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Reply 20
My point is that weight will be small in all cases. The weight of a feather is very small on earth and will be less on the other planets.

I think the effect of the atmosphere may greater than gravity - viscous drag is one atmospheric factor and I think upthrust may be another. Pressure drag could be another.

Unfortunately all we have to go on in the question regards the atmosphere is the pressure. neither of us has been capable so far of saying qualitatively how this effects those other things.

Su, until someone cleverer than either of us comes along, we are left with gut feeling. Mine is still for A, MEV.
Reply 21
rsk
I think the effect of the atmosphere may greater than gravity - viscous drag is one atmospheric factor and I think upthrust may be another.

Dude, I know upthrust has nothing to do with it. Trust me, I'm in my 3rd year of a physics degree here.

All we need to know further is the viscosity of each atmosphere.

Now, I know that viscosity is basically independent of pressure in standard ranges. So with this in mind, the answer is 'B' because the gravity is the only relevant factor affecting the terminal velocity.

My query - the query that we need to answer - is whether a dependence creeps in at low pressures, such as that of Martian atmosphere. If it does creep in, then we need to know to what extent it affects the terminal velocity. We don't need someone 'clever' to find this out, we just need some information.

I personally think it is a bad question. Either the person who wrote it is expecting you to assume that viscosity is independent of pressure, OR they have made a mistake and think that it is proportional to pressure or something.

You can't just say that you think it is 'A' on gut feeling when I am making a reasoned argument that it is 'B'.
Reply 22
Worzo
Trust me, I'm in my 3rd year of a physics degree here.

:adore:


You can't just say that you think it is 'A' on gut feeling when I am making a reasoned argument that it is 'B'.


You've admitted that your reasoned argument is incomplete because we cannmot directly connect pressure and viscosity.


So I stick with gut feeling.
Worzo
Trust me, I'm in my 3rd year of a physics degree here.


You know she is a teacher right? (IIRC)

I'm going to base my argument on little physics whatsoever, which is to say that this is (presumably) an A-level question, therefore complications such as the dependence of viscosity on pressure are as far as I am aware a mile off the syllabus since we don't know it either (at low pressure) and that therefore it is assumed that you neglect viscous effects and assume it independent of pressure. Therefore, my gut instinct, based entirely on exam technique and what one could be expected to know is B... ie that only the gravity matters.
Reply 24
Sorry guys, I jsut can't accept it, out-numbered or not.

I make a living by simplifying physics and neglecting-air-resistance, but I just can't accept that here we can ignore the effect of the atmosphere. Maybe Worzo is right and whoever wrote the question was thinking of something else. I'd love to know what answer they expect. Unfortunately it comes from some kind of University entrance test and we´re unlikely to ever know.
Reply 25
Right, I will try to show this:

To find terminal velocity, equate gravitational force with Stokes viscous drag:
Mg=6πηrv[br]Mg=kηrv[br]v=Mgkηr\\Mg = 6\pi\eta rv[br]\\Mg = k\eta rv[br]\\v = \frac{Mg}{k\eta r}

where:
η\eta = dynamic viscosity
kk = some constant of proportionality
rr = some characteristic length of the feather
and the other symbols have their usual meaning

From kinetic theory, the formula for viscosity of an ideal gas is:
η=13nmλ<c>\eta = \frac{1}{3}nm \lambda <c>

where:
nn = density of molecules
mm = mass of molecules
λ\lambda = mean free path of molecules
<c><c> = mean speed of molecules

Now:
λ1nσ\lambda \propto \frac{1}{n\sigma}, where σ\sigma is the cross-section for the collision of two molecules - a constant.

So subbing this in:
ηnm(1nσ)<c>[br]ηmσ<c>\\\eta \propto nm (\frac{1}{n\sigma})<c>[br]\\\eta \propto \frac{m}{\sigma} <c>

So, the viscosity is independent of nn and hence independent of pressure. Therefore, from our first equation, so is the terminal velocity is also independent of pressure.

