Newton's law of universal gravitation

[Anonymous No.10965167]

Heavier objects fall faster.
Why did all my high school teachers lie about this?

[Anonymous No.10965174]

>>10965167
>Heavier objects fall faster
Heavier objects will experience a greater force due to gravity than a lighter object at the same height, but both objects will experience the same acceleration and thus will fall at the same rate.

[Anonymous No.10965177]

>>10965167
They experience more force from gravity, but they also have more inertia, meaning they require more force to accelerate. This happens to even out, which causes everything to fall at the same speed.

[Anonymous No.10965181]

>>10965167
no they don't

[Anonymous No.10965191]

>>10965167
No they don't.
[eqn]a_1=\frac{F}{m_1}=G\frac{m_1m_2}{r^2}/m_1=G\frac{m_2}{r^2}[/eqn] which shows the acceleration (i.e. how fast it falls) is independent of the object's mass

[Anonymous No.10965197]

>>10965167
Your teacher was an idiot ? , Is that what you wanted to hear ?

[Anonymous No.10965203]

>>10965167

Are you even trying?

[Anonymous No.10965207]

>>10965167
i just downloaded a bunch of bait response photots but i dont feel like using one right now

[Anonymous No.10965212]

>>10965174
>>10965177
aren't you ignoring the acceleration of the earth towards the object? gravitation applies to both objects, but people usually ignore the movement of the earth because it's usually so small.

[Anonymous No.10965306]

>>10965212
Yes, but this is on order a proton radius difference or smaller, so in any real experiment there will be many far larger sources of error

[Anonymous No.10965357]

>>10965191
That's very similar to the Coulomb's law and the Electric field. So [math]a_1[/math] it it is really a field not an acceleration.
[eqn]
\vec{F} = k \dfrac{q_1q_2}{r^2} \qquad
\vec{E} = \dfrac{\vec{F}}{q_2} = k \dfrac{q_1}{r^2}
[/eqn]
Also what does [math]r^2[/math] apply to after you divide by [math]m_1[/math]? The distance between [math]m_2[/math] and what?

[Anonymous No.10965360]

>>10965357
Just distance from m_2. You're solving for the gravitational field in space.

[Anonymous No.10965632]

>>10965167
It's the acceleration that tells you how fast it falls, not the force
F = ma
a = GM/(R^2)

[Anonymous No.10965937]

>>10965212
Yeah but then you're using an earth centered reference frame.
Rookie mistake if you ask me.

[Anonymous No.10966453]

Actually if there is two planets that varies in weight, the heavy planet will "fall" slower

[Anonymous No.10967500]

>>10965632
ding ding ding. but he wants to talk about minute movements of the earth as caused by the object. Retarded considering the negligible magnitude of such a calculation

[Anonymous No.10967514]

>>10965167
Dude, wtf, it's right there in your (pic related).
If you can understand M1 is Earth mass, and everything else is absolutely nothing compared to it.

[Anonymous No.10968174]

>>10967514
Yes but if r^2 <<<1 for example 0.00000001 that will result in a HUGE force, trust me. It would mean if you lift an object by just a fraction of a mm off the earth surface, the resulting force will crush you. Which doesn't really happen but this formula doesn't explain why.

[Anonymous No.10968227]

>>10968174
r^2 is squared distance from Earth's center, not the surface. Newton's formula for gravitational force only really works straight out of the box for point masses. To properly calculate the gravitational field of an object with nonzero size, you need to integrate Newton's formula over the whole volume of the object, replacing the mass with density. However, Newton used calculus to prove the shell theorem, which says that if you're outside of a spherical symmetric mass (like the earth), you can treat it as a point mass located at the center of mass. That's why r^2 is the square of the radius of the earth, not distance to the surface.

[Anonymous No.10969398]

>>10965167
F = ma2