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It seems like it would be like: a(m)=km and may be a(m1,m2)=K(m1-m2) Am I doing any sense?

Btw I'm no negationist, nor I'm trying to create a negationist movement here, I just wonder how physics would be If we lived on different physical environment in order to understand better the physics we have now.

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  • $\begingroup$ youtube.com/watch?v=h96SW0PfQcg&t=33s $\endgroup$
    – jelly ears
    Commented Aug 24, 2022 at 7:45
  • $\begingroup$ it's not really possible because aristotle is inconsistent. there is a book by paul nahin. maybe "in praise of simple physics" where he shows this, i believe. the main problem is that aristotle was sort of unaware of acceleration. he didn't know how to describe it so he didn't even really know what a force was. $\endgroup$
    – jelly ears
    Commented Aug 24, 2022 at 7:51
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    $\begingroup$ @jellyears F=m*v ok what free falling objects and gravittation? I'm sorry that I skewed through the entire video, now tell me to watch it complete to find my other answer and I will. $\endgroup$
    – Lerian Acosenossa
    Commented Aug 24, 2022 at 8:01
  • $\begingroup$ @Lerian_Acesenossa it isn't much different than asking for the neanderthal form of "newton's laws" tbh . he was very very wrong. $\endgroup$
    – jelly ears
    Commented Aug 24, 2022 at 21:04
  • $\begingroup$ @jellyears they already gave nite equations that woulkd reflect Aristoteles thought, and if I'm not getting it wrong the only error he had was not discounting friction from the equation. $\endgroup$
    – Lerian Acosenossa
    Commented Aug 25, 2022 at 1:29

3 Answers 3

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If acceleration due to gravity depended on mass then you have the following conceptual obstacle (first pointed out by either Galileo or Newton, I think).

Two objects each of mass $m$ will fall with a certain acceleration. If you join them together into an object of mass $2m$ then they fall with a different acceleration. But suppose they are joined by a long thin rope that is not under tension, so not applying any force to either object. What is it that changes the acceleration of the objects from the mass $m$ acceleration to the mass $2m$ acceleration ? If we cut the rope in the middle then we have two separate objects again, and their accelerations should change back to the mass $m$ acceleration. But how does cutting the long rope in the middle change the acceleration of either object ?

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When solving Newtons equation for a free falling object the mass cancels out on both sides, which implies that two objects of the different mass fall with the same acceleration $$ m a = F = - m g \quad \Rightarrow \quad a = -g $$ For the acceleration to be mass dependent you would need to change either side of the equation. For example, you could say that the laws of gravity are different, say $F = -m^2 g$, which leads to $a = -m g$, i.e. the acceleration increases linearly with its mass. You could also change the other side of the equation ($F = m a$), i.e. how force is related to acceleration. In this case you are changing the Newtonian laws of motion entirely.

As pointed out in the earlier answer, the implications of these equations being different are of course absurd.

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  • $\begingroup$ I supose that g would be measured in units of m/(s*g)? Also what about universal gravitation? Gm1m2/d²=m²g ? $\endgroup$
    – Lerian Acosenossa
    Commented Aug 24, 2022 at 18:36
  • $\begingroup$ $g$ has units of acceleration. You can have say $F = G m_1^2 m_2^2/r^2$ (it should probably be symmetric under $1 \leftrightarrow 2$), then $g = G m_1^2/R_E^2$, where $R_E$ is the earth radius. $\endgroup$
    – jkb1603
    Commented Aug 25, 2022 at 7:31
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You can define it in multiple ways. Let's say near the earth, the weight is not

$$-mg$$

(pointing downwards)

rather:

$$-m^2g$$

Then from Newton's second law:

$$ma=-m^2g \implies a=-mg$$

So objects with greater mass accelerates more.

But you can also define the gravitational force as:

$$-me^{m/k}g$$

for a constant k, and then the acceleration would be:

$$a=-e^{m/k}g$$

which increase as mass increases. After that, it's just a matter of finding a function that fits with reality

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  • $\begingroup$ would g be of units of m/gs ? How are the dimensions solved? $\endgroup$
    – Lerian Acosenossa
    Commented Aug 24, 2022 at 18:47

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