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Gravatational attraction

  1. Mar 12, 2009 #1
    1. The problem statement, all variables and given/known data

    If you're standing on the ground 17 m directly below the center of a spherical water tank containing 6.0×106kg of water, by what fraction is your weight reduced due to the gravitational attraction of the water?

    2. Relevant equations

    F = GMm/r^2

    3. The attempt at a solution

    not sure where to start.
  2. jcsd
  3. Mar 12, 2009 #2
  4. Mar 13, 2009 #3
    So that equation gives you the gravitional attraction force. You can use it to calculate the force between u and the water tank, between u and the earth, etc.
    The tank and earth are both spheres, so u can use Gauss's law and symmetry..
  5. Mar 14, 2009 #4
    Still a little confused.

    I get I need to calculate the force between the tank and I . but then I am confused on where to go.
  6. Mar 14, 2009 #5
    You can calculate the force between you and the water tank, then do the same thing between you and the earth (which should essentially be F=ma) and compare. The centre of the water tank directly opposes the centre of the earth so they are opposite forces. The effect of the water tank will be very minor, but still interesting to calculate. If what I said does not make sense let me know and I will try to reword it all. Also, weight is a measure of force, so by comparing the two forces you can solve.

    I don't have my mechanics text near me for the constants, but if you calculate the force exerted by the tank and subtract it from the force exerted by the earth then the resulting force is your weight in newtons
  7. Mar 14, 2009 #6

    F_g_t = GMm/r^2

    = G(6.0*10^6) M_p / 17^2


    F_g_e = GMm/r^2


    G(5.98*10^24)M_p / (6378.1)^2

    is this the correct step so far?
    If so then what should I do next?
  8. Mar 14, 2009 #7


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    It might have been easier to solve it as variables first.

    For instance they want the ratio of the forces, so ...

    F/Fo = (Gm(M_tank)/d2)/Gm(M_earth)/R_earth2)

    Simplifying you have

    Ratio = M_tank*R_earth2/M_earth*d2

    Then just calculate.
    Last edited: Mar 14, 2009
  9. Mar 14, 2009 #8
    M_tank = 6.0 *10^6
    M_earth = 5.98 * 10^24
    d = 17m
    r_e = 6.37 * 10^6

    my answer came out as 5.58 * 10^-8
    But I guess this is wrong
  10. Mar 14, 2009 #9
    forget it this is solved. I miscalculated. Thanks
  11. Mar 14, 2009 #10
    1 question. Why is it the force of earth over the force of the spherical tank?

    Why does that give you the fraction of you weight reduced?
  12. Mar 14, 2009 #11


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    Look again. The Mass of the Tank is in the numerator, earth below. Do the algebra for dividing the expressions for gravity and you will see.

    As to the fraction ...

    they want ΔF / F

    ΔF = Ft
    Last edited: Mar 14, 2009
  13. Mar 14, 2009 #12
    alright thanks, but I am not sure how f1/f2 gives you the fraction of your reduced weight.

    Although I get the math, I don't get the concept of this problem. Would you
    mind explaining why a force over another force gives you a fraction of one's reduced weight?

    F_t = (m_t)(a_t)

    F_e = (m_e)(a_e)


    I get that this gives you a fraction. And that this fraction gives you how
    much you weight relative to the earth ....something?

    but how does it tell you how much less you weight.

    Can you add some context to the math that you did above?
  14. Mar 14, 2009 #13


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  15. Mar 14, 2009 #14
    I am sorry I am still not getting it.
  16. Mar 14, 2009 #15


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    Consider you are measuring some mass m's weight.

    F = m*g

    Well that is also G*M*m/r2

    That's weight.

    Now they also set you underneath the old town water tank. That has a prodigious mass. And as we know masses have attractiveness to each other and that attractiveness in this case is up.

    So ... if it is going up that means it is reducing the weight measured down.

    What is the difference in that weight as a fraction of the weight? Well I'd say it's the weight (force) up divided by the weight down. The weight up is your difference in weight, and the weight down is as before nothing changed from weighing the mass in an open flat parking lot.
  17. Mar 14, 2009 #16
    OH, nice. Thanks a lot. You know I don't just want to memorize formula's. I want to understand it so thats why I ask. I am thankful for you explanation.
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