Tonight I have to answer one of my Physics teacher's random questions: Karen has a mass of 62 kg, and is standing on the earth. What is Karen's mechanical energy?
The equation I plan to use is E(littlem) = 1/2mvv - (GMm/R)
First, is this the correct one? The reason there are two v's is because v is squared, it is not a w.
Is the R the radius of the earth?
Also, the speed I am using is how fast the earth is turning, but this site confused me.
Help!
[ November 19, 2003, 08:29 PM: Message edited by: Teshi ]
Posted by rivka (Member # 4859) on :
Back up a sec. Her mechanical energy relative to WHAT?
The earth's axis? The moon? The sun? The center of the Milky Way? Epsilon Eridani III? Posted by Teshi (Member # 5024) on :
I don't know! That's part of the reason why I'm confused.
Let's say relative to the rest of the solar system ASSUMING the earth is still (as in not rotating around the sun) and is only rotating itself.
EDIT: or I mean relative to something hypothetically still, next to the rotating earth.
EDIT #2: or, perhaps the earth's axis is a good choice.
[ November 19, 2003, 08:36 PM: Message edited by: Teshi ]
Posted by fugu13 (Member # 2859) on :
Yeah, for mechanical energy to be valid you need to choose a zero. Usually the zero is the surface of the earth, and the rotational motion of the earth is disregarded. So her mechanical energy would be zero under the classic circumstances.
Posted by Teshi (Member # 5024) on :
So assuming I use the rotation of the earth has her speed, am I right? What speed is the one I should use?
Posted by fugu13 (Member # 2859) on :
I'd answer 0. That's based on the only generally used frame of reference for beginning physics. Anything else is pure speculation.
Posted by Teshi (Member # 5024) on :
I don't think that's the answer my teacher wants, he did mention that she was moving, because this came up in class.
It that case, am I right?
Posted by imogen (Member # 5485) on :
Hey Teshi... another response to make you feel better.
I actually did a physics degree - but it seems I've lost all the knowledge I acquired, cos I have no help.
On the other hand, I know a lot about extraordinary chickens... (see the dobie)
Posted by fugu13 (Member # 2859) on :
Just phrase it in terms of velocity vectors and gravitational potential energy relative to the center of the univers .
Posted by Teshi (Member # 5024) on :
I was considering doing it relative to the sun, and then the galaxy, and then the centre of of the universe, but I decided against it. Posted by fugu13 (Member # 2859) on :
The trick is, you can't calculate most of those. So just right it in terms of the relevant vectors and constants .
Posted by Bokonon (Member # 480) on :
If your teacher is trying to be tricky, I'd give her mechanical energy as being related to the earth's REVOLUTION speed (that is, speed around the sun), with the corollary that this will vary (see below)
Alternatively, simply compute her mechanical energy relative to the earth's axis and the earth's revolution speed around the sun, independently of each other (I reccomend this method).
In reality, the mechanical energy of Karen relative to the earth's revolutionary center (roughly the sun) would be a sinusoidal equation, depending on at what point the earth is in it's rotation (with a maximum at REV Energy + ROT Energy, and a minimum at REV Energy - ROT Energy, and a whole lot of intermediary vector energies in between).
Hope this helps.
-Bok
[ November 19, 2003, 09:43 PM: Message edited by: Bokonon ]
Posted by Morbo (Member # 5309) on :
Bok, why should the answer be relative to the center of the Sun? The Sun is not even mentioned in the question, and there are no privledged viewpoints. Choosing the center of mass of the galaxy or of the local galactic supercluster would be equally valid, but needlessly complicated.
I would keep it simple and assume a zero either at the surface of the Earth or at the center of the Earth.
quote: So assuming I use the rotation of the earth has her speed, am I right? What speed is the one I should use?
Teshi. Actually, her rotational speed would vary as a function of her distance from the nearest pole or latitude, with v= zero at either pole and a maximum at the equator.
Since her latitude is not given in the problem, you are free to choose it. Place her at the North Pole and her velocity with respect to both the surface and the center is zero, so her mechanical energy is zero, like fugu said.
Unless I'm missing something, I hate tricky word problems . . .
Posted by JonnyNotSoBravo (Member # 5715) on :
I think I can answer at least one of the questions from your first post. The R in GMm/R is the distance from the center of gravity of the earth to Karen's center of gravity, which is approximately the distance to the center of the earth. The question probably wants to know the mechanical energy relative to the axis of the earth, so you need to get Karen's velocity relative to the Earth's axis. You may already have her speed, you didn't really say, but if you need to figure it out, if you have the earth's radius, then you can get the circumference of the earth, and you know the earth makes 1 revolution every 24 hours and thus can get her velocity.
I really don't think the equation you wrote is the correct one. The total mechanical energy of a system is defined as the sum of the kinetic energy and the potential energy of the system at any moment. Karen has no potential energy since she is standing on the surface of the earth. This means your equation simply includes E(mechanical) = (1/2)mv^2. (v^2 is v squared, with v as simply the magnitude of the velocity vector. Seems like a pretty simple question. What physics class is this for?
Posted by Morbo (Member # 5309) on :
quote: You may already have her speed, you didn't really say, but if you need to figure it out, if you have the earth's radius, then you can get the circumference of the earth, and you know the earth makes 1 revolution every 24 hours and thus can get her velocity.
Jonny. You need the latitude to calculate rotational speed, as I pointed out.
Posted by JonnyNotSoBravo (Member # 5715) on :
Yup, I agree with you Morbo. But since one isn't given to us by Teshi, either Teshi hasn't told us what latitude was given or the problem didn't tell Teshi what latitude was given. So you get to assume one. I assume the teacher probably told them the equator 'cuz Teshi was already talking about the "speed of the Earth" and I'm guessing it wouldn't be zero or that would make the problem boring. I'm also wondering why gravitational potential energy was put in that form for that equation she used. Most of the time "mgh=U" is used (although "g" changes with latitude). Also, Karen could be standing at other than sea level which might change things. Somehow I don't think the prof is giving Teshi a complicated question, though.
Posted by Morbo (Member # 5309) on :
Yeah, assuming either a pole or the equator makes sense. You're right, the teacher did say she was moving.
Posted by ssywak (Member # 807) on :
Hmmm...the units match (kg*m^2/s^2), but not the concepts.
The second half of the equation is Newton's law of gravitational attraction--almost.
Newton stated that: F = G*M*m/R^2 F = force due to gravity G = Gravitational constant (units are N*m^2/kg^2) M,m = Mass of the LARGE and small objects R = Distance between their centers
But your equation is missing an "R" in the denominator. Multiplying out Newton's equation gives you:
E_m = F*R
Oh, wait! I get it! It's your potential energy for being the distance "R" above the center of the Earth.
The other term (1/2*m*v^2) is your kinetic energy (from rotational velocity around the Earth's center, and/or from orbiting the Sun, and/or from moving through the galaxy, from moving through the universe, etc.)
But why do you reduce your overall energy by your potential energy? Typically, you would add the energies. This all appears to be a measure of energy vs. some ill-defined "rest state" (such as Iowa or South Dakota ).
--Steve
Posted by Teshi (Member # 5024) on :
I knew there were physics people out there somewhere Thank you everyone... my mark is going UP! (well- not by much...but it is going up!)
Turns out he wanted the speed of the rotating earth at my latitude (approx. 45% above the equator) but compared to the hundreds of billions of gravitational potential energy, kinetic energy was negligible (if you count millions negligible...) so my answer, taken from the speed of a spot on the earth, is technically right!
EDIT: Potenentiol area indeed. What am I going on about?
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