g{D^2}}{G}$$. Since the mass of the Earth is constant, then where does D come in here?

Or is my way of doing this totally wrong ? !

Thanks..

 

[dextercioby]

I was wondering something about gravity and I think someone has posted this before here but i think the thread is gone. Galileo's experiment showed that everything falls onto the Earth at equal acceleration.

So, i messed around and, say if I equate $$F=ma$$ with $$F=\frac {Gm_1m_2}{D^2}$$, the masses actually cancel out. Is this the right way to do it? and to explain why the mass of an object doesn't matter when free falling on the Earth's gravitational field?

but also if we take $$m_1$$ as the mass of the particular object, would $$m_2$$ be the mass of the Earth? If so, i would get, $$m_2 = \frac {g{D^2}}{G}$$. Since the mass of the Earth is constant, then where does D come in here?

Or is my way of doing this totally wrong ? !

Thanks..

Nothing you're doing is wrong,just the assumption that D is anything different that Earth's mean radius...or the distance between Earth's center and the object itself...
So,yes,Earth's mass is equal with the product between gravitational acceleration and Earths's mean radius squared devided by Cavensdishs' constant,as long as the object is assumed at the Earths'surface and Earth a spherical body...