Now, know that <c>T12<c> \propto T^{\frac{1}{2}} and you get ηT12\eta \propto T^{\frac{1}{2}}

Hence, going back to our terminal velocity equation
vgT12\\v \propto \frac{g}{T^{\frac{1}{2}}}

If not affected by the pressure, the viscosity will be affected by the cooler temperature on Mars (about 200K) relative to that of Earth (300K), so you would expect to see the terminal velocity change as:
vMvE=gMTM12gETE12[br]=200122.6×30012[br]vM0.5vE\\\frac{v_M}{v_E} = \frac{g_MT_M^{-\frac{1}{2}}}{g_ET_E^{-\frac{1}{2}}}[br]\\= \frac{200^{-\frac{1}{2}}}{2.6\times300^{-\frac{1}{2}}}[br]\\v_M \approx 0.5v_E

A similar calculation for venus reveals:
vVvE=700121.1×30012[br]vV0.6vE\\\frac{v_V}{v_E} = \frac{700^{-\frac{1}{2}}}{1.1\times300^{-\frac{1}{2}}}[br]\\v_V \approx 0.6v_E

So from this, the order of landing goes EVM, hence B.

I'm not 100% sure that this is right, of course. If anyone can pick a hole in it, that would be great because rsk's 'simplification' is an attractive argument. I want to believe it, but this physics tells me otherwise. This is assuming the atmospheres all have a similar mean molecular weight, which isn't that bad an approximation, I think. It's also assuming that feathers are the same size, shape and mass.
Reply 26
Uff, now that's impressive. It's surpassed my critical limit for Greek letters and at this time of night I can't be sure I've understood it properly. It does though, look far more convincing that the "we-don't-know-what-to-do-about-atmosphere-so-let's-assume gravity-is-the-important-factor" school of thought and I have no reason to doubt that it's absolutely logical and correct.

BUT, and it's a big BUT, it depends on Temperature. Which, unless these candidates already know the temperatures of the atmospheres of the 3 planets, or have on a data sheet, means that this can't be what the examiners are thinking - even if it's 100% correct.

Sorry Worzo, first I thought you were over-simplifying, now I think it's too complicated. I just don't know where the happy medium is here.
Reply 27
rsk
BUT, and it's a big BUT, it depends on Temperature. Which, unless these candidates already know the temperatures of the atmospheres of the 3 planets, or have on a data sheet, means that this can't be what the examiners are thinking - even if it's 100% correct.

I know it was obscenely complicated, but I was just trying to derive a reasonable answer.

But I know what you mean. Why would they even bother providing the pressure data if you didn't need to use it? This suggests that you can't just ignore the atmosphere. But then if they've provided pressure data, what on Earth do they expect you to do with it in relation to terminal velocity?

I too would love to know what the examiners were thinking when they set it. I personally believe that they thought "higher pressure means more molecules in the way, so it's going to increase the air resistance". What they forget is that not only does the feather get slowed by more molecules from below when it falls, it gets accelerated by more molecules hitting it from the top as well, so the pressure effect cancels out.
Wow, that's some seriously decent procrastination there Worzo! :biggrin:

I think, what our result seems to be is that the OP should not be at all discouraged that they can't answer this question since it has a teacher and 2 3rd year uni students fairly confused also. I'm still thinking B, but I agree, it seems bizarre to give you values that you do not use... another bit of exam technique - always use all the values given.

rsk --> Could you run your argument for A by me again? As far as I see it you're going on gut instinct rather than any physical basis per se, but there must be a physical reasoning in there somewhere?
Reply 29
F1 fanatic
Wow, that's some seriously decent procrastination there Worzo! :biggrin:

Ohhh, yes. No atomic getting done for me today....
Reply 30
I have no real argument.

It's true that sometimes there is irrelevant information in a Q to throw candidates, bit in this we HAVE to know something about the atmosphere.

