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10965167 No.10965167 [Reply] [Original] [archived.moe]

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

>> 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.

>> 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.

>> No.10965181

>>10965167
no they don't

>> 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

>> No.10965197

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

>> No.10965203

>>10965167

Are you even trying?

>> No.10965207

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

>> 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.

>> 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

>> 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?

>> No.10965360

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

>> No.10965632

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

>> No.10965937

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

>> No.10966453

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

>> 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

>> 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.

>> 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.

>> 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.

>> No.10969398

>>10965167
F = ma2

>>
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