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No One Can Solve This

  1. Jun 13, 2008 #1
    1. I have a spacecraft hovering over the Earth. Therefore the weight of the spacecraft is cancelled by the thrust, i.e. the spaceraft is at rest in respect to the surface of the planet. Since there's no net force, there's no centripetal force, and the ship should stop moving in a circular trajectory (along with the Earth's rotation) and go off on a tangent. But that does not happen in reality. Why?

    2. An elevator lifts a ball at a constant velocity. Therefore, there's no net force on the ball, or the velocity wouldn't be constant. However, the ball is gaining energy (potential energy) since it gains height, so there is work being done on the ball. If the net force is zero, what force is doing the work?

    If you don't know the answer don't post.
     
  2. jcsd
  3. Jun 13, 2008 #2
    spaceship

    I'll try #1, but it seems too obvious, so maybe wrong :smile: . When the ship was put in orbit it had to have been given (just the right amont of) angular momentum. That angular momentum has to be conserved which would maintain the rotational part of the trajectory. The up/down forces will have no effect.
     
  4. Jun 13, 2008 #3

    malawi_glenn

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    If you don't have any of your own thougts, then don't post.

    If you don't know how to post only once, then don't post at all (Iam referring to your double posting)

    On the question #2, there is work done on the ball in the gravitational field due to the motion upwards, the force of gravity is always acting on the ball.

    ¤ Let's have a look at the situation:
    -Let the ball travel in positive z-direction: [tex] \vec{v}_{\text{ball}} = v\hat{z} [/tex]
    -The forces acting on the ball is: Gravity [tex] F_G = -mg\hat{z} [/tex]
    Normal force from the floor of the elevator: [tex] F_N = mg\hat{z} [/tex]
    The net force on the ball is zero, since the elevator is not accelerating.
    The force associated with potential energy is Gravity, hence increase in potential energy is: [tex] U = mg\cdot \Delta z [/tex]
     
    Last edited: Jun 13, 2008
  5. Jun 13, 2008 #4

    BobG

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    You don't need any thrust at all to "hover" over the same place on Earth. You just need to be in a circular orbit with a radius of about 42164 km. Your mean motion has to match the rotation rate of the Earth (which is a WGS-84 constant in case you can't figure it out on your own). You also have to be located above the Equator.

    Mean motion equals [tex]\sqrt {\frac{\mu}{r}} [/tex] where [tex]\mu[/tex] is the universal gravitational constant times the mass of the Earth (it's also a WGS-84 constant).

    For any radius smaller or larger than 42164 km, the thrust has to reduce or increase the net force to the same gravitational force that would exist at 42164 km.

    Doing this for the Earth's orbit around the Sun is more interesting. This is the principle that allows placing a solar observing satellite between the Sun and the Earth (ACE, SOHO, etc).
     
    Last edited: Jun 13, 2008
  6. Jun 13, 2008 #5

    Dale

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    I know the answers!
     
  7. Jun 13, 2008 #6

    malawi_glenn

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    :rofl: best answer today!
     
  8. Jun 13, 2008 #7
    I'll go for the #2 then :rolleyes:

    Since the elevator lifts the ball, it DOES the work as it has to use a force exactly the same as mg.
    Wow... that was a terrible explanation :yuck:
     
  9. Jun 13, 2008 #8
    As for #2,

    Note that the total work done on the body, equals the change in its *kintetic energy*. In your case, since the ball remains in constant speed, its kinetic energy is obviously unchanged. Moreover, considering the fact that gravity did do work on the ball, its potential energy would necessarily not remain constant.
     
  10. Jun 13, 2008 #9

    rcgldr

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    Although I'm not "no one", I too can answer this.

    If the spacecraft's thrust is equal to it's weight as stated, then it is hovering, and not circling the earth or moving in a "circular trajectory". If the spacecraft were moving in a circular pattern around the earth, then it's thrust would be less than it's gravitationally induced weight, by M V2 / R.
     
  11. Jun 13, 2008 #10

    rcgldr

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    As for #2, assume the elevator is attached to a cable that runs up and over a pair of frictionless pulleys, back down to a weight with the same amount of mass as elevator and ball. The elevator and ball are moving upwards at a constant speed, and the weight downwards at the same speed. In this system of weight, cable, elevator, and ball, there is no net work done, and no net change in potential energy (assuming that vertical displacement is small enough that the pull of gravity is not affected by the altititudes invovled).
     
  12. Jun 13, 2008 #11

    Doc Al

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    :rolleyes:

    In the future, post questions like these in the HW Help forum, along with your work.
     
  13. Jun 13, 2008 #12
    In reality, it does happen. Rocket scientists take advantage of the rotation of the earth to give rockets extra speed. It is the reason why rocket bases are often located in tropical regions.
     
