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Gravitational attraction

  1. Apr 18, 2008 #1
    The mass of the moon is 7.35*10^22Kg. At some point between earth and the moon, the force of earth's gravitational attraction on an object is cancelled by the moon's force of gravitational attraction. If the distance between earth and the moon (centre to centre) is 3.84*10^5 Km, calculate where this will occur, relative to earth.

    required: r1

    analysis: Fge=Fgm


    therefore: (6.67*10^-11)(5.98*10^24)/(6.38*10^6)^2 = (6.67*10^-11)(7.35*10^22)/r^2

    (5.98*10^24)(3.84*10^8)/(6.38*10^6)^2(7.35*10^22)= r^2

    I got 282m which is clearly wrong can i please get some assistance with the steps? thank you very much
     
  2. jcsd
  3. Apr 19, 2008 #2

    Dick

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    Look at the post for tascja. Are you in the same class? You posted the same question.
     
  4. Apr 19, 2008 #3
    yes i see it. i posted before him, but it's ok. i still don't understand how to do the steps,the explaination on the posts are still too vague. i understand that Fge=Fgm as well as the universal formula. my problem is how to do the steps fully. the answers i keep getting are too small.
     
  5. Apr 19, 2008 #4

    HallsofIvy

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    Here's a tip: say what you are doing: I can guess that the numerator your first line is GM for earth and moon respectively but what distances are you using in the denominator?
    If "r" be the distance from the object to the center of the earth and "R" the distance between moon and earth, then the distance from the object to the center of the moon is R-r. Letting Me and Mm be the masses of the earth and moon respectively, The two gravitational forces are
    [tex]\frac{GM_em}{r^2}= {GM_mm}{(R- r)^2}[/tex]
     
  6. Apr 19, 2008 #5
    [tex]\frac{GM_em}{r^2}= {GM_mm}{(R- r)^2}[/tex]
    do i plug in the value for r^2 on the left hand side which would be the distance of the earth? and leave r on the right alone?
     
  7. Apr 19, 2008 #6
    can anyone actually help me with the calculations?
     
  8. Apr 19, 2008 #7

    Dick

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    It should be pretty obvious Hall's missed a '/' on the right hand side. Now just multiply both sides by r^2*(R-r)^2 to clear out the fractions. Then you have a quadratic equation. Solve it for r. r is the distance to the earth, R is the total distance earth to moon so (R-r) is distance to moon.
     
    Last edited: Apr 19, 2008
  9. Apr 19, 2008 #8
    this what i got.

    GMem/r^2=GMmm/(R-r)^2 therefore r^2GMmm=GMem(R-r)^2

    (R-r)=(6.38*10^6)^2(5.98*10^24)/7.35*10^22

    (R-r) = 2.43*10^38/7.35*10^22

    (R-r) = square root 3.31*10^15 = 5.75*10^7

    -r = -3.84*10^8 + 5.75*10^7 = -3.27*10^8

    r = -3.27*10^8/-1

    r = 3.27*10^8

    please let me know if thats the distance that the earth and moon's gravities cancel each other out. Thanks
     
  10. Apr 19, 2008 #9

    Dick

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    The first line is good, but now you have a quadratic equation. Mm*r^2=Me*(R-r)^2. Then I have no idea what you are doing but it doesn't look like legal algebra. You want to solve for r. Expand it out. You should get an equation like ar^2+br+c=0. What are the a, b and c in terms of Mm, Me and R?
     
  11. Apr 19, 2008 #10
    if i was expected to use quadratic formula for this question then it should have been mentioned in the lesson. there should be another way to do this
     
  12. Apr 19, 2008 #11

    Dick

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    There is if you dislike the quadratic equation. Take the square root of both sides of Mm*r^2=Me*(R-r)^2.
     
  13. Apr 19, 2008 #12
    can you help me with the first step on squaring both sides? i'm a bit confused with that. if it was just one side then i would know what to do. thanks
     
  14. Apr 19, 2008 #13

    Dick

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    Don't square, square root. As in, if you have 4*9=36, then it's correct to say sqrt(4*9)=sqrt(36). sqrt(4*9)=sqrt(4)*sqrt(9)=2*3. sqrt(36)=6. 2*3=6. Do the same thing with your expression. sqrt(r^2)=r, e.g.
     
  15. Apr 19, 2008 #14
    ok so. sqrt(r^2)=r , sqrt(R-r)^2=R-r? then Mm*r=Me*R-r?

    Mm*r/Me then subtract R =-r then then divide by negative 1? then i'm left with the answer for r? is this right?
     
  16. Apr 19, 2008 #15

    Dick

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    No offense, but your algebra skills don't seem to be up for this. Are you in the same class as tascja? Because neither of you seem very prepared for this question. sqrt(Mm)*r=sqrt(Me)*(R-r).
     
    Last edited: Apr 19, 2008
  17. Apr 19, 2008 #16
    i can do math it's just that i can get confused when things don't get properly explained to me. but not everyone is good at what they do.

    oh and offense taken.
     
  18. Apr 19, 2008 #17

    Redbelly98

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    No.

    You need to take the square root of all terms on each side of the equation:
    Mm*r^2=Me*(R-r)^2
    Not just the (r^2) and (R-r)^2 terms.
     
  19. Apr 19, 2008 #18

    Dick

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    Ok, then it's time to finish this. sqrt(Mm)*r=sqrt(Me)*(R-r). Now it's a linear equation. Can you solve for r?
     
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