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Help with gravitational problem

  1. Jun 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Given that we know the mass of the moon and the earth and the distance between their centers as the moon orbits the earth, if the earth's angular velocity about its own axis is slowing down from say some initial given omega to a final omega (due to tidal friction in reality), find out what happens to the orbital distance between the earth and moon as a consequence
    2. Relevant equations

    [itex] F_g = \frac{Gm_mm_E}{r^2} [/itex]


    3. The attempt at a solution

    My first guess is that this is an angular momentum conservation problem becuase if you make the origin the center of the earth, there are no net external torques.

    soo...
    [itex] L_i = L_f [/itex]
    [itex] r_{imoon} x p_i + I_{earth} \omega _i = r_{fmoon} x p_f + I_{earth} \omega _f [/itex]
    [itex] r_im_m \frac {v_{imoon}}{r_i} + I\omega_i = r_fm_m \frac{v_{fmoon}}{r_f} + I\omega_f [/itex]

    obviously this is wrong, as the r's cancel out and we arent left with the variable we are trying to solve. How would I go about solving this problem and where am I going wrong?
     
  2. jcsd
  3. Jun 18, 2015 #2

    haruspex

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    What is x?
    Where do the divisors ri, rf come from?
    You need another equation, one relating the moon's orbital radius to its speed (linear or angular).
     
  4. Jun 18, 2015 #3
    The x is supposed to be a cross.... To multiply vectors. Ri is given the distance from earth to moon, and we are trying to find how Rf changes.

    Is the relationship just [itex] v_{earth} + \omega _{moon} = v_{moon} [/itex] ... so [itex] v_{earth} + \frac{v_{moon}}{R} = v_{moon} [/itex] ?
     
    Last edited: Jun 18, 2015
  5. Jun 18, 2015 #4

    haruspex

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    To show a cross product in Latex use \times.
    I'm asking how you get ri and rf divisors in the last line of your original post. What are you substituting for pi and pf?
     
  6. Jun 18, 2015 #5
    p=mv... so but the velocity of the moon isnt linear as its rotating around the earth, so i said I tried to figure out how to model the velocity, but i went wrong there.
     
  7. Jun 18, 2015 #6

    haruspex

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    So do that substitution, correctly this time, and post the equation you get. As I mentioned, you need a second equation to relate the orbital radius to the speed. Think centripetal.
     
  8. Jun 20, 2015 #7
    yeah i wasnt sure, how to do the substitution. i completely lost on this problem. can you explain to me how to proceed?
     
  9. Jun 20, 2015 #8

    haruspex

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    You have ##r_{im} \times p_{im} + I_{e} \omega _{ie} = r_{fm} \times p_{fm} + I_{e} \omega _{fe}##, ##p_{im}=m_mv_{im}##, ##p_{fm}=m_mv_{fm}##. What's hard about doing the substitutions of the p*m?
    What is the centripetal force required to keep the moon orbiting at radius r? Where is that force coming from? What equation does that allow you to write?
     
  10. Jun 20, 2015 #9
    ok so [itex] bm_{moon}v_{imoon} + I_e \omega _{ie} = bm_{moon}v_{fmoon} + I\omega_ {ef} [/itex]

    entripetal force: [itex] F_g = m \frac{v^2}{r} [/itex] .... the force is coming from the gravitational force from the earth.... so this equation give me the velocity of the moon (because of the v of in the equation).. so i can plug this v value into the v value in the momentum equation... am i correct?
     
  11. Jun 20, 2015 #10

    haruspex

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    What's b? What happened to ##r_{im}## and ##r_{fm}##?
    It gives you two equations, one for each of initial and final radius. Post them.
     
  12. Jun 20, 2015 #11
    oh my bad... i should have written: r_{im}m_{moon}v_{imoon} + I_e\omega_{ie} = r_[fm}m_{moon}v_f{moon} + I\omega_{ef} [/itex]
    so the 2 equations are

    [itex] v_i = \sqrt {\frac{Gm_e}{r_i}} [/itex]
    [itex] v_f = \sqrt {\frac{Gm_e}{r_f}} [/itex] ... right? ... then its a straight plug and chug? and see the relationship as [itex] \omega [/itex] decreases right?
     
  13. Jun 20, 2015 #12

    haruspex

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    Yes.
     
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