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lack of air resistance VS lack of gravity |
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| Apr2-05, 05:55 PM | #1 |
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lack of air resistance VS lack of gravity
a small question, on the moon, the small amount of gravity prevents sudden falling like on earth, however would the lack of gases allow somebody jumping on the moon a much faster falling speed than on earth, because they do not have resistance from the electrons in usual earthly gases.
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| Apr2-05, 06:02 PM | #2 |
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I don't know what you mean by "resistance from the electrons" but yes, if you jumped off a really tall building on the moon, you'd end up going a fair bit faster. Because there wouldn't be any forces opposing your falling motion, you would simply carry on accelerating (until splat-down) rather than achieving a terminal velocity as on earth.
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| Apr2-05, 06:22 PM | #3 |
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so you mean that in long distances, yes, but you would still have the slow-mo effect when jumping short distances. |
| Apr2-05, 06:45 PM | #4 |
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lack of air resistance VS lack of gravity
Sorry, I was talking on the macroscopic scale (people jumping etc).
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| Apr2-05, 06:57 PM | #5 |
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It depends on how high you jump. Without air resistance there is no terminal velocity on the moon. However on earth, a person typically has a terminal velocity of about 150mph on earth.
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| Apr2-05, 07:14 PM | #6 |
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An interesting question for someone to figure out. (I may in a bit)!
How far must you fall on the moon to reach and exceed the terminal velocity of a fall on earth? (Assume terminal velocity of 50 m/s (nice round number which is close) I'll give some else a chance at it. Then, in a bit, when I get some time I will post a solution. |
| Apr2-05, 08:00 PM | #7 |
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You want the height of fallin for an impact velocity of 50m/s
v = a t 50 = (3.4)(t) t = 14.705s Constant acceleration of 3.4m/s^2 for 14.705s without air resistance gives: [tex] \Delta X = \frac{at^2}{2} = \frac{(3.4)(14.7)^2}{2} = 367.35m [/tex] high. Anything above this will breach 50m/s |
| Apr2-05, 08:34 PM | #8 |
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I tried this using energy conservation:
2GM/R = V^2 [tex] R = 2GM/V^2 = (2*6.67*7.36*10^(22-11))/2500 = (98.18x10^11)/2500 = 3.927 x 10^9 [/tex] from the moon's center (1737400m radius) Thats MUCH bigger, but I think im just on the wrong track over all. any help? |
| Apr2-05, 08:55 PM | #9 |
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According to my CRC Handbook of Chem and physics [itex] g_m = 1.62 \frac m s [/itex]
As you did I started with v= at or [tex] t = \frac v a [/tex] also [tex] x = \frac 1 2 a t^2 [/tex] so [tex] x = \frac 1 2 a ( \frac v a )^2 [/tex] [tex] x = \frac {v^2} {2 a} [/tex] let v = 50 [itex] \frac m s [/itex] and a = 1.62 [itex] \frac m {s^2} [/itex] x ~ 770m From energy considerations we have [tex] mgh = \frac 1 2 m v^2 [/tex] or [tex] h = \frac {v^2} {2 g} [/tex] This is the same as my final expression above. |
| Apr2-05, 09:49 PM | #10 |
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Stupid me, I was doing my problem for mars for some reason.
Is it wrong to use energy conservation in this way: [tex] KE = -GPE [/tex] [tex] \frac{mv^2}{2} = -\frac{GMm}{R} [/tex] The kinetic energy gained will equal the loss in gravitational potential energy. |
| Apr2-05, 09:56 PM | #11 |
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[tex] R_1 - R_2 [/tex] Where [itex]R_1[/itex] is the starting point and [itex] R_2 [/itex] the ending. But if you just let [itex] h = R_1 - R_2 [/tex] it is the same. |
| Apr2-05, 10:02 PM | #12 |
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Im trying to accomodate this but I can't get it. Does GMm/R simplify to mgh?
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| Apr2-05, 11:21 PM | #13 |
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My apologies I should have caught this right off the bat.
[tex] G \frac {mM} {r^2} [/tex] is not energy that is gravitational FORCE. No, that expression will not get you what you want. |
| Apr3-05, 12:06 AM | #14 |
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in order to get what you want you need to do this. The potential energy at the moon's surface would be [tex]P_em =-\frac{GMm}{R_m}[/tex] At height h above the surface: [tex]P_eh =-\frac{GMm}{R_m+h}[/tex] [/tex] We want the difference so we get: [tex]\Delta P_e = \frac{GMm}{R_m}-\frac{GMm}{R_m+h}[/tex] This is what will equal the kinetic energy of the falling mass. thus: [tex]\frac{GMm}{R_m}-\frac{GMm}{R_m+h}= \frac{mv^2}{2}[/tex] [tex]GMm \left( \frac{1}{R_m}-\frac{1}{R_m+h} \right) = \frac{mv^2}{2}[/tex] [tex]\frac{1}{R_m}-\frac{1}{R_m+h}= \frac{mv^2}{2GMm}[/tex] [tex]\frac{1}{R_m}-\frac{1}{R_m+h}= \frac{v^2}{2GM}[/tex] [tex]\frac{1}{R_m}-\frac{v^2}{2GM}= \frac{1}{R_m+h}[/tex] [tex]\frac{1}{\frac{1}{R_m}-\frac{v^2}{2GM}}= R_m+h[/tex] [tex]\frac{1}{\frac{1}{R_m}-\frac{v^2}{2GM}}-R_m= h[/tex] If R is very large as compared to h, then this answer comes out very close to that you would get using mgh. |
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