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Universal gravitation 11- determine the gravitational force of attraction

  1. Jun 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Two bags of apples, each containing 20 apples of equal mass, experience a gravitational force of attraction of 200 units when separated by a distance of 25.0cm. If 10apples are removed from one bag and placed into the other bag, and the two bags are separated by the same 25.0 cm distance, determine the gravitational force of apples now. Note that the magnitude of the gravitational force should be expressed in units.


    2. Relevant equations

    I have made a list of equations that are relevant for this entire module on universal gravitation. So although there are many of them does not mean that they all apply in this circumstance. The ones relevant to this question will be placed in bold.

    Kepler's 3rd law: (Ta/Tb)2=(Ra/Rb)3

    motion of planets must conform to circular motion equation: Fc=4∏2mR/T2

    From Kepler's 3rd law: R3/T2=K or T2=R3/K

    Gravitational force of attraction between the sun and its orbiting planets: F=(4∏2Ks)*m/R2=Gmsm/R2

    Gravitational force of attraction between the Earth and its orbiting satelittes: F=(4∏2Ke)m/R2=Gmem/R2

    Newton's Universal Law of Gravitation: F=Gm1m2/d2

    value of universal gravitation constant is: G=6.67x10-11N*m2/kg2

    weight of object on or near Earth: weight=Fg=mog, where g=9.8 N/kg
    Fg=Gmome/Re2

    g=Gme/(Re)2

    determine the mass of the Earth: me=g(Re)2/G

    speed of satellite as it orbits the Earth: v=√GMe/R, where R=Re+h

    period of the Earth-orbiting satellite: T=2∏√R3/GMe

    Field strength in units N/kg: g=F/m

    Determine mass of planet when given orbital period and mean orbital radius: Mp=4∏2Rp3/GTp2


    3. The attempt at a solution
    Fg=200N
    d=25cm=0.25m

    I used Fg=(6.67X10-11)(30)(10)/(0.25)2=3.2X10-7N

    I have a strong feeling this answer is wrong, but if someone could point me in the right direction it would be greatly appreciated! Thank you so much in advance.
     
    Last edited: Jun 15, 2012
  2. jcsd
  3. Jun 15, 2012 #2
    Your feeling is correct :tongue2:

    First off, you do not know the mass of apples in each bag, so your equations are incorrect.

    Let the mass of 10 apples be m, the mass of 20 would be 2m. Now set up an equation for both bags,

    [tex]F_{1} = \frac{G\cdot 2m\cdot 2m}{r^2}[/tex]

    Try making a similar equation for the 10 apple case. Now without the need to substitute the values of G and r, as they remain same, just divide the equations to get an equation for F2.
     
  4. Jun 15, 2012 #3
    You were told that 10 apples were removed from one bag. But you were not told that they were transferred to the other bag.
     
  5. Jun 15, 2012 #4
    why is it 2m*2m? im confused
     
  6. Jun 15, 2012 #5
    grzz the question actually did state that they were transferred into the other bag, I just edited the problem! Thanks for noticing
     
  7. Jun 15, 2012 #6
    I explained why in my previous post... The mass of one 20 apple-bag is 2m, the mass of the other 20 apples bag is 2m. What would the gravitational force between them be(without substituting numerical values)??
     
  8. Jun 15, 2012 #7
    F=4.2688x10-9*m2?
     
  9. Jun 15, 2012 #8
    Uhh I asked without substituting the numerical values, its easier that way :wink:
     
    Last edited: Jun 15, 2012
  10. Jun 15, 2012 #9
    I left m for the mass... but I used the value for G and r...
     
  11. Jun 15, 2012 #10
    so you mean like 4m2=F1*r2/G?
     
  12. Jun 15, 2012 #11
    Yes! Can you make a similar equation for the case with 10 and 30 apple-bags?
     
  13. Jun 15, 2012 #12
    3m2=F2*r2/G for the second?

    But do I use 200N for the force in both equations to determine the mass?
     
  14. Jun 15, 2012 #13
    Yep. :approve:

    Nope. You don't need to find the mass. You need to find the force F2, which ofcourse is not equal to 200N(F1). Try dividing the equations and see how you can find F2..
     
  15. Jun 15, 2012 #14
    dividing the F1 equation by the F2 equation?
     
  16. Jun 15, 2012 #15
    Yes.... :rolleyes:
     
  17. Jun 15, 2012 #16
    F2=266.67 N? Does this seem right?
     
  18. Jun 15, 2012 #17
    Nope. Recheck your equations. What expression did you get after division?
     
  19. Jun 15, 2012 #18
    3/4=F1/F2... because everything else cancelled out
     
  20. Jun 15, 2012 #19
    Umm no. Without much trouble, that should be 4/3=F1/F2, dont you think? :wink:
     
  21. Jun 15, 2012 #20
    ops haha so the answer should be 150 N right?
     
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