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Satellite moving around the Earth

  1. May 9, 2016 #1
    1. The problem statement, all variables and given/known data

    ?temp_hash=893f5bf000dda95fa55272f74ff77f80.png

    2. Relevant equations

    Force due to gravity ##\frac{GM_em}{r^2}##

    3. The attempt at a solution

    C) is incorrect as there would be no torque on the satellite about the center of earth .

    I think B) is correct . But I am not entirely sure about the reasoning .Here is what I think -

    In the frame of reference of satellite ,the two forces acting would be the force due to gravity ##\frac{GM_em}{r^2}## towards the center of earth and the centrifugal force ##\frac{mv^2}{r}## away from the center of earth. As the force due to gravity decreases there would be an unbalanced force in radially outward direction which would move the satellite in orbit of larger radius .

    Is the reasoning correct ?

    Thanks
     

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    Last edited: May 9, 2016
  2. jcsd
  3. May 9, 2016 #2

    mfb

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    Correct.
    You can also show that constant angular momentum requires a larger orbital radius if G is smaller.
     
  4. May 9, 2016 #3
    Ok. Thanks .
    Could you explain a bit more .
     
  5. May 9, 2016 #4

    mfb

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    For a given angular momentum and gravitational constant, there is exactly one distance where the satellite can have a circular orbit. You can calculate this distance. It will increase when G decreases, and you already figured out that angular momentum stays constant.
     
  6. May 9, 2016 #5
    ##\vec{L}## = ##\vec{r}## x ##m\vec{v}## and Force due to gravity = ##\frac{GM_em}{r^2}## . How do we relate angular momentum and gravitational constant ?
     
  7. May 9, 2016 #6

    PeroK

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    The way I would look at this question is that a change in G is no different from any other change in the central force. A larger force would pull the satellite further in, possibly causing a collision. And a smaller force would allow the satellite to move away, possibly out of orbit.
     
  8. May 9, 2016 #7

    mfb

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    What is the speed of an object in a circular orbit, for a given r, m, M and G?
    While the conclusion is right, I'm not sure if you can get away with that so easily. For every G, every radius has a stable orbit (in Newtonian gravity), and you have to consider the transition period.
     
  9. May 9, 2016 #8

    PeroK

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    If it starts in a circular orbit it must move out into an elliptical orbit. I don't see how it could move out into another circular orbit.

    Trickier than at first sight, perhaps.
     
  10. May 9, 2016 #9

    mfb

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    Make the change of G slow enough. I guess mathematically you won't end up in a perfect circle, but arbitrarily close to one.
     
  11. May 9, 2016 #10
    can you quote/display the approprate mathematicas...equations etc
     
  12. May 9, 2016 #11

    PeroK

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    I nearly added something to that effect, as I've been imagining a more significant change in G.

    But, I wonder whether the new orbit will always intersect the original? And whether, even with a slow change, the orbit will eventually become more eccentric?
     
  13. May 9, 2016 #12

    mfb

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    Force equilibrium, Newton's law of gravity, centripetal acceleration, one line of calculation, but I won't do other's homework here.

    @PeroK: If G changes faster than the orbital period, the orbit will be eccentric, sure. But then there is no point in the "decreasing at a constant rate" information, a sudden reduction would have a similar effect. With a very slow change, the orbit will stay roughly circular.
    I would expect that any finite time for the change leads to a final orbit that does not intersect the original orbit.
     
  14. May 9, 2016 #13

    rcgldr

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    Velocity decreases as radius increases since angular momentum remains constant, but as long as G decreases, there will be an imbalance between gravitational force and the fictitious centrifugal force.

    As the object spirals outwards, a component of the objects path is radially outwards, which is opposed by gravitational force, so the object is slowed down as it moves outwards, but angular momentum remains conserved, since the only real force is radial.

    An alternative example would be an object initially sliding in a circular path on a frictionless surface, attached to a massless string that goes through a hole in the surface at the center of the initial orbit. Then the tension in the string is reduced at a constant rate.

    I'm wondering if there's a reasonably simple equation (polar coordinates?) to describe the path of the object with G decreasing at a constant rate. Velocity would decrease at a constant rate ## v = G \ M \ / \ (r_0 \ v_0) ## and radius would increase at constant rate ## r = (v0 \ r0)^2 / (G \ M)## .
     
    Last edited: May 9, 2016
  15. May 10, 2016 #14

    mfb

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    Assuming the change is slow enough: v^2 r = GM, and we know ##vr=v_0r_0## is constant. Everything apart from M is a function of time. Then ##v(t) = G(t) \frac {M}{vr}## and similarly for the radius. I can confirm your formulas.

    Note the interesting result that the satellite never escapes, as long as the change if G is slow enough. For a fast change it can escape.
     
  16. Jun 1, 2016 #15

    haruspex

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    Really? When G reaches zero, no matter how slowly, the satellite will move in a straight line, no? And it must still have angular momentum about the Earth.
     
  17. Jun 1, 2016 #16

    mfb

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    If G reaches zero, then the relative change of G will diverge, which does not satisfy the "slow enough" condition any more.
     
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