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Centripetal acceleration of Earth around Sun?

  1. Dec 17, 2004 #1
    New question: centripetal force?

    I can't seem to get the right answer...

    R = 1.50 x 10^11

    and I figure that 1 day/rev = 86400 sec/rev

    So: [tex]\frac {2 \pi 1.50 * 10^{11}}{86400}[/tex] is the velocity, and then plug that into [tex]\frac {v^2}{r}[/tex] but I keep on getting 700 something..what am I doing wrong?

    Question 2: At what minimum speed must a roller coaster be traveling when upside down at the top of a circle if the passengers are not to fall out? R = 8.6m

    What I think you have to do is
    [tex] mg + F_c = m * \frac {v^2}{R}[/tex]

    but I don't know what F would be...or am I approaching this the wrong way?
    Last edited: Dec 17, 2004
  2. jcsd
  3. Dec 17, 2004 #2


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    Homework Helper

    Whoever told you that the Earth circles (on an ellipse :tongue2: ) the Sun in one day (86400s) is a WACKO!!!!!!!!!!!!! :surprised :surprised


    PS.I hope u see the error. :tongue2:
  4. Dec 17, 2004 #3
    Wow. I'm dumb :tongue2:
  5. Dec 17, 2004 #4
    *Bump for new question*
  6. Dec 18, 2004 #5
    Hmm - Newton's second law(the mass cancels)... so, the centripetal acceleration has to be bigger than the gravitational acceleration.

    \frac{v^2}{r} > -g
  7. Dec 18, 2004 #6

    Doc Al

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    So far, so good. What you call [itex]F_c[/itex] is the force exerted by the coaster on the passenger. The minimum speed of the coaster is that which will make [itex]F_c = 0[/itex]. When that force goes to zero, it means that the passengers are beginning to lose contact with the seat.
  8. Dec 18, 2004 #7
    Ah, thank you very much!
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