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How fast does the Earth need to move to provide centripital acceleration

  1. Sep 19, 2015 #1
    Tltr: The quesion d makes no sence to me and I dont know what my teacher mean.
    1. The problem statement, all variables and given/known data

    The Earth turns once around its axis in 24.0 hours. We will assume that it is perfectly spherical, with radius 6400 km. The mass of the Earth is taken to be 6.00 × 1024 kg. The gravitational acceleration on the surface is taken to be g = 9.80 m/s2 .

    3cb4a377a8b6059507d674328953603f.png

    a) What is the centripetal acceleration of a person at the equator? How big a fraction of the gravitational acceleration does this correspond to? Remember to draw a sketch of the situation.

    b) How fast would the Earth have to turn for the centripetal acceleration to be exactly equal to gravity? Give the answer in revolutions per day. Remember to draw a sketch of the situation.

    c) Define θ to be the latitude, so that θ = 0 corresponds to the equator, and θ = 90◦ is the North Pole. What is the centripetal acceleration of a person standing on the surface at a given value of θ? Remember to draw a sketch of the situation.

    d) Taking into account that the centripetal acceleration is towards the rotation axis, and gravity is towards the centre of the Earth, how fast should the Earth now turn for gravity to be just enough to provide the necessary centripetal acceleration (as function of θ)? Remember to draw a sketch of the situation.

    e) Using the force of gravity |Fg| = G (Mm)/r^2 , where M is the mass of the Earth, m the mass of the orbiting object. Find the required speed vorbit, as a function of r, so that the object performs uniform circular motion around the Earth, under the influence of gravity. How long does it take to go round once in an orbit 900 km above ground? Remember to draw a sketch of the situation.

    2. Relevant equations
    a=v^2/r
    v = (2*pi*r)/T , where T is the time it takes for the earth to pass one round around it's own axis (24*60*60)s

    3. The attempt at a solution
    A)
    I used a=v^2/r and found it to be 0.0338 m/s^2
    9.8/0.0338 = 290 thus the acceleration is 1/290 of g
    B)
    By setting 9.8 = (V^2)/r and solving it for v i got 7920 m/s
    Bearing in mind that T=(2*pi*r)/v I got 5078 sec. or 84.6 min.
    (24*60)/84.6=17 rounds per 24 hours
    C)
    This one is difficult to explain here but I'll try my best.
    Thinking that the radius for the person gets smaller as one gets further north (because the earth is smaller there) I found that the "new" radius is r*cos(tetta) where r is the original radius
    Putting this into the a=(v^2)/r gives this witch I hope is the correct one :
    (4*pi^2*r*cos(tetta))/T^2
    I've allso solved e but the problem is for me d

    "how fast should the Earth now turn for gravity to be just enough to provide the necessary centripetal acceleration (as function of θ)?"

    I thought that i could decompose g in x- and y-direction because as the task sais, a is in horizontal direction and g is in both x and y since it towards the center of earth. I tried to find then when the x-component of g (gcos(tetta)) is equal to my answer from task c. This resulted in the angle dissaparing completly from my equation and so it must be wrong.

    Quite francly I don't understand the question. How can g produce the necessary centripetal acceleration when that should come from the movent of the earth around its own axis. Those two accelerations has nothing to do with eachother as far as I can understand. Also what is the necessary centripital acceleration?
     
  2. jcsd
  3. Sep 19, 2015 #2

    haruspex

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    I see no reason to assume your answer to (d) is wrong. It looks right to me. If you think of the centripetal acceleration in the form ##\omega^2r## you can see that the higher the latitude the less acceleration is needed.
    The centripetal acceleration does not "come from" the rotation of the Earth. It is that component of acceleration required to keep a body rotating about the Earth (at the given rate and distance).
     
  4. Sep 19, 2015 #3
    I see, but the reason I thought it was wrong is because the task ask for the answer as a function of θ and when I solve the equation (4*pi*r*cosθ)/((2*pi*r)/v)^2=gcosθ for v the θ dissapairs and thus I haven't answered the task. But mabye I did it wrong. I'll do it again to check
     
  5. Sep 19, 2015 #4

    haruspex

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    If I write that y is a function of x, y=y(x), and later arrive at a solution that y is a constant, that is not a contradiction. It is just a trivial function of x.
     
  6. Sep 19, 2015 #5
    Okey, well if I do that i get that v = sqrt((r^2*g)/pi) Does that mean that the angle is irrelevant? Like if I'm at the equator or further north it dosn't matter?
     
  7. Sep 19, 2015 #6

    haruspex

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    Yes.
     
  8. Sep 19, 2015 #7
    Awesome! Thank you soooo much! :)
     
  9. Sep 19, 2015 #8
    Do the units work out correctly here?
     
  10. Sep 19, 2015 #9

    haruspex

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    Sorry, I didn't notice you had r2 there. Go back to your equation in post #3 and work it through again.
     
  11. Sep 20, 2015 #10
    I forgot to check that. Also see that I forgot to write pi^2 in the equation. But anyways I did it again an got v = sqrt(g*r) , an answer that seems really simple for me considering the complexity of the task. Does this answer make sense to you guys? At least now the units are correct.
     
  12. Sep 20, 2015 #11

    haruspex

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    That's what I got yesterday, if I recall correctly.
     
  13. Sep 20, 2015 #12
    In your answer c. shouldn't also rcos(tetta) be (rcos(tetta))^2?

    I get the same as you in d, only that then it's: v=sqrt(g*r^2)
     
  14. Sep 20, 2015 #13
    In c this is my calculation:
    824d4c6731b7d0d823b8b9f90319750d.png
    When I put this equal to gcosθ i get v = sqrt(g*r) and I've tried several times with same result. Still don't think its the correct answer though.
     
  15. Sep 20, 2015 #14
    If you have v=sqrt(g*r^2) your answer isn't in m/s so it makes no sence
     
  16. Sep 20, 2015 #15
    I agree with you for c). For d), doesn't your answer give v at the equator? I would think you're looking for an the answer for any location on the Earth, i.e., solve for revolutions per day as in b).
     
  17. Sep 20, 2015 #16
    Seems it went a bit to fast when I did it. Then I get the same as you two
     
  18. Sep 20, 2015 #17
    Thats what I'm concerned about, acording to haruspex earlier in this thread it doesn't matter if your on the equator or not, but to me that seems weird. if it was like sqrt(g*r*cosθ) i would actually have a function of θ which is what the task asks for. Honestly I don't know anymore. The task makes little sense to me :P
     
  19. Sep 20, 2015 #18
    Well, if it's independent of theta, then isn't the answer the same as b) or 17 revolutions/day?
     
  20. Sep 20, 2015 #19
    Maby, but the bigger θ is the smaller the radius for which the centripital acceleration is conserned gets. in other words gcosθ gets smaller and thus the centripital acc. also gets smaller. Then the necessary acc. produced by g should get smaller.
    Also from the way my teacher has formulated the question I find it hard to belive that its the same as b. Plus his asking for speed not revolutions/day
     
  21. Sep 20, 2015 #20
    It appears he is asking for revolutions/day, just like in b).
     
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