arkie87

You said you cannot apply it to an open system. What are you talking about then?

My integrals are analytical. If w doesnt change, we have an analytical match, so i dont know what you are talking about. What? Ever heard of transport equation? You can apply conservation of energy for open and closed systems, it just has slightly different form. sum(delta_energy) = 0 vs sum(delta_energy) = IN - OUT

I wasnt ignoring it. There are real-world situations where the term is negligible, which do not violate physics. I'm not ignoring the system as a whole, i am just not presently analyzing it. You can apply conservation of energy at any level you choose, and it should be satisfied at all levels. Here, i am interested with the energy of the astronaut/person. As he climbs towards the center, he inputs work and loses kinetic energy. I already did the math, and presented it here (above), and it works out. Here it is again: The energy required for the astronaut to climb to the center is: integral(F,dR,R,0)=integral(w^2*R,R,0)=1/2 w^2 R^2 = 1/2 V^2, which is the same form as kinetic energy, so the energy required to pull inwards is equal to the change in kinetic energy. Now, granted, in reality, the angular velocity changes proportional to the ratio of masses. If we computed the change in angular velocity, and then summed energy from the person and the rod, we would also have energy balance.

I'm not stipulating away conservation of momentum. It just that angular velocity for all practical purposes doesnt change for that case. Stop wasting our time with trivial distinctions. And yes, on a system level, the energy goes into a minute increase in angular momentum. But our reference frame is clearly the person, not the system as a whole. On the level of the person, the energy require to "climb" towards the center is used to reduce his kinetic energy. I never claimed energy was destroyed. What? The mass doesnt retract. The body does...

I clearly stipulated in the OP that the angular velocity doesnt change. Since there is a way for this to happen in real life--if the mass of the rod is much greater than the mass of the body, your minute technical point is both obvious and trivial. Being technically correct is cancer.

Oh god. I specifically said the rod is rotating at constant angular velocity (i.e. the mass of the rod is much larger than the mass of the human). Also, you clearly missed the point of the thought experiment....

On second thought, if we imagine someone holding onto a rigid rod spinning around at a constant angular velocity about a central axis floating in space. If the person decided to pull themselves closer to the center, that would take energy. At the same time, they would lose kinetic energy (since the rod does not speed up). The kinetic energy formula and the centrifugal energy formula are equivalent, thus one is exchanged for the other. In this case, there is no gravitational potential, and centrifugal and kinetic energy are exchanged, thus, it appears my explanation above is incorrect. Thus, half of the energy is due to centrifugal forces/kinetic energy (its the same thing in this case), and the other half could be due to the gravitational well of an oblate spheroid.

Ok, i think i see what you were implying here. Since at the pole, you are zero distance from the axis of rotation, the distance covered is not (R_equator - R_pole), but rather, just R_equator. Interestingly enough, your equation is exactly equal to kinetic energy, which itself is also equal to half the potential energy difference. Thus, the centrifugal and kinetic energy terms sum to ~100% of the potential energy difference due to height, which implies that the gravitational potential of an oblate spheroid has nothing to do with it. Sorry, Snark. Does that seem right to everyone?

I really dont think mass concentrations compare to the 20km radius difference from equator to pole. I get the governing physics. I want some math to quantitatively show where this seeming 20km height difference went, since 20km of potential energy is a lot.

I understand the physics. I want a calculation that would show what "level" is or equivalently, the surface of constant potential energy.

The first half is the centripetal acceleration; the second is the height, so analogous of g*h g = V^2/R h = (R_equator-R_pole) What you are computing is not centripetal acceleration, but kinetic energy. Yeah. It's not a perfect sphere, but can be approximated as an ellipse. Yes, its not a perfect ellipse, but it has minor deviations. Each subsequent approximation gets more and more accurate. But that has no bearing on the question.

Thanks for this post. I'll need some time to digest this...

The errors you showed were on the order of 80m, compared to 20,000m. Before we can understand the 80m, we need to understand the 20,000m error.

You should notice that i have not put a centripetal term at the pole, but there is at the equator... i never said it had anything to do with radius. The problem is that this term is orders of magnitude off (approx 600 J/kg compared to 200 kJ/kg). I'm not sure what to do with this...

Arg. We don't seem to be communicating well. Anything to say about the second half of my response?

I'm sorry if my wording was poor/unclear, let me rephrase: Sea-level is at constant potential energy. Sea level at the poles is 20 km closer to earth's center than at the equator. Considering gravitational potential energy alone, as PE = -GMm/R, we can see they are at different energy potentials due to the different radii (i mention radii to indicate different gravity potentials, not to imply that they are spherical or anything to do with a sphere). If i do a calculation like what you are describing, i dont get an energy balance, so that explanation doesnt work (counter-intuitively): -GMm/R_pole = -GMm/R_equator + mV^2/R_equator*(R_equator - R_pole)

Well yeah, potential energy is complicated so lets just use equations. What you are describing is something like this: -GMm/R_pole = -GMm/R_equator + mV^2/R_equator*(R_equator - R_pole) If you plug in the numbers, they are not equal, so my guess is the gravitational potential of an oblate sphere is slightly different than a perfect sphere in an amount that would cause an energy balance. Oh, you said kinetic energy, which comes out to 101 kJ/kg, which is about half of the deficit (but still much more important than centripetal forces)

Sea level is also level, meaning constant potential energy. So how does that work, given that as you travel to the pole, you are getting closer to the center.

So i was watching a flat earth video (because its fun to laugh at crazy people), and one of their questions got me thinking: given that the radius of the earth at the equator is 20km longer than at the poles, how does sea level (as a radius from center) work from the equator to the poles? Obviously, 20km of potential energy is a lot, so where is that energy harnessed? Does it have something to do with the m*v^2/R term changing as you go from the equator to the pole (since this mv^2/R term is what creates the oblateness to begin with), or is the gravitational well itself of an oblate spheroid different to account for this 20km of potential energy. TIA (and people who know what they are talking about only, please)

Hey, Matt, i have a challenge for you (inspired by your SSTO to Eeloo video): given that in this mission you use Kerbin as a gravity assist and Kerbals complain about everything, it stands to reason that the Kerbals wouldnt have had to make the preceding interplanetary journeys, and might have insisted on boarding the SSTO during the gravity assist at Kerbin. So, i challenge you to dock with a craft that is on an escape trajectory at interplanetary speeds

Preepahre tu bea terhminayted!

Raydeo Sihgnahls weel deegrayde Ovah Tyme!

Are you sure that the relative velocities matter compared to the speed of light?

Now we can debate if 210 makes sense. It seems like 210 is the expected number of civilizations to have evolved over the lifetime of our galaxy. Even if that number is correct, assuming they all still exist and are all just starting to broadcast signals is ridiculous IMO I was mostly making a joke. Everyone knows Kerbal broadcast towers are such high power. That's why Kerbals are green. And when they accidentally aim them at your ship, you get kracken attacks.

What about Kerbal broadcast towers?