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Velocity in a gravitational field

  1. Jan 16, 2015 #1
    1. The problem statement, all variables and given/known data
    a) A mass ##m##, initially at rest, is released from infinity, and is attracted towards a planet of mass ##M##, displacing towards it in a direction parallel to the gravitational field lines present between the masses. Assuming that the only force acting on ##m## is the gravitational force of attraction, and provided ##M>m##, find an expression for the velocity ##V_0## of the mass ##m## at a separation distance of ##x## from ##M## in terms of ##G##,##M## and ##x##.
    b) Given that the velocity required for an object to be in circular motion at a distance ##r## is given by ##\sqrt{\frac{GM}{r}}##, determine if ##m## will enter a circular orbit around the planet.
    2. Relevant equations
    ##a=\frac{GM}{r^2}##
    ##\delta PE = -\delta KE##

    3. The attempt at a solution

    I tried finding an expression for ##\int_x^∞ a## ## dt## by letting it equal to ##v \frac{dv}{ds}## (since there is no ##t## in the formula for ##a##), but was unable to find an expression for ##\frac{dv}{ds}## as well(s=displacement).

    Then I tried using the law of conservation of energy to solve it, and I used the equation ##KE = \frac{GMm}{x} = \frac{1}{2}mv^2## (deriving it from the second relevant equation) and came to the expression ##v = \sqrt{\frac{2GM}{x}}##, but I have my doubts about this.
    As to the second part, I'm lost.

    Any help is appreciated.
     
  2. jcsd
  3. Jan 16, 2015 #2

    Orodruin

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    Your approach with energy conservation is reasonable. You could in principle use something like the first, but that would amount to essentially the same as solving the differential equation for the scenario and it is much easier to use energy conservation.

    For part 2, why dont you just compare the velocities?
     
  4. Jan 16, 2015 #3

    BiGyElLoWhAt

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    Well here's my deal with this. Your conservation of energy equation uses x as a distance traveled from infinity, which is not necessarily the same as the separation distance of the two masses. Part 2 I think is a trick question. You could possibly end up with a circular orbit if m started with some non zero velocity which was not directed radially toward the center of M, but I just see m and M crashing together.

    I agree with orodruin on part 1, I would probably try CoE, but I still need to think about how I would actually approach it. I still think part 2 is a trick question. You need some non radial force acting on m.
     
  5. Jan 16, 2015 #4

    BiGyElLoWhAt

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    I guess for 2 I just kind of automatically assumed M was not accelerating and was not in relative motion perpendicular to the radial line between the two masses.
     
  6. Jan 16, 2015 #5

    haruspex

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    I think the equation was used correctly. The distance travelled from infinity would be infinity.
    I agree.
     
  7. Jan 16, 2015 #6

    gneill

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    The original problem statement says that M > m, not that M >> m. Are we to take it that both masses will be accelerating towards each other? If so, a CoE approach will have to take this into account, and the frame of reference in which Vo "measured" needs to be specified.
     
  8. Jan 16, 2015 #7

    haruspex

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    Yes, that bothered me. But then, why mention M > m at all?
     
  9. Jan 16, 2015 #8

    gneill

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    Perhaps to break the symmetry of the problem so one would have to consider a bit more deeply how the KE is to be divided between the masses?
     
  10. Jan 16, 2015 #9

    Orodruin

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    It bothered me too. I took it as M >> m and tried to be as brief and to the point as possible not to confuse the OP further with the details that arise if you look at what the actual quoted problem was.

    I did the same for problem 2, obviously there needs to be a non-zero angular momentum for the masses to have circular orbits and without any non-radial force, the orbit is either a hyperbola, ellipse, or parabola (or as in the case with zero angular momentum, a line). However, the formulation of the problem suggests that you can argue for the non-circularity of the orbit just from the velocity at a given radius (i.e., the circular orbit velocity is lower than the escape velocity by a factor ##\sqrt 2##).
     
  11. Jan 17, 2015 #10
    Thank you all for replying. I'm sorry, I forgot to one thing that was there in the original question: the masses never collide.
    Additionally, there was a typo by me - it is ##M>>m## , not ##M>m##. Since this is the case, I believe that the kinetic characteristics of ##M## are supposed to be ignored for this question(please correct me if I'm wrong). There was an accompanying diagram with the question, which I unfortunately do not possess at this moment. I can, however, provide a description:
    ##M## was drawn out to be a planet, the separation between ##M## and ##m## was labelled as ##x##, and a vector arrow for ##V_0## was drawn from ##m## that was not pointed straight towards ##M## (it was in the same general direction though, and I can reasonably estimate the angle between the vector and a line joining the centers of the masses to be not more than ##\frac{π}{4}## ##rad##). No other object was drawn other than these two masses. Also, the answer space for part b) had quite a few lines provided (it was for 4 marks), so I think it requires a fair bit of mathematical explanation.
    The question provided no further details.
    @BiGyElLoWhAt I think you misunderstood something - ##x## is just used to denote the general separation distance between the masses.
    @gneill Frames of reference are not specified; I believe the motion of the masses are being considered from the point of a hypothetically stationary observer.

    I tried comparing the velocities, and as Orodruin correctly pointed out, their relationship appears to have a factor of ##\sqrt{2}## . If I let the velocities be equal to each other, then ##x=2r## , and I have no idea what this could imply. I still didn't get one thing - is my expression for ##V_0## correct (for part a) )?
     
  12. Jan 17, 2015 #11

    Orodruin

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    Do not stare yourself blind on letters. Think about what they represent.
     
  13. Jan 17, 2015 #12
    The velocity of ##m## is never equal to the velocity required to maintain circular motion at any particular distance...?
     
  14. Jan 17, 2015 #13
    I must add that all this is being considered in two dimensions.
     
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