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Acceleration given Earth's radius

  1. Dec 4, 2012 #1
    Given that the Earth has a radius of 6.4x10^6 m, what is the acceleration of a Trojan Badger launched from a catapult when it is 8.5x10^8 m above the surface of the Earth?

    I'm not sure how to go about this question. All help is appreciated: formulas, explanations, answers, etc.
     
  2. jcsd
  3. Dec 4, 2012 #2

    berkeman

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    I believe it's just a gravity type question. Remember that a projectile's horizontal motion is at a constant velocity, neglecting air resistance. And the vertical motion is influenced by the acceleration of gravity.

    What is the graviational equation that relates the attractive force between two masses and the separation distance? You should be able to use that to figure out the gravitational acceleration at that altitude. Please show your work.
     
  4. Dec 4, 2012 #3
    The formula I know of is Fg = (G x m1 x m2)/ r^2 . However, I don't have the mass of the badger.
     
  5. Dec 4, 2012 #4
    Here's what I've tried.
    acceleration of gravity = [Radius of earth/Distance between objects]^2 x 9.8m/s/s

    g = [6.4 x 10^6 m / 8.5 x 10^8 m]^2 x 9.8m/s/s
    g = 5.56 x 10^-4 m/s/s

    Is this correct?
     
  6. Dec 4, 2012 #5

    haruspex

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    Distance between objects? What does that mean in this context?
     
  7. Dec 4, 2012 #6
    Distance between the Trojan Badger and Earth. In other words, distance from earth.
     
  8. Dec 4, 2012 #7

    haruspex

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    Think again. What answer would that give if it were 1mm from the surface of the earth?
     
  9. Dec 4, 2012 #8
    Yeah, yeah, 4 x 10 ^ 20 m/s/s
    I'm stumped. How would you solve this?
     
  10. Dec 4, 2012 #9

    berkeman

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    You don't need the badger's mass. What is the equation relating force to mass and acceleration?

    And for the equation above, treat both masses as point masses for the purposes of calculating distance.....
     
  11. Dec 4, 2012 #10
    F = ma

    But, I don't know why I'd need to use the above equation to calculate distance...I already have distance.
     
  12. Dec 4, 2012 #11

    berkeman

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    That equation is why you don't need the mass of the badger. Combine the two equations, and be careful to think about what separation distance to use...
     
  13. Dec 4, 2012 #12
    So it'll be...
    ma = (G*m*m)/r^s simplified to a = (G*m)/r^2 ?
    where...
    a = acceleration of gravity
    G = gravitational constant
    m = mass of earth
    r = distance from earth (8.5x10^8 m)
    Is this correct?

    And if the above is correct, then why would've my teacher included the radius of the earth into the problem?
     
    Last edited: Dec 4, 2012
  14. Dec 4, 2012 #13

    berkeman

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    You are on the right track. And the radius of the Earth was in your OP problem statement...
     
  15. Dec 4, 2012 #14

    berkeman

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    And for the distance of separation.... think centers of mass...
     
  16. Dec 4, 2012 #15
    Now I understand why the Earth's radius was in the original problem; it's because (as you hinted) the real distance is 8.5x10^8 m plus the radius of the earth which is 6.4x10^6 m.
     
  17. Dec 4, 2012 #16

    haruspex

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    Distance from which bit of earth, exactly? Go back and think about why my 1mm example gave a silly answer.
     
  18. Dec 4, 2012 #17
    Is it this:
    acceleration of gravity = [Radius of earth/(Distance from surface + Radius of earth)]^2 x 9.8m/s/s
     
  19. Dec 5, 2012 #18

    SammyS

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    Yes.
     
  20. Dec 5, 2012 #19
    g = [Radius of earth/(Distance from surface + Radius of earth)]^2 x 9.8m/s/s
    g = (GM)/(Distance from surface + Radius of earth)^2

    So from this thread, I can conclude that both of the above formulas can be used to get an answer. Thankfully, they both yield 5.47 x 10^-4 m/s/s, confirming this to be the answer.

    Thanks to everyone for the help.
     
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