Joe I am on your side but say again ?JGMEERT said:
That acceleration changes as free falling bodies get closer to Earth ?
Acceleration is g. Speeds is the one that gets greater, acceleration remains the same !!!!
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Joe I am on your side but say again ?JGMEERT said:JM: (since the variability is the same for both on the way down g slightly changes for both en-route because of the changing r).
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Joe Meert
The variabilty of g isn't that important. If you take all the accelerations to be constant it's even easier to show that the two different objects take different times. However what you're saying is still not quite true the variabilty of g does have a quantitive effect on the results (which is why I included it, so their could be no arguement about it)JGMEERT said:JM: This is an error w/respect to the bodies in question. Small g is simply GMe/r^2. Since the bodies are both starting at the same r and ending at the same r, there is no need to worry about the variability in small g as they move toward the surface of the earth (since the variability is the same for both on the way down g slightly changes for both en-route because of the changing r).
Cheers
Joe Meert
No for a start a = 2h/t^2 for constant accelartion (from rest), see the 1-D kinematic equations, but as I said before the acceleration isn't constant.The Son of Him said:It is very simple, you refuse to see it, and it does not change if you want to define a=dv/dt and also v=dh/dt implies a=dh/dt.dt
It does not change a thing
You are mixing two views, and in the process confusing your frame of reference and refuse to see it
My friend, you truly do not understand what you are integrating either,Aeschylus said:The variabilty of g isn't that important. If you take all the accelerations to be constant it's even easier to show that the two different objects take different times. However what you're saying is still not quite true the variabilty of g does have a quantitive effect on the results (which is why I included it, so their could be no arguement about it)
JM: Now you are being inconsistent with your own argument. You earlier claimed the accelerations are not constant (which is correct); however that makes no difference in the end.Aeschylus said:The variabilty of g isn't that important. If you take all the accelerations to be constant it's even easier to show that the two different objects take different times. However what you're saying is still not quite true the variabilty of g does have a quantitive effect on the results (which is why I included it, so their could be no arguement about it)
No it is not a Gaussian intergral, so far I have considered only the vacuum outside the Earth. Ask yourself this: is g the same for someone who is a million miles away in spaceship as it is for someone standing on the Earth's surface?The Son of Him said:My friend, you truly do not understand what you are integrating either,
The integral you are using is an ever increasing sphere , taking as point of origin the CENTER of EARTH , after that integral has surpassed the raduis of Earth , g remains constant , your "g" changes only in the range of the integral inside Earth beacuse you are taking in account more and more of Earth , But after the integration goes beyond The surface of earth ,the amount of mass becomes constant so g becomes constant.
That is the reason Newton could put F=G M.m/r.r !!!!!!!!!!!!!!
No the accelrations aren't constant, I said that if you assumed that the change in r was small and approximated constant accelartion it's easier to show that the two objects take different times to hit the ground.JGMEERT said:JM: Now you are being inconsistent with your own argument. You earlier claimed the accelerations are not constant (which is correct); however that makes no difference in the end.
Cheers
Joe Meert
Because Newtons law States F=G M.m/ r.rAeschylus said:No it is not a Gaussian intergral, so far I have considered only the vacuum outside the Earth. Ask yourself this: is g the same for someone who is a million miles away in spaceship as it is for someone standing on the Earth's surface?
g = F/m soThe Son of Him said:Because Newtons law States F=G M.m/ r.r
Not becuse the accelerations (g) decreases but because F is inversaly proportional to r.r !!!!!!!!!!!!!!
I knew you would see it otherwise I wouldn't of bothered postingJGMEERT said:JM: Whew, I think I have it. To Aeschylus, I say well done, you are correct and I am wrong (told you I coud do this). Whilst I won't demand a refund for my education, the math you have presented is solid and when I sketched out why this is so, it becomes obvious. Still no answer from my physics friend, but I don't think he would disagree with you. Told you I could be graceful. I owe you a drink. Son of Him, he is correct.
Cheers
Joe Meert
JM: Yup, i should have sat down and done the maths before arguing so vociferously. I will happily buy you a drink if our paths cross.Aeschylus said:I knew you would see it otherwise I wouldn't of bothered posting
It's just a case of examing the problem in detail, becuase the exact solution isn't as obvious as one might think. As I said earlier I think part of the problem was poor presentation on my part and not properly explaining all the terms I used.
So what are u saying my friend , that both balls will hit earth at different times ?JGMEERT said:JM: Whew, I think I have it. To Aeschylus, I say well done, you are correct and I am wrong (told you I coud do this). Whilst I won't demand a refund for my education, the math you have presented is solid and when I sketched out why this is so, it becomes obvious. Still no answer from my physics friend, but I don't think he would disagree with you. Told you I could be graceful. I owe you a drink. Son of Him, he is correct.
Cheers
Joe Meert
JM: The more massive ball will take less time (though small). It's easier to convince yourself of the fact (at least with me it was) to consider masses with smaller differences than the earth. The solution is counter-intuitive to what we previously thought as a counter-intuitive solution! Live and learn, but despite my vociferous arguments, I was wrong and they are correct.The Son of Him said:So what are u saying my friend , that both balls will hit earth at different times ?
Please, Please see this article in the NASA site:JGMEERT said:JM: The more massive ball will take less time (though small). It's easier to convince yourself of the fact (at least with me it was) to consider masses with smaller differences than the earth. The solution is counter-intuitive to what we previously thought as a counter-intuitive solution! Live and learn, but despite my vociferous arguments, I was wrong and they are correct.
Cheers
Joe Meert
Please take a quick look at NASAs site :Aeschylus said:I knew you would see it otherwise I wouldn't of bothered posting
It's just a case of examing the problem in detail, becuase the exact solution isn't as obvious as one might think. As I said earlier I think part of the problem was poor presentation on my part and not properly explaining all the terms I used.
JM: Son, I repeat that I made an error and our intuition is quite wrong in this case. I understand what you are arguing (did it myself), but if you sketch it out, do the math, you will see we are wrong (even though it is counter-inutitive to what we are taught). I am looking for a web explanation that will help explain it. I worked the equations in MathCad.The Son of Him said:Please, Please see this article in the NASA site:
http://quest.arc.nasa.gov/galileo/About/galileobio.html
Take a look where it says LEANING TOWER OF PISA
Please
Did you read NASA's site ?JGMEERT said:JM: Son, I repeat that I made an error and our intuition is quite wrong in this case. I understand what you are arguing (did it myself), but if you sketch it out, do the math, you will see we are wrong (even though it is counter-inutitive to what we are taught). I am looking for a web explanation that will help explain it. I worked the equations in MathCad.
Cheers
Joe meert
JM: Yup, I found a couple of web based explanations (they are somewhat poorly done). have a look and try it youself.The Son of Him said:Did you read NASA's site ?
Nah, despite NASA's incomplete explaination I've followed the proof here. I'd have massive problems constructing it on my own, but laid out like this its screamingly obvious. Good job Aeschylus!The Son of Him said:Did you read NASA's site ?
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