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Graphical analysis work problem

  1. Apr 4, 2013 #1
    1. Problem statement and all known/given variables

    2sbr47q.jpg

    2. Related equations
    W = fd
    Newtons kinematic equations

    3. Attempt at solution

    For part a) since the magnitude of the work done by gravity to 2 a.u is 2500GJ I use the equation W = fd and I solve for f by dividing work by the distance 2500 = f(2a.u) (after I convert the units).

    For part b) I just looked at the graph and saw that the change in gravitational energy was 4500GJ so I'm assuming that's what they are asking for?

    For part c) since the change in work done by gravity at 6.5au is 6000, i use the equation w = fd, setting w = 6000, change f to ma, solve for a, plug that acceleration into newtons kinematic equation (vf^2 = vo^2 + 2ad) with vo being 0 , the distances cancel out and I'm left with vf = sqrt(2a) .. or see below

    bi5ohe.jpg


    Thanks for any input!
     
  2. jcsd
  3. Apr 5, 2013 #2
    For part A, I think it would be more appropriate if you were to find the slope of the graph at distance =2 A.U. Because the infinitesimal work done by gravity dW when the satellite is moved from 2 A.U. by a very small distance dx is dW = F.dx and hence F=dw/dx at that point (which is simply the slope)

    Part B is all right.

    Part C has a minor glitch. You see, in order to go to 6.5 A.U. the satellite must overcome a potential energy barrier of more than 7000 GJ at around 5 A.U. If the velocity cannot overcome that potential energy barrier (i.e. If you don't provide it enough kinetic energy to cross 5A.U.) the satellite will never be able to go across to 6 A.U and hence it cannot escape.
     
  4. Apr 5, 2013 #3

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    The way I interpret the graph, freshcoast, you are off by 500 GJ for part B's magnitude.

    Secondly, is that value the work done on the satellite by gravitational forces, or is it the other way around? (Is the change in the satellite's position the same direction as the force, or in the opposite direction?) (Still in other words, pay attention to the sign. If gravitational forces are pushing the satellite in the same direction the satellite moves, the work done by the gravitational forces is positive. If the gravitational forces pull the satellite in the opposite direction the satellite moves, the work done by the gravitational forces is negative.)
     
  5. Apr 5, 2013 #4
    Oh right. Sorry about that. I got the sign of the work done wrong.
     
  6. Apr 5, 2013 #5
    Oh I see thank you guys for your help, so for the first part I just find the slope from the starting point to the point at 2au, part b the work would be in the positive direction since it went from negative work to positive work and for c I pretty much use the same equation as I did above but use the work at 7000gj?
     
  7. Apr 5, 2013 #6

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    Yes, the slope is the important part. But pay special attention to the direction.

    If the force were uniform, you could say [itex] W = Fd [/itex]. But in this problem, the force is anything but uniform. But over a small enough [itex] \Delta d [/itex] you can approximate a linear portion of the curve to find [itex] \Delta W [/itex]. So even though the forces are not uniform, you can still claim that [itex] \Delta W = F \Delta d[/itex] for a small range [itex] \Delta d [/itex]. Solve for the force.

    And once again, pay special attention to the direction. Are [itex] F [/itex] and [itex] \Delta D [/itex], over this particular portion of the curve, in the same direction or opposite directions?
    You might want to rethink that.

    The graph is plotting the satellite's gravitational potential energy. As the satellite traves from 3 to 3.5 AU, it gains potential energy.

    • What does that tell you about the satellite's kinetic energy? (Think conservation of energy.)
    • And what does the satellite's kinetic energy tell you about the satellite's speed? (Is it speeding up or slowing down?)
    • What does that tell you about the direction of force compared to the direction of motion?

    It looks to me though that the peak is not 7000 GJ, but something more like 7250 GJ.

    In your previous attempt on part c), you had kinematics equations for uniform acceleration. But like I alluded to previously, the satellite's acceleration is not constant. It's not even close to constant (it's all over the place). Your approach still works though because you found the average acceleration.

