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Gravitational Fields - Spacecraft approaching planet

  1. Feb 17, 2016 #1
    Hi, as part of my physics assignment, I have the following question:

    As a spacecraft approaches a planet, the following measurements of velocity and radius were taken. From these values, a graph of v2 (y axis) and 1/r (x axis) was plotted. Use this graph to obtain a value for the mass of the planet.

    Now I have used the graph to obtain a gradient however I am unsure as to what to do with this. As the gradient is v2 x 1/r then this equates to m2s-2m-1 which simplifies to ms2. Now I am assuming that this is acceleration, however I have no idea how to use this to calculate mass.

    I know that I need to get a force value from somewhere, however I keep confusing myself using circular motion equations and I know that is not what I need.

    I am thinking I need to use potential energy and kinetic energy and probably gravitational potential to get my answer, but I am constantly confusing myself.

    I don't want the answer, I want to work it out for myself but I need a push in the right direction or someone to tell me I am going down the wrong road.

    Any help would be greatfully received.

    Thank you
     
    Last edited by a moderator: Feb 17, 2016
  2. jcsd
  3. Feb 17, 2016 #2

    berkeman

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    Welcome to the PF.

    Please be sure to use the Homework Help Template that you are provided when starting a new thread here in the schoolwork forums. It will help to organize your post to make it easier for us to help you.

    One section of the HH Template asks that you list the Relevant Equations for the problem. Can you please list them for us? You have alluded to some of them, but it helps to write them down explicitly. What is the equation for the gravitational attraction force between two masses versus the distance apart? What are the equations for the kinetic and potential energy of a mass as it approaches a much larger mass under gravitational acceleration?...
     
  4. Feb 17, 2016 #3
    My thoughts equations wise are:

    F=GMm/r2
    K.E.=1/2mv2
    Ep=dV/dR
    Vg=GM/r

    My main thinking was that 1/2mv2=GMm/r which simplifies to v2=GM/r
     
  5. Feb 17, 2016 #4

    berkeman

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    Great! So what do you get for the slope of your graph then? :smile:
     
  6. Feb 17, 2016 #5
    The slope of the graph gives me a value but I thought this was acceleration as it is v2 x 1/r. As v2 is m2s-2 and 1/r is m-1 this gives me a value which is ms-2 which is acceleration. Now I know this could also be g but it would not be a constant and therefore I really don't know where to turn.
     
  7. Feb 17, 2016 #6

    berkeman

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    It's G, which is indeed a constant. g is not a constant, and depends on distance...
     
  8. Feb 17, 2016 #7
    That is what I thought. So I have my constant of v2 x 1/r which I believe is the acceleration of the spacecraft itself.

    But as I have no force values and no mass values, I can't see where to use it.

    Is it as simple as my gradient = GM son therefore gradient/G = mass of planet?
     
  9. Feb 17, 2016 #8

    gneill

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    Consider that the total mechanical energy of the system must be constant. If the mass of the spacecraft is negligible compared to that of the planet then it's convenient to look at what's called the "specific mechanical energy" which is really just the energy per unit mass of the spacecraft, like J/kg, and is at heart just the sum of the kinetic and potential energy divided by the mass of the craft.

    ##\xi = \frac{v^2}{2} - \frac{GM}{r}##

    M is the mass of the planet, G is the universal gravitational constant, and ##\xi## is the specific mechanical energy.

    Does this give you any ideas?
     
  10. Feb 17, 2016 #9
    So by mechanical energy, you are referring to potential and kinetic energy, as potential energy decreases, kinetic energy increases (as the spacecraft approaches the planet) but the two sums remain the same?

    Am I understanding you correctly?
     
  11. Feb 17, 2016 #10

    gneill

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    Yes.
     
  12. Feb 17, 2016 #11
    I am still confused here.

    Looking at your equation, I still have 2 unknown variables which are M and ξ.

    From calculating the gradient, I have a value that I am still not certain as to what it is for, or even necessary.

    I know I am missing something blindingly obvious, but as much as I look I confuse myself further. I know this is not particularly difficult but I cannot for the life of me see where I am supposed to go from here.

    I can't calculate ξ without knowing either the gravitational potential or mass of the planet and I don't see any equation which I can manipulate to get there, even though I know there is one.
     
  13. Feb 17, 2016 #12

    gneill

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    Start by rearranging the equation to isolate ##v^2##. You should end up with something that looks a lot like the equation of a line: y = mx + b, if you interpret ##v^2## and ##\frac{1}{r}## appropriately. After all, you do have a plot of ##v^2## versus ##\frac{1}{r}##, right? What's the slope term in your version?
     
  14. Feb 17, 2016 #13
    Ok, so the slope is my gradient.

    v2=GMr-1+2ξ

    (It is late and I am very tired so have I rearranged this correctly?)

    If so, GM is my gradient so I can say that GM = v2 x 1/r ? Am I on the right path here?

    That would mean that M=v2 x r-1/G correct?
     
  15. Feb 17, 2016 #14

    gneill

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    You've dropped a factor of 2 on the first term in your rearrangement.

    You should have the slope already, as measured on your plot. Just equate that slope with its corresponding expression from the equation.
     
  16. Feb 17, 2016 #15
    So it's v2=2GMr-1+2ξ

    Therefore 2GM=gradient so M=gradient/2G?
     
  17. Feb 17, 2016 #16

    gneill

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    Sounds about right :smile:
     
  18. Feb 17, 2016 #17
    Thank you so much for all of your help. It has been greatly appreciated!
     
  19. Feb 18, 2016 #18

    vela

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    You made a mistake with the units. Slope is ##\Delta y/\Delta x##, not ##\Delta y\times\Delta x##, so the units of the slope are going to be (m/s)2/(1/m) = m3/s2. If you work out the units of GM, you'll see they match.
     
  20. Feb 18, 2016 #19
    Thank you Vela, you are absolutely right. I noticed that this morning.

    Thank you to everyone for your help on this.
     
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