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I saw in a piece of writing (that I no longer seem to be able to aquire) on dimensional analysis that:

$$t \propto h^\frac{1}{2} g^\frac{-1}{2}.$$

How can $t$ be proportional to $g$ when $g$ is a constant? I ask this because as $t$ rises surely $g$ can neither rise nor fall, which makes the thought that it cannot be proportional to $g$ spring to mind immediately.

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Can someone please merge this question to this account: physics.stackexchange.com/users/8082/olly-price –  Olly Price Apr 3 '12 at 18:04
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"surely g can neither rise nor fall" Try running the experiment (dropping a weight, I suppose, though you didn't say) on the moon. –  dmckee Apr 3 '12 at 18:55
    
To add to dmckee's comment: even with g constant this proportionality can work as $h=h(t)$. –  Alexander Apr 3 '12 at 19:52
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4 Answers

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$g$ is not necessarily a constant if you consider it as "the gravitational acceleration at point A on Earth", and more so if you consider other planets.

$g$ varies around the earth--since the distance from the center of the earth varies. $g$ at mount Everest is lesser than $g$ elsewhere.

Aside from that, $g$ on the moon is approximately one-sixth of $g$ on Earth. So $g$ can vary.

Anyway, one can include relevant dimension-ed constants while doing dimensional analysis. In fact, one has to do so. Otherwise, with dimensional analysis, you will get the wrong expression--since by multiplying/dividing by a power of the dimensioned constant (which you will have to do sooner or later to make it dependant on the constant), the dimensions of the result change. Aside from this, you may have a two-equations-three-variables moment.

A more intuitive reason for why we include dimensioned constants--you can imagine that they changed, and predict the result based on that. In most cases anyway, the constant is not truly a constant, like $g$. The only "true" constants are $G,c,\hbar,R$, and parameters of various bodies. And some other stuff I probably forgot.

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$g$ is an acceleration.

So consider another similar situation with an acceleration: suppose you have a car starting at rest but which has a steady acceleration $a$ and you want to know how much time $t$ it takes to travel a given distance $s$. Exactly the same dimensional analysis would lead to $$t \propto s^\frac{1}{2} \;a^{-\frac{1}{2}}.$$

The signs here at least are intuitively obvious: if you have to travel further then you expect the time to be longer, but if your acceleration is greater then you expect the time to be shorter.

And then you come to dmckee's point: if gravity is less then (ignoring anything like air resistance) the time increases compared with what you are used to on Earth, as you can see in the slowness of the feather and hammer Moon drop: that video also demonstrates that if there is nothing other than gravity operating then mass does not affect time. The Moon's gravitaional acceleration at the surface is about one sixth of the Earth's so using the dimensional analysis you can predict that the time for the hammer to fall the same distance is about two-and-a-half times as long as it would be on Earth.

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@OllyPrice Not sure this is what you want, or even if it will work here. New to this???

Yes g is a constant, so can be dropped from the equation.

So t is proportional to the sq.root of h

or h is proportional to t squared

I.E. Distance h,is proportional to the square of the elapsed fall time, t.

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Yeah I know that, all I want to know is how t is proportional to a constant –  Olly Price Apr 4 '12 at 10:27
    
For all near Earth tests, small g, is a constant, which is known as “standard gravity” (Google this) and its value is: g=35.30394 (km/h)/s (≈32.174 ft/s2). Because you are 16, I assumed you would be using standard gravity. I did say in your other question, that I was surprise your teacher had left g within the square root. It could have been included in the constant C, and would then, not have caused you confusion. Once you get further away, from Earth, variations in g become greater. A big G is use for this new variable. This can also cause confusion if g and G are mixed up. –  Clive Ballard Apr 4 '12 at 14:37
    
Note that the local value of $g$ can easily vary by several times $10^-3$ from place to place, and that this variation is used by mining and petroleum engineers during initial surveys as a guide to where it would be worth setting up more expensive instrumentation. –  dmckee Apr 4 '12 at 14:45
    
Yes, but there are so many things that can cause a variation in g, this is why "Standard Gravity" was introduced, or should I say has always been used for near Earth tests. Even Standard Gravity assumes that the body, is falling in a vacuum. For general use near Earth's surface, the errors are low if Standard Gravity is used. As the questioner was only 16 I assumed that he would not go beyond Standard Gravity. I agree with you, that using a constant g would probably not be applicable, when drilling deep into the Earth's surface. Thanks for your comment. –  Clive Ballard Apr 4 '12 at 15:09
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$y={gt^2}/2$ so $t=(2y/g)^{1/2}$ so $t$ is proportional to $g^{-1/2}$.

$g$ is not constant. It falls off the higher you go above the surface of the earth.

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And is different on other planets! –  Martin Beckett Apr 4 '12 at 2:19
    
The questioner is 16, and unlikely to be involved beyond the Earth's near surface, where g is a constant known as standard gravity. I.E. g=35.30394 (km/h)/s. see my comment, above. –  Clive Ballard Apr 4 '12 at 14:43
    
@Clive: You're right of course, and so is everyone else. When I was 16 (I wasn't always old) I considered myself pretty smart :) So I hope OP would not be derailed by a little challenge. –  Mike Dunlavey Apr 4 '12 at 15:48
    
Didn't we all, consider ourselves pretty smart, when we were 16. Age wise, I'm a little bit ahead of you. I live in the UK, and I consider myself lucky to have been 16 when I was. In my opinion, life was a lot freer, easier, and in fact, better, then, than it is, for 16-year-olds today. I will probably be in trouble for this, though I suppose you could say, it's social history, which verges on physics.Take care. –  Clive Ballard Apr 4 '12 at 16:27
    
@Clive: Hey, I'm always on the edge of being in trouble. It's more fun that way. You too. –  Mike Dunlavey Apr 4 '12 at 16:48
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