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Evaluate the magnitude of the gravitational field at the surface of the planet

  1. Feb 26, 2010 #1
    1. The problem statement, all variables and given/known data

    The trajectory of a rock thrown from a height with an initial speed of 16.9 m/s is shown in the figure below. Evaluate the magnitude of the gravitational field at the surface of the planet. The planet has no atmosphere.

    The graph is as follow:

    http://i995.photobucket.com/albums/af79/huybinhs/plot-1.png



    2. Relevant equations

    g = G*M/R²

    a = v^2 / r

    3. The attempt at a solution

    I know using g = G*M/R² to find magnitude of the gravitational field. In order to find R using a = v^2 / r; but I dont know how to find a from the graph. Please help! Thanks!

    Note: I'm not sure if my tried solution is correct or not. Please advise!
     
  2. jcsd
  3. Feb 26, 2010 #2

    tiny-tim

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    Hi huybinhs! :smile:

    Sorry, but you're going completely berserk on this. :redface:

    You don't know the mass of the planet, or its radius, so you can't use the first equation; and the second equation is for centripetal acceleration

    All you're being asked is to measure the acceleration on the graph (you can assume it's vertical, and a constant). :wink:
     
  4. Feb 26, 2010 #3
    "All you're being asked is to measure the acceleration on the graph (you can assume it's vertical, and a constant)" => you mean a = 10 m/s^2 (when I look at the vertical on the graph) ??? I'm confused...
     
  5. Feb 26, 2010 #4
    Still dont get this. Please help!
     
  6. Feb 26, 2010 #5
    Any one?
     
  7. Feb 27, 2010 #6
    I haven't looked at your graph, but acceleration in any equation is the second derivative. First derivative is velocity, second is acceleration.

    So if your graph shows you something happening at the surface of the planet, take the second derivative of the function at that place on the chart.
     
  8. Feb 27, 2010 #7
    I understand. My graph showing Vertical Positive Above Surface (m) and Horizontal Position (m). What can I do?
     
  9. Feb 27, 2010 #8
    Tell me, between which of vertical position and horizontal position does gravity affect? Also this equation may help you

    v^2=v0^2+2as
     
  10. Feb 27, 2010 #9
    The gravity absolutely affects on the rock. Using your formula, what about the v?
     
    Last edited: Feb 27, 2010
  11. Feb 28, 2010 #10
    Extremely stuck with this one. Anyone????
     
  12. Feb 28, 2010 #11
    Suppose you throw away a rock from an alitude of 10m with a speed v=16.9m and with an angle of [itex] \phi [/itex] with the horizontal on a planet with acceleration of gravity a.
    you should be able to calculate the maximum altitude and the range as a function of a and [itex] \phi [/itex]

    Now you can read the maximum altitude and the range from the graph, so you get to equations
    for a and [itex] \phi [/itex] that you can solve.
     
  13. Feb 28, 2010 #12
    Got ya. Do you have any ideas about equation calculator here? That's would be great ;)
     
  14. Feb 28, 2010 #13

    tiny-tim

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    Hi huybinhs! :smile:

    From the graph, measure the initial angle, and use that to find the initial components of velocity (you know the magnitude is 16.9).

    Then read off the horizontal distance travelled when the ball returns to its original height (10m).

    Call the time when that happens "t".

    The use the standard constant acceleration equations (with a = 0 and a = -g) for the x and y directions …

    that gives you two equations, from which you can eliminate t and find g. :wink:
     
  15. Feb 28, 2010 #14
    Hello,

    How can I find intial angle on the graph???? guessing?
     
  16. Feb 28, 2010 #15

    tiny-tim

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    Measure it! :smile:
     
  17. Feb 28, 2010 #16
    Ok. Taking x = 0; x =10 ; y = 10 ; y = 20. So:

    tan(theta) = o / a = 20 /10 = 2 => angle = 63.4 degree. Correct? then what?

    what formula to calculate the v?

    I got v^2 = tan(theta) * r*g but lack of r and need to find g ???????
     
    Last edited: Feb 28, 2010
  18. Feb 28, 2010 #17

    tiny-tim

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    (have a theta: θ :wink:)
    Then use vx = 16.9 cosθ , vy = 16.9 sinθ :smile:
     
  19. Feb 28, 2010 #18
    Using vx = 16.9 cosθ (i) , vy = 16.9 sinθ (ii)

    (i) => vx = 16.9 * cos (63.4) = 7.57 m/s

    (ii) => vy = 16.9 * sin (63.4) = 15.1 m/s

    then "Then read off the horizontal distance travelled when the ball returns to its original height (10m)." = 124 m

    "Call the time when that happens "t"."

    Using the standard constant acceleration equations (with a = 0 and a = -g) for the x and y directions:

    [tex]\Delta[/tex]x = v0x * t (1)

    [tex]\Delta[/tex]y = v0y - 1/2 gt2 (2)

    How can you cancel the t because the equation (2) having t2 ; what's value for [tex]\Delta[/tex]x and [tex]\Delta[/tex]y ?? (60 and 124 ????)
     
  20. Mar 1, 2010 #19
    Still waiting for the answer!!!
     
  21. Mar 1, 2010 #20
    Are you sure your initial angle is correct(looks like 45 to me). Delta y=0 as there is no change in height and delta x=124.
     
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