Gravity Questions: Explaining Why Mass Doesn't Matter When Free Falling

[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...

 

[GSXR750]

Your approach to equating F=ma with F=\frac {Gm_1m_2}{D^2} is correct. This equation, known as Newton's law of universal gravitation, states that the force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

In the case of free falling, the only force acting on the object is gravity. This means that the acceleration of the object is solely determined by the gravitational force, which is dependent on the mass of the object and the mass of the Earth. However, as you correctly pointed out, the mass of the Earth is constant and the distance between the object and the Earth's center (D) is also constant. This means that the acceleration of the object will be the same, regardless of its mass.

To understand this concept further, we can look at the equation for acceleration, a=\frac {F}{m}. Since the gravitational force (F) is constant, the acceleration (a) will also be constant. This means that objects of different masses will experience the same acceleration due to gravity when free falling.

In summary, the mass of an object does not affect its acceleration when free falling because the gravitational force is dependent on both the object's mass and the Earth's mass, but the other factors such as distance and the gravitational constant (G) remain constant. This is why Galileo's experiment showed that all objects fall at the same rate, regardless of their mass.