Does g change at higher altitutes?

[coolioyo]

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I got this question from a Kaplan Full length?

What happens to the velocity of water pouring from a hole in a bucket of water when it is moved to a higher altitute.

The formula is square root of 2gh. So, I though g changes and thus v. But they mentioned nothing about that and say velocity is constant. Please help. thank you

 

[hoss22]

g is a constant 10 m/s^2 anywhere near the surface of the earth

 

[themule]

F=(G)(m1*m2)/r^2

Where G=6.67*10^-11 Nm^2/kg^2

This is the force of gravity between 2 objects. Plug in the mass of the earth (6*10^24 kg), for m1 the mass of your second object for m2 and the distance between the two objects, radius of the earth is 6.4*10^6m, for any object touching the earth, and you get 9.8 m/s^2.


TM

 

[majik1213]

F:=ma=GMm/r^2; cancelling the m on each side we get that a=GM/r^2. We ask, what is a at earth's surface?

At the surface of earth, r takes on the precise value of about 6.37*10^6 meters, while M becomes the mass of Earth. In this special condition, we adopt the seemingly and perhaps truly bizarre convention whereby we re-denote the a, the acceleration of an object at the surface of the earth, as g, for because of the spherically symmetric nature of the gravitational force, the acceleration at the surface of earth becomes constant. In the course of all things considered, we find a lot of physical applications take place at or near the surface of earth. Examples of such phenomena are roller coasters, basketball shots, baseball games, soccer games, simply running, tennis, etc. Because a number of physical applications require the consideration of the force of gravity on an object at or near the earth's surface, rather than solving for the acceleration at this level every time, we solve it once, and relabel a as g. We could write out the value of g, 10 m/s^2, every time we considered gravity near the surface, but the act of writing a simple letter, g, is more succint and manipulatable algebraically than is writing out the value g represents, 10m/s^2, so we keep this letter around often when performing a series of calculations to arrive at perhaps a complex solution.

Let me recap: we find life easy if, instead of always solving for a, we simply memorize a constant called g, which equals GM/r^2 when at the surface of Earth. There is nothing special other than that. g cannot ever change, because we decidly chose that it would represent the acceleration due to gravity for only a very specific region of earth, namely the set of all points on the surface shell.

 

[detvar]

but the higher up (the farther up) you go, g would decrease. since it's inversely proportional to r. of course, the altitude would have be to be an appreciable distance.

as such, the velocity of the water coming out should decrease?

 

[FlowLimited]

Let me recap: we find life easy if, instead of always solving for a, we simply memorize a constant called g, which equals GM/r^2 when at the surface of Earth. There is nothing special other than that. g cannot ever change, because we decidly chose that it would represent the acceleration due to gravity for only a very specific region of earth, namely the set of all points on the surface shell.

Alright, let's clarify some things. You've mistaken standard gravity with the local acceleration due to gravity.

Standard gravity, "g0" is a constant (9.8 m/s^2) and is defined as the acceleration due to gravity at the Earth's surface at sea level.

"g" without the subscript, which is the "g" we are referring to, is the local acceleration due to gravity and it varies based on position.

Therefore, the "g" at a higher altitude is less than the "g" at the Earth's surface.

 

[apoptos]

I don't see this as a gravity problem so much as a pressure problem.
When they tell you it's water, it's probably because they want you to know the density, no?

For an open container at ground level,

P = density*gh + 1 atm

shouldn't this be the starting equation? I hope someone can clarify this. I hate physics.

 

[artaxerxes]

I don't see this as a gravity problem so much as a pressure problem.
When they tell you it's water, it's probably because they want you to know the density, no?

For an open container at ground level,

P = density*gh + 1 atm

shouldn't this be the starting equation? I hope someone can clarify this. I hate physics.

this is a good candidate question for over analyzation

Pressure is a function of gravity,
gravity decreases as altitude increases as per the gravitation force equation, although for all points on earth this is negliable, I'm guessing you don't just assume its constant for this question
unfortuantely, atm pressure decreases as altitude increases, but I believe this is overanalyzation.
SO, the answer is it depends on which is greater effect, atm pressure decrease or gravity. But for the purpose of the MCAT, I think you can say here that atm pressure is constant, so velocity will decrease as altitude increases.....but this is the type of question that I hate.

 

[artaxerxes]

F:=ma=GMm/r^2; cancelling the m on each side we get that a=GM/r^2. We ask, what is a at earth's surface?


Let me recap: we find life easy if, instead of always solving for a, we simply memorize a constant called g, which equals GM/r^2 when at the surface of Earth. There is nothing special other than that. g cannot ever change, because we decidly chose that it would represent the acceleration due to gravity for only a very specific region of earth, namely the set of all points on the surface shell.

incorrect. g due to earth at surface is approximately 9.8m/s^2, but this is NOT A CONSTANT, it varies from area to area.....in fact gravitational variation is how they find oil underground. As the equation states, g is dependent upon GMm/r^2, where M and r are variables. Additionally, local mass distribution also causes variation in g, but thats beyond the scope of the MCAT.

 

[lsk87]

i think you guys are missing the point. the op stated that the answer is that velocity is constant. i still don't understand why, so if someone could explain it, i'd appreciate it.

 

[detvar]

****, i didn't even notice that. lsk's right tho, u guys are missing the point

i suppose if they're saying velocity doesnt' change, that means g isn't changing either.

because if g changes, v has to change.

 

[theman120]

so........they say to predict what happens to the velocity.....and they say that velocity remains constant..??? ok...lol

 

[apoptos]

so........they say to predict what happens to the velocity.....and they say that velocity remains constant..??? ok...lol

I didn't even notice that.
So I don't necessarily suck at physics... I might just be dyslexic. Hooray!

Anyway since it's not a physics problem but some kind of pseudo-wittgensteinian tautology, and philosophy is not on the mcat, maybe we should retire this thread.

 

[rhizotummy]

I got this question from a Kaplan Full length?

What happens to the velocity of water pouring from a hole in a bucket of water when it is moved to a higher altitute.

The formula is square root of 2gh. So, I though g changes and thus v. But they mentioned nothing about that and say velocity is constant. Please help. thank you

First post, woo hoo!
I havent taken any Kaplan FLs, so I don't know their detailed explanation behind this problem. I'll take a stab at it:

There could be several ways to reason through this:

1) this one is probably what they wanted:
Kaplan probably wants this to be a conservation of energy problem and they didn't mention to take gravity as a constant of 10m/s^2. So, the take home message is: treat gravity as a constant unless they specifically tell you otherwise.