Worzo is right - my gut instinct has no scientific basis - just that Venus has an atmosphere around 9000 times that of Mars which seems more significant that the approx 3 times gravity or whatever. But given that atmospheric pressure is simply the weight of the atmosphere bearing down on the planet, a low-pressure atmosphere could be a high-density but not very deep one, I don't see how it can help here.

I can't help feeling there's something blindingly obvious. Ask you Uni mates tomorrow, both. I'm just glad I'm not applying for University in india this year.


[Edited to add - this is the kind of argument I like - and way simpler than Worzo's, but relies on density which we can't assume anything about from pressure alone. Can we?]
Reply 31
Worzo is a great person to bug about anything, because eventually he will just get of his chair and show you some really great physics. :smile:

Seriously just bug him always worth it :smile: Sorry Worzo but it is true. And I'm sure you love going into this stuff, no doubt that's why you know so much, well that and the fact that you've done the time just learning the ins and outs.

Being wrong is never such a pleasure with the real experts here, of which there are a few :biggrin:

No one can pick a hole in it, because few know enough about gas dynamics to do so. I'm sure there's a PhD student rattling around that can analyse the gas laws from fundamentals :smile: but why would they?
Reply 32
rsk
[Edited to add - this is the kind of argument I like - and way simpler than Worzo's, but relies on density which we can't assume anything about from pressure alone. Can we?]

For something large and fast like a falling rocket, the fluid flow is not considered viscous, and becomes non-linear, which (unusually) is an easier situation to understand. Then most of the drag comes not from viscosity, but from turbulent flow (eddies and such like around the object). This turbulent drag can be modelled with a dependence on easy things to understand like area, a constant drag coefficent, and fluid density. Viscosity does not need to be considered.

I believe the situation of the falling feather involves a linear flow with very little turbulence, and so viscous drag is the most important force to consider, which means you must tackle viscosity.

Sidhe
Sorry Worzo but it is true. And I'm sure you love going into this stuff

Oh, for sure I do. Trying to apply things I've learnt to real world situations is a relaxing break from trying to learn hard, new things. (Seriously, this problem has kept me off atomic physics revision all day...)
Reply 33
Have a look at the question paper this came from

yes, viscosity is mentioned so some knowledge must be assumed. but also there are questions about planets so some other knowledge (like the T necessary in Worzo's calculations) must also be assumed. Look at 14 and 17 for example.


[Edited - some of the other questions are also open to misinterpretation
A car moves from zero velocity increasing linearly to 10m/s velocity in 4 seconds.
What distance it has moved after 8 seconds?


What, if it continues to accelerate? If it then moves with constant speed? 8s counted from t=0 or from the end of the first 4 seconds?

The one about the weight at the poles also comes from here.
rsk
Have a look at the question paper this came from

yes, viscosity is mentioned so some knowledge must be assumed. but also there are questions about planets so some other knowledge (like the T necessary in Worzo's calculations) must also be assumed. Look at 14 and 17 for example.


[Edited - some of the other questions are also open to misinterpretation


What, if it continues to accelerate? If it then moves with constant speed? 8s counted from t=0 or from the end of the first 4 seconds?

The one about the weight at the poles also comes from here.

In other words it would seem to be generally written by someone who isn't shall we say excellent in writing clear and precise questions. I had a long discussion over lunch about this question with another physicist and we couldn't reach a conclusion either. We were trying to consider what differences there would be in for example an atmosphere of water compared to that of air. For example it seems likely that the feather would float. However, we decided this was probably down to the viscosity, which still doesn't give you any relation to the pressure as we have already ascertained. I'm starting to think this question is virtually unsolvable, at least in terms of a definite answer rather than a bit of guess work.
Clearly the viscous drag of a vacuum is zero - so at some point viscosity cant remain independent of pressure.

I suspect the question expects you to simply assume that drag is proportional to pressure which means that the time of fall would just depend on the ratio of P:g

Crap question.
edit:crap paper.

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