arkie87

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About arkie87

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  1. arkie87

    driving at orbital velocity

    If you are allowed downward thrust i.e. thrust perpendicular to your velocity vector, it should be possible. Otherwise, as you approach orbital velocity, your weight drops to zero, and your wheels loose traction
  2. arkie87

    Sea-Level on Oblate Spheroid

    ok, so it only applies to perfect spheres. no other shape?
  3. arkie87

    Sea-Level on Oblate Spheroid

    Could be. 351 J out of 107,559 J is about 0.326% , which is comparable to the oblateness of the spheroid 22 km out of 6371 km (~0.345%). Though, on the other hand, if you used the value of g at the equator and poles, though, you might reduce the error as well...? However, i thought that there was some law that says that the law of gravitation -- GMm/R^2 -- applies to point masses as well as solid bodies, as long as the distance is between centers of mass. Based on that, i'm not sure why an oblate spheroid would have any effect?
  4. arkie87

    Sea-Level on Oblate Spheroid

    Well, i think i have figured it out. Thanks to all for your help. The main mistake we were making is the assumption that gravitational force scales with the inverse square law-- this is only true outside the body; but inside the body, it increases linearly. My solution is as follows: The gravitational acceleration inside the earth is: g=g_0*r/R where g_0 is the gravitational acceleration on the surface at R. The specific energy required to ascend from the center of the earth up to the north pole is: E_p/m = integral(g_0*r/R_p*dr,0,R_p)= 1/2 g_0 *r^2/R_p (@r=0,R_p) = 1/2*g_0*R_p Similarly, the specific energy required to ascend from the center of the earth up to the equator is: E_e/m = integral((g_0*r/R_e-w^2*r)*dr,0,R_e)= 1/2*(g_0/R_e-w^2)*r^2 (@r=0,R_e) = 1/2*g_0*R_e - 1/2*w^2*R_e^2 Plugging in values for g_0 = 9.81 m/s2 R_p = 6356 km R_e = 6378 km w = 2*pi/(60*60*24) = 7.272e-5 rad/s We get E_p/m = 31.17618 MJ/kg E_e/m = 31.17653 MJ/kg Alternatively, we can write: 1/2 (w*R_e)^2 = 1/2*g_0*(R_e-R_p) 1/2 (w*R_e)^2 = 107,559 J/kg 1/2*g_0*(R_e-R_p) = 107,910 J/kg So we have energy balance within 351 J, the error of which is probably due our assumption of g_0. Thus, the integral of the potential energy from the center of the earth contributes the needed factor of 1/2 to make centripetal energy and gravitational potential energy balance ( @m4v @sevenperforce). Accordingly, it appears the gravitational potential energy of an oblate spheroid has nothing to do with it. Sorry again, @Snark
  5. arkie87

    Sea-Level on Oblate Spheroid

    So to summarize so far: (1) Ignoring gravitational potential, the energy required to walk from the equator to the pole (equivalent to pulling oneself to the center of a spinning object) is exactly equal to the kinetic energy due to rotation, and that kinetic energy is half of the potential energy difference due to the difference in height. (2) The other half is supposedly due to different gravitational potential of a perfect ellipsoid vs. a perfect sphere. Does anyone know the formula for the gravitational force of an ellipsoid. A linearization when a~=b would probably be sufficient to predict the difference of potential for 20km/6000km.
  6. arkie87

    Sea-Level on Oblate Spheroid

    Are you referring to me?
  7. arkie87

    Sea-Level on Oblate Spheroid

    What's your point? I am talking about A and you are talking about B and C.
  8. arkie87

    Sea-Level on Oblate Spheroid

    I'm not sure when i said that delta_energy needed to be zero... just that energy needed to be conserved.
  9. arkie87

    Sea-Level on Oblate Spheroid

    You said you cannot apply it to an open system. What are you talking about then?
  10. arkie87

    Sea-Level on Oblate Spheroid

    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
  11. arkie87

    Sea-Level on Oblate Spheroid

    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.
  12. arkie87

    Sea-Level on Oblate Spheroid

    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...
  13. arkie87

    Sea-Level on Oblate Spheroid

    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.
  14. arkie87

    Sea-Level on Oblate Spheroid

    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....
  15. arkie87

    Sea-Level on Oblate Spheroid

    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.