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PeterDonis
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Again, read my post #52.jeff einstein said:different masses dropped at a certain height do reach the ground at different times?
Again, read my post #52.jeff einstein said:different masses dropped at a certain height do reach the ground at different times?
No. You need to think more carefully about exactly what the different claims involved mean. I am hoping that my post #52 will help. At the very least, it should give you more of a framework for thinking about the problem.jeff einstein said:so you give up sir? and i can go around with my "unproved" misunderstanding on this topic
This is the part that you can't just handwave in this scenario. You need to show the OP the actual math, since he is only in 11th grade and probably has not had the opportunity to work such problems in detail. I know he said originally that he wasn't interested in the math, but the fact remains that the math is there and is what we all are relying on in making our posts, and without it we're just asking the OP to take claims on our authority, which he is quite justified in refusing to do, since no one should ever take scientific claims as true just on someone else's authority.PeroK said:Making some basic assumptions ... The 2kg mass hits the ground at time ##t_2##. After time ##t_2##, when the experiment is repeated with the 1kg mass, that mass is still ##1 \times 10^{-24} m## above the ground. It takes a further ##1 \times 10^{-25} s## for the ##1 kg## object to hit the ground.
I missed post #52 entirely in the wave of posts from the OP.PeterDonis said:I gave the math for a simpler example in post #52, but the basic ideas I gave there are applicable to this scenario.
The accelerations will be of opposite signs (assuming we're just looking at radial motion, as we are in this thread, otherwise we have to do full vector addition), so there should be a minus sign in your last equation, not a plus sign.hmmm27 said:##F = \frac{Gm_1m_2}{r^2} \Rightarrow##
##a_1+a_2 = \frac{G(m_1m_2)}{r^2m_1} + \frac{G(m_1m_2)}{r^2m_2} = \frac{G(m_1 + m_2)}{r^2}\Rightarrow##
##a \propto m_1 + m_2##
The force on both objects has the same magnitude, which depends on the product of the masses, so yes, the mass of both objects affects the magnitude of the force.jeff einstein said:my physics teachers say that when an object is dropped at a certain height both objects "pull" on each other with the same "force" meaning that even the very very tiny amount of gravitational force the object has compared to the earth has an effect in the collision
Yes.jeff einstein said:objects "pull" on each other with the same "force" meaning that even the very very tiny amount of gravitational force the object has compared to the earth has an effect in the collision. all I am asking is are we ignoring the object's gravity cause it's too small?
jeff einstein said:OK, maybe I am wrong about the theory of equivalence. what I actually meant was the idea that all masses fall to the earth at the same velocity. and I have been asked about if know the equation relating to this, Yes I do but I clearly mentioned that I do not want equations used for this doubt clearance. after all, I want to clarify one thing, Mercury (larger mass relative to the moon) "would fall to the earth" faster than the moon (smaller mass) both positioned at an equal distance, am I right?
On the contrary; it is the very thing the OP is now asking about, if you follow their most recent posts.Cutter Ketch said:...the Earth’s acceleration is not the same in the two cases ... I think that just clouds your issue ... and we should stick with comparing bowling balls and feathers so we can ignore the Earth’s response.
i understand this part really well and you have now further solidified my understanding by confirming it but even if this effect was very very very very microscopic I say that larger masses accelerate either slower or faster compared to smaller masses. I say slower or faster because I have a reason let me post an image just waitCutter Ketch said:There is a caveat here in that these masses are large enough to make the Earth’s acceleration significant, and the Earth’s acceleration is not the same in the two cases. There is no problem dealing with that, but I think that just clouds your issue (particularly if you don’t like math) and we should stick with comparing bowling balls and feathers so we can ignore the Earth’s response.
To be fair (I almost hate to do this) I think I’ll have to explain. To be more precise, at the moment the two cases are at the same distance from the Earth they both experience the same acceleration (relative to the fixed stars, say). As described in previous posts, both the Force and the inertia are proportional to mass and the mass cancels. So, in that sense, they do not fall at different rates.
However, the Earth is also being pulled and has its own significant acceleration. The Earth’s acceleration is different in the two cases, so Mercury would get to the Earth sooner. So in that sense you could say it falls faster.
Or, to put it another way, in center of mass coordinates the acceleration of the second mass is not independent of the second mass. However, the dependence is $$ \left( \frac {m_1} {m_1 + m_2} \right) ^2 $$ so when m1 is significantly greater than m2 this quickly becomes 1.
They take different times. The variable mass falls with the same acceleration in each case, but the non-variable mass has a different acceleration in each case. If they start with the same separation then the time will necessarily vary with longer times for slower accelerations of the non-variable mass.jeff einstein said:Now let's consider that all 3 collisions take the exact same time for collision. the distance between the objects in all three cases is the same.
sorry my bad I had posted the whole writing but sir @Dale said it was hard to read so i cropped the image and I cropped the wrongDaveC426913 said:Your diagram only shows two cases.
Yes, you are correct.jeff einstein said:...we can see that the m1 object has a larger distance to cover compared to the m2 object. Because of this, I conclude that acceleration has to be different in both cases.
so sir i was right from the start i have tried to prove my point that different masses have different accelerations when dropped at same heightDaveC426913 said:Yes, you are correct.
You can convert F=ma to a=F/m.
For a given F: If m is small, then a is large. If m is large, then a is small.Looked at another way: Newton's first law says the F is equal and opposite, so
m1a1 = m2a2
And again, if m1 is smaller than m2, then a1 will have to be larger than a2
but if this is true....DaveC426913 said:Yes, you are correct.
You can convert F=ma to a=F/m.
For a given F: If m is small, then a is large. If m is large, then a is small.Looked at another way: Newton's first law says the F is equal and opposite, so
m1a1 = m2a2
And again, if m1 is smaller than m2, then a1 will have to be larger than a2
Sorry, we are now mixing up two things. I may be culpable for that.jeff einstein said:disprove this i guess...
jeff einstein said:but if this is true....
You have been instructed by multiple people over multiple pages carefully explaining the situation. Carefully and repeatedly explaining where your original statements were wrong and also where they were almost right. With multiple corrections from multiple people carefully explaining the physics and the source of the confusion.jeff einstein said:disprove this i guess...
PeterDonis said:The acceleration of both objects, however, is not the same, because it is the force divided by the mass of the object.