    Conservation of energy still applies though. Perhaps a more graceful approach is to realize that the satellite's initial kinetic energy is [itex] KE = \frac{1}{2}mv_0^2 [/itex] and equate that to the necessary gravitational potential energy to overcome. Then solve for v.

    Whatever the case, yes, your equation of [itex] v = \sqrt{\frac{2 W_{workgrav}}{m}} [/itex] is correct.
     
  8. Apr 5, 2013 #7

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    By the way, here's a bit of advice about the sign (positive or negative) of work.

    If you are pushing pushing an object -- applying a force in a given direction -- and a component of object's position moves along that direction, you are doing positive work.

    On the other hand, if a competent of the object's position moves in the opposite direction to the force you are applying, you are doing negative work. The object is doing work on you!

    If work was easy, they wouldn't call it work! :smile:
     
  9. Apr 5, 2013 #8
    I see, its starting to make more sense. So for part a) I found the slope from point 0.5au to 2au which gave me a positive term meaning that the work being done on the satellite by the center of star group is pushing the satellite away from the center.

    Part b) since energy is conserved, I set the change of potential energy equal to the change in kinetic energy which I can set to the work equation of fd and I solved for f which yielded a negative term where I can interpret as a force from the satellite pushing towards the center.

    Part c) to use the conservation of energy, I know that the initial potential energy would be 0 but would that also mean that this initial kinetic energy is 0? If so, then I try to solve for kinetic energy but it gives me a negative number under the radical (-PEf) = 1/2mvf^2
     
  10. Apr 5, 2013 #9

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    You should re-think this one again too.

    When the satellite is at 2 AU, and increases a tiny amount away from the center, its potential energy increases. So what does that say about its kinetic energy (conservation of energy again) as it moves away from the center in this region?

    In other words, when the satellite is around the vicinity of 2 AU and moves a little away from the center, will it speed up or slow down in the process?

    What direction of a force applied by gravity, on the satellite, would cause it to do that (i.e. would cause it to speed up or slow down)?

    Yes, the answer is a negative value.

    That's because the question asked for the "net work done upon the satellite by gravitational forces."

    Imagine some guy named "Gravity" pushing on the satellite as it moves from 3 to 3.5 AU. Does the satellite move (starting at 3 AU ending at 3.5 AU) in the same direction that Gravity is pushing? If so, Gravity did positive work. If the satellite moved in the opposite direction Gravity is pushing, Gravity did negative work.

    (Just remember, 'If work was easy they wouldn't call it 'work.'")

    the satellite definitely does not have an initial kinetic energy of 0. The question is essentially asking you to find the minimum initial kinetic energy (well, its minimum initial velocity that is, which is directly related). So no. The initial kinetic energy is not zero.
    Conservation of mechanical energy:

    [tex] KE_i + PE_i = KE_f + PE_f [/tex]

    In order to make it over that hump, the [itex] KE_f [/itex] must be [itex] \geq 0 [/itex] when the satellite as at location where that hump is (that big hump Sunil Simha mentioned at position slightly over 5 AU.)

    From there, solve for [itex] KE_i [/itex], and ultimately [itex] v_i [/itex]

    [Edit: fixed formatting mistake and added additional clarification at the end.]
     
    Last edited: Apr 5, 2013
  11. Apr 5, 2013 #10
    Ahh much clearer now, for part a, as potential energy increases, kinetic energy decreases which means satellite is slowing down and the force that's slowing it down is pointed towards the center

    For part b, negative work means that the gravity is trying to pull the satellite towards it but the satellite is still moving in the opposite direction

    Part c, I do conservation of energy where the final kinetic energy is 0 and i solve for initial velocity

    Here's what I got for the answers
    Don't mind the direction for part b, I forgot work was scalar

    2q857aq.jpg
     
  12. Apr 6, 2013 #11
    Looks good so far. Also remember that a conservative force acts in the direction of _decreasing_ potential energy as per the class discussion...
     
  13. Apr 6, 2013 #12

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    ... and the mass of the satellite is 735 kg (not 750 kg). :smile:
     
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