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Finding thing due to gravitation

  1. May 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Assume that the orbital radius of the earth decreases from R to R-ΔR due to air drag, provided the change is very small compared to the radius of earth.
    Then what is change in orbital velocity, change in kinetic energy, change in potential energy and work done by air resistance?
    2. Relevant equations
    F = Gm1m2/r2
    Kinetic energy = 1/2 mv2
    Potential energy = mgh


    3. The attempt at a solution
    Work done by air resistance is Gm1m2Δr/r2 ?
     
  2. jcsd
  3. May 15, 2015 #2

    BvU

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    Hello Raghav,

    No A, B, C, D this time ? Where does this exercise come from ?
    Can you help me imagine the situation ? This about an earth moving around the sun in air instead of in a vacuum ?

    My impression (more like a proven hypothesis) was always that central forces conserve angular momentum and I don't see anything non-central in this.

    Are you assuming circular orbits ? There is no law the orbit should be circular, as far as I know.

    (In fact earth orbit excentricity is some 3 million km, much more than earth radius !)
     
  4. May 15, 2015 #3

    ehild

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    Potential energy of what??? Is this equation relevant for the problem?

    How do you get the work of a force? Is Gm1m2/r2 the air resistance???
     
  5. May 15, 2015 #4

    ehild

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    Well, not "air" resistance, but the space is not empty, there are molecules and dust particles which interact with the Earth and exert some force of resistance, reducing its speed. But such forces are opposite to the velocity. The velocity is not parallel to the radius of the orbit. The resistive force has got torque and changes the angular momentum.
     
  6. May 15, 2015 #5
    Well this was a matrix match type question
    And I don't know how to make a matrix with latex.
    The second column was
    ## \frac{-GmMΔR}{R^2} ##

    ## \frac{-GmMΔR}{2R^2} ##

    ## \frac{GmMΔR}{2R^2} ##

    ## \frac{ΔR}{2} \sqrt{\frac{GM}{R^3}} ##

    This exercise was of a practice, not directly from an exam.
    Yeah, there is no air in space and orbit is approximately elliptical
    Well I was looking potential energy in Wikipedia and they say it is because of some field.
    They have written for capacitor ½ CV2 is potential energy. Why it is called potential energy! Because it is of electric field?
    Whenever someone calls kinetic energy ,the first thing comes to my mind is ½mv2.
    But I know it is energy due to motion.
    I have not applied it in other scenarios.

    The work expression is wrong for air resistance, I know. I was not able to imagine it properly.
    Work done = Force * displacement cosθ
     
  7. May 15, 2015 #6
    I think one can use dimensional analysis to match options only helping for orbital velocity.
    But what is the correct logical way?
     
  8. May 15, 2015 #7

    ehild

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    Is it real that you were not taught about kinetic and potential energy in the school???
    About conservative forces? Energy of a planet orbiting about the Sun? Kepler's Laws?
    You can assume circular orbit and assume that the one with smaller radius is also a stable orbit. Knowing the radius, you get the speed with comparing the centripetal force to the force of gravity.
    The potential energy around a central mass M is -GmM/R.
    Knowing potential energy and kinetic energy on both orbits, you can find the change of the total energy, due to the "air resistance".
     
  9. May 15, 2015 #8
    I was taught about it in school.
    But I have not applied it in space, means I have applied it for objects on ground level.
    What is the difference when we go in space about these energies?
    Don't know energy of a planet orbiting sun.
    Is it Gm1m2/R where R is distance and m1 and m2 masses of sun and earth?
    Kepler laws I know.

    So Mv2/R = GMm/R2 M is mass of earth and m of sun.
     
  10. May 15, 2015 #9

    gneill

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    Presumably this is meant to be the orbital radius of an Earth-orbiting satellite, yes?
     
  11. May 15, 2015 #10
    Yes I think so. As earth is so big and air drag on it does not make sense.
     
  12. May 15, 2015 #11
    So orbital velocity = ## \sqrt{\frac{GM}{R}} ##
    How to calculate Δv ?
    Should I differentiate with respect to M or R?
    But mass of earth and radius of earth are constant.
    If we take R as satellite orbital radius.
    Then ## \frac{dv}{dR} = \frac{-1}{2R} \sqrt{\frac{GM}{R}} ##
    so
    ## Δv = \frac{-ΔR}{2} \sqrt{\frac{GM}{R^3}} ##
    But in options - sign is not there.
     
    Last edited: May 15, 2015
  13. May 15, 2015 #12
    Okay, got it,
    Here dr = R -ΔR - R = -ΔR.
    So I will get correct option.
    Now I can easily use K.E = 1/2mv2 and get things.
    Also for potential energy = -GMm/R
    Work energy theorem for air drag.
    It would be a lot of latex to show but I have got it.
    Thanks BvU, ehild and gneill.
    I'm now on a long break.
     
  14. May 15, 2015 #13

    BvU

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    You deserve it !
     
  15. May 15, 2015 #14

    berkeman

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    I wanted to try to fix up the title "Finding thing due to gravitation", but I'm still not sure what to change it to...
     
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