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# /sci/ - Science & Math

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Heavier objects fall faster.

 >> Anonymous Wed Sep 11 17:51:04 2019 No.10965174 >>10965167>Heavier objects fall fasterHeavier 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 Wed Sep 11 17:51:38 2019 No.10965177 >>10965167They 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 Wed Sep 11 17:52:26 2019 No.10965181 >>10965167no they don't
 >> Anonymous Wed Sep 11 17:53:41 2019 No.10965191 >>10965167No 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 Wed Sep 11 17:55:04 2019 No.10965197 >>10965167Your teacher was an idiot ? , Is that what you wanted to hear ?
 >> Anonymous Wed Sep 11 17:56:56 2019 No.10965203 >>10965167Are you even trying?
 >> Anonymous Wed Sep 11 17:57:34 2019 No.10965207 >>10965167i just downloaded a bunch of bait response photots but i dont feel like using one right now
 >> Anonymous Wed Sep 11 17:58:56 2019 No.10965212 >>10965174>>10965177aren'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 Wed Sep 11 18:28:25 2019 No.10965306 >>10965212Yes, 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 Wed Sep 11 18:50:30 2019 No.10965357 >>10965191That's very similar to the Coulomb's law and the Electric field. So $a_1$ 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 $r^2$ apply to after you divide by $m_1$? The distance between $m_2$ and what?
 >> Anonymous Wed Sep 11 18:53:30 2019 No.10965360 >>10965357Just distance from m_2. You're solving for the gravitational field in space.
 >> Anonymous Wed Sep 11 20:19:43 2019 No.10965632 >>10965167It's the acceleration that tells you how fast it falls, not the forceF = maa = GM/(R^2)
 >> Anonymous Wed Sep 11 22:23:57 2019 No.10965937 >>10965212Yeah but then you're using an earth centered reference frame.Rookie mistake if you ask me.
 >> Anonymous Thu Sep 12 02:27:42 2019 No.10966453 Actually if there is two planets that varies in weight, the heavy planet will "fall" slower
 >> Anonymous Thu Sep 12 14:07:03 2019 No.10967500 >>10965632ding 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 Thu Sep 12 14:10:10 2019 No.10967514 >>10965167Dude, 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 Thu Sep 12 18:57:26 2019 No.10968174 >>10967514Yes 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 Thu Sep 12 19:16:08 2019 No.10968227 >>10968174r^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 Fri Sep 13 10:02:10 2019 No.10969398 >>10965167F = ma2
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