  14. Jun 14, 2008 #13

    Dick

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    2. The force lifting the elevator is doing the work. Force times distance equals work. Duh. I think this should be moved back to some confusing discussion forum and out of the legitimate homework area. It doesn't deserve to be here.
     
  15. Jun 14, 2008 #14

    BobG

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    Just a correction, mean motion is:

    [tex]\sqrt {\frac{\mu}{r^3}} [/tex]

    (I inadvertantly deleted my power during editing.)
     
  16. Jun 14, 2008 #15
    Thanks for the answers

    First of all, I'd like to apologize for the arrogant attitude. I was just trying to create a challenge, so I would get good answers, and you can't deny it worked. Sorry for the double posting as well, I didn't know I wasn't supposed to do that, and that's my fault.

    THESE ARE NOT HOMEWORK QUESTIONS. They're questions that occurred to me while studying Classical Mechanics.

    What bothered me in these questions was the apparent fact that even though the NET FORCE on the objects was zero, they would behave as if it wasn't (moving in a circle on the first case and gaining energy on the second). They are not that elementary, and do deserve some considerations. Even the Homework Helper Award Winner Super Big Shot of 2007 didn't grasp the point.

    What I conclude from the answers (correct me if I'm wrong):

    - In the first case, if an object is hovering above a certain point of the surface of the Earth, then the net force on it is not zero, i.e. the thrust is not equal to the weight (If it's at Clarke's Belt there's no thrust at all). This is very interesting, but it generates another question: if a helicopter is hovering in a perfectly still manner above a helipad, and it stays that way for a long time, then we can conclude its weight is not being completely cancelled by the aerodynamical force?

    - In the second case, there is no work done, since you would have to lower a counterweight in order to lift the ball. Very interesting again, but what if I had a different kind of elevator, which wouldn't rely on counterweights, what would the answer be then?

    Thanks a lot to Jeff Reid and the other guys who gave enlightening answers.
     
  17. Jun 14, 2008 #16

    Dick

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    The Homework Helper Award Winner Super Big Shot of 2007 stands by his point that 2) IS that elementary.
     
  18. Jun 14, 2008 #17

    rcgldr

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    If an object is "hovering" above a certain point of the surface of the earth, then technically it's moving in a very large circle and not "stationary" relative to the center of the earth. The difference gravitational force - thrust force is equal to the inertial reaction force to the net centripetal force (gravity-thrust) accelerating the object to maintain a circular patch, M V2 / R as posted previously.

    Assuming no losses, you'd be converting one form of energy to another. For example, if the source of energy was a lossless rechargable battery and motor / generator setup, then changes in eletrical potential energy from the battery would result in the opposite change in gravitational potential energy of the elevator and ball.
     
  19. Jun 14, 2008 #18
    Yes, that's conservation of energy. So we can say the "net work" on the universe is always zero, right? But when you consider an individual system, like the ball, there is work being done on it. I understand the work has to be done by the force which is lifting the ball, but if instead of calculating with the individual forces (like weight and lifting force) we calculated with the net force acting on the ball, which is zero, we would be forced to conclude that the work is zero, since, as Dick said, work is force times distance. That's the question.

    I'll put it in a more technical way. A body moves upwards with a constant speed in a gravitational field. Since there's no acceleration, the net force is zero. Since Work = Force x Displacement, Work = 0 x Displacement, Work = 0. But the body's energy is changing, so work must be different than zero. Where's the mistake?
     
  20. Jun 14, 2008 #19

    dynamicsolo

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    Well, not often, but it's just desirable. The ESA has its launch site where it does (4º N. latitude) only because the French still controlled some land on the north coast of South America. The closest the U.S. could come was Florida (which, interestingly, Jules Verne, understanding this aspect of launch velocities, chose as the launch site in From the Earth to the Moon). The closest to the Equator the Soviet Union could arrange was Tyuratam. The Japanese, Chinese, and Indian launch sites are also at mid-latitudes.

    This isn't always a necessity for launch sites, however. Sun-synchronous satellites (as used in Earth surveillance for, ahem, various purposes -- and as seen on GoogleEarth, etc.!) require a large orbital inclination; the Russian launch site at Semipalatinsk is almost ideal for the purpose...
     
    Last edited: Jun 14, 2008
  21. Jun 15, 2008 #20

    Dick

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    Work is force times distance. The work pulling the elevator up is mg*distance, since the gravitational force balances the pulling force. Not coincidentally, the potential energy gained is mg*distance. The work done by one force shows up as the potential energy gained relative to the other force. Now if you ask has the whole system gained energy, yes. It's gained mg*distance. They are just different bookkeeping systems for counting the same thing. Sorry if I was dismissive of your original question, it was possibly the tone of the inquiry.
     
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