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What if...
Daedalian Member

 Posted: Tue Apr 29, 2003 6:49 pm    Post subject: 41 Whether or not the units are important depends on whether you just want an answer, or you want the method to be sound. It is a matter of opinion in my view, although I think the mathematicians are very bad and should consider another career. Whether or not the Earth-Sun thing works depends on whether "closer" is relative to the center of mass or surface of the body.
pk
Icarian Member

 Posted: Wed Apr 30, 2003 7:33 am    Post subject: 42 But there is "an answer" and there is "the right answer"; an unsound method will not necessarily give the right answer. For example, what is the area of a rectangle with sides 3 by 4? 12? Who knows? If the sides were 3 cm and 4 cm then the answer would be 12 cm^2, but you don't know that - it could be 3 cm by 4 inches, in which case the answer of 12 would be in units of cm.inches (whatever that means). Also, 12 cm^2 is the same area as 0.0012 m^2, yet these are numerically quite different. Which numerical answer would be correct? This question makes no sense, because you MUST give the units for the numerical value to mean anything, and the unit of the answer depends on the units of the bits that were multiplied to get that answer. PS it doesn't matter whether it's the centre or the surface (for the other one), because they're in a straight line [This message has been edited by pk (edited 04-30-2003 03:42 AM).]
Rollercoaster
Daedalian Member

 Posted: Wed Apr 30, 2003 1:01 pm    Post subject: 43 pk, in general I agree, but the problem is much worse than just a unit 'conversion' problem. The fact is, 12 cm-inches makes perfect sense. It is a unit of area because it represents a unit of length squared, and it is usable in any other calculation. The torque density of a car engine could be quoted in "FOOT-POUNDS per CUBIC INCH", for example. Sure, it could be simplified by converting units, but it isn't necessary. The issue with the original problem is that the FUNDAMENTAL UNITS OF MEASUREMENT are incorrect (a unit of length being equated to the square of time, for example). And that (to me) is a bigger problem. As "What If" said (and was the crux of my previous post), the correctness of the answer depends on the interpretation. My take is that both scientists derived WRONG equations which lead to the RIGHT answer.
pk
Icarian Member

 Posted: Wed Apr 30, 2003 2:15 pm    Post subject: 44 I agree entirely with your comment about units. However, I don't understand what you mean by WRONG equations that give the RIGHT answer. They do both get the same numerical answer, but that doesn't make it the right one. The answer (a distance) is a physical quantity and MUST therefore have a unit.
What if...
Daedalian Member

 Posted: Wed Apr 30, 2003 7:06 pm    Post subject: 45 What do you mean "in a straight line"? Depending on whether the distance is to the center of mass, or to the closest point of the object, you can have different answers. In the Earth-Sun example, the object is farther from the Earth, which is the closest other body, so if you just use shortest distance to another body it wouldn't work. However, it does get closer to the center of mass. Look at reply 16. Also, the question asks for two-body answers (although by imagining an Earth-Sun-like system with unusual connecting strings, the mentioned answer would still work).
pk
Icarian Member

 Posted: Thu May 01, 2003 11:19 am    Post subject: 46 In a system with the Earth and the Sun, I would have thought whether the distance is to the surface or to the C of M would be irrelevant because the masses would be virtual points in such a system. I emphasised straight line because I thought someone was thinking about tangents etc. which is not what I meant. ps the question doesn't say there must only be two bodies.
CrystyB
Misunderstood Guy

 Posted: Sat May 03, 2003 3:54 pm    Post subject: 47 I agree with the phrase "wrong equations with the right answer(s)". But it's not very wrong, it's just a bit incomplete: if we'd know the units were m/s and s, we could adjust the equations as "d=(p/2)v2 / (1m/s2)" and "d=(p/2)(t/2)2 x (1m/s2)" respectively. [This message has been edited by CrystyB (edited 05-03-2003 11:54 AM).]
pk
Icarian Member

 Posted: Wed May 07, 2003 7:21 am    Post subject: 48 Could we? Where do those units come from?
What if...
Daedalian Member

 Posted: Wed May 07, 2003 6:50 pm    Post subject: 49 What Crysty B is saying is that, provided you know that s and m/s are the units of time and velocity, you can reason that since the velocity is the time times 0.5 m/s2, the equations are correct if you multiply or divide by the m/s2, or any other unit of distance over time squared, which is the only thing the "mathematicians" left out. Edited to fix what I missed. [This message has been edited by What if... (edited 05-07-2003 02:53 PM).]
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