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I have read that it is the equality of inertial and gravitational mass that causes all objects to fall at the same rate irrespective of size. But if this were not true, what effect would we see? Would inertial mass being half g-mass mean an object twice as massive would accelerate twice as fast? If so, why?

EDIT: Here is where it is asserted that the equality is responsible for all objects falling at same rate:

Why do two bodies of different masses fall at the same rate (in the absence of air resistance)?

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Jeff
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    There is always a constant of proportionality, like G in gravity. So, if you introduce another constant like "2" it won't make a difference. It would just imply that the force equations have a different constant. – Peter R Aug 08 '16 at 19:12
  • So the equivalence of g-mass and e-mass is not responsible for all objects falling at same rate? – Jeff Aug 08 '16 at 19:13
  • Imagine a baloon. The difference would be that it would work even in vacuum. – peterh Aug 08 '16 at 19:13
  • What do you mean by 'equality of inertial and gravitational mass'? – Obliv Aug 08 '16 at 19:14
  • Indeed, when physicists refer to gravitational mass and inertial mass being the same, they mean there is a constant of proportionality relating them. For them to be not equal, it would mean more massive objects have less inertia, or have a non-linear relationship between gravity and inertia – Jim Aug 08 '16 at 19:14
  • @Obliv inertial mass is defined by force / acceleration. Gravitational mass is the inertial mass if the force is gravitational. – peterh Aug 08 '16 at 19:16
  • @Obliv Sorry for the sudden switch from Newton to GR, but as I know, the energy denstiy of the gravitational waves doesn't gravitate. And, the cosmological constant in the Einstein Field Equations behaves as if it would be an antigravitating matter. Yes, I admit "gravitational force on gravitational waves" sounds a little bit surreal (and mixes the theories). – peterh Aug 08 '16 at 19:29
  • Who says that gravitational waves don't gravitate? They carry energy, they gravitate. It's part of the reason the gravitational field is nonlinear and non-renormalizable. In fact, in sort of non-covariant way, you can take the stress energy of the gravitational waves and place it on the rest of the stress energy tensor to get conservation of energy. That gravitational wave stress energy can be thought of as gravity inducing. Read up any good intro to GR book. – Bob Bee Aug 09 '16 at 02:36

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Objects would be accelerated by gravity at a uniform rate regardless of their mass provided that inertial mass and gravitational mass were proportional. There's no requirement for them to be identical, from that perspective.

Marco Ocram
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Say you have Earth's mass as $M$ and your object's mass and distance from Earth is $m$ and $r$

Using Newtonian gravity we have the Force experiences by the object to be $F_g = \frac{GMm}{r^2}$ where $G$ is a constant.

Now the force experienced by earth due to the object's mass is $F_{g2} = \frac{GMm}{r^2}$

Acceleration of both objects, $F_1 = ma_1 \to \frac{GMm}{r^2} = ma_1$ so $a_1 = \frac{GM}{r^2}$

$F_2 = \frac{GMm}{r^2} = Ma_2 \to a_2 = \frac{Gm}{r^2}$

The acceleration that the earth experiences is negligible in comparison. The acceleration the object experiences is only due to Earth's mass. (Note that for considerably large objects the acceleration due to the object's mass would make it seem like the object is accelerating quicker towards Earth than other objects because Earth is moving towards the object fairly quickly too)

If inertial mass $F = ma$ was different from the gravitational mass you would just have a different acceleration for each object BUT every object would still be accelerated at the same magnitude.

ex: $m_g = \frac{m_i}{2}$ then $F_g = \frac{GMm_g}{r^2} = m_ia$ so that $a = \frac{GM}{2r^2}$ and you'd have halved acceleration.

Obliv
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Assume that Force by each if $W = \frac{GM_gm_g}{R^2}$

Where $m_g$ and $M_g$ are the gravitational masses of both body under test and earth respectively Suppose that $m_i$ is the body inertial mass Assuming $m_i=km_g$

Where k is not 1 but a proportionality constant

Newton 2nd law:

$W = \frac{GM_gm_g}{R^2}=k m_g a$

Where $a$ is the acceleration

==>

$a=\frac{G M_g m_g}{k R^2}$ ==> independent of the both values of $m_i$, and $m_g$, but different than the case when:

k=1

The acceleration is different in its absolute value but independent of the inertial/gravitational mass of the falling body

Ahmed Kamal Kassem
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