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Finding the Semimajor Axis

  1. Oct 1, 2013 #1

    VU2

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    A planet in another solar system orbits a star with a mass of 4.0 ×1030 kg. At one point in its
    orbit it is 250×106 km from the star and is moving at 35km/s. Take the universal gravitational
    constant to be 6.67 × 10−11 m2/s2 · kg and calculate the semimajor axis of the planet’s orbit.
    The result is:
    A. 79 × 106 km
    B. 160 × 106 km
    C. 240 × 106 km
    D. 320 × 106 km
    E. 590 × 106 km

    I used the formula for the orbital speed v^2 = GM [2/r - 1/a].

    Solving for a, I got 1/a=2/r-v^2/GM. Plugging in the given data I got 293x10^6 km. None of the options is relevant with mine. Can someone show me what I did wrong?
     
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  3. Oct 1, 2013 #2

    Simon Bridge

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    $$\frac{1}{a}=\frac{2}{r}-\frac{v^2}{GM}$$
    ... check your arithmetic.

    [edit]got a relation wrong.
     
    Last edited: Oct 1, 2013
  4. Oct 1, 2013 #3

    VU2

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    Simon Bridge,

    I first converted everything to meters, and my final answer came out to be 2.93x10^11 meters which equates to 293x10^6 km. I don't know what is wrong.
     
    Last edited: Oct 1, 2013
  5. Oct 1, 2013 #4

    Simon Bridge

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    You converted to meters because G has meters, and it's harder to convert G to km^2. Fair enough.

    OK - number crunching.........
    You have 293 M(km) ... which is what I got for the same method.

    Either the method is wrong or there is something wrong with the question.
    Do the units pan out for the equation you used? Try a dimensional analysis ...
     
  6. Oct 1, 2013 #5

    VU2

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    Simon Bridge,

    Yeah, for the part v^2/GM, instead of G as Nm^2/kg^2, I used 6.67m^3/kgs^2. So it will cancel the kg and s^2 leaving meters on the bottom like the part 2/r. Am I wrong, or is the online homework question is making a mistake?
     
  7. Oct 1, 2013 #6

    Simon Bridge

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    Ah - there's a typo in post #1.
    So perhaps there is an assumption in the equation you are using that does not apply here?
    Do you know that's the right equation or did you just look for one that had all the right variables in it?

    You may also want to contact anyone else doing the problem to see how they handled it.
     
  8. Oct 1, 2013 #7

    VU2

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    Simon Bridge,

    Thanks for helping. The equation I found was from a yahoo recipient, who had the same problem. I will ask my professor for assistance today.
     
  9. Oct 2, 2013 #8

    Simon Bridge

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    You should use your own class notes more.
    Make sure you understand the limitations on the equations you use.

    It is possible that there is a mistake in the question - it's just unlikely since such questions tend to get reused over the years and someone was bound to have spotted it before now. Still... we have spotted a few such mistakes in these forums. If you've found one, it should up your brownie points with your professor.
     
  10. Oct 2, 2013 #9

    VU2

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    Simon Bridge,

    I talked to my professor today, and he said I went ahead. That's why I couldn't figure it out, so I will try to attempt it again today. Thanks for your help.
     
  11. Oct 2, 2013 #10

    Simon Bridge

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    Sorry - I don't know what that means.

    Have you tried looking at the lessons you've done recently to see if the physics can tell you anything?
    Are there are particular physical principles that the orbit must follow for instance?

    But from what you've written in post #1 - I don't see why the equation shouldn't work.
    It's the Vis-Viva equation.
     
    Last edited: Oct 2, 2013
  12. Oct 2, 2013 #11

    arildno

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    Simon Bridge:
    I believe OP "went ahead", meaning that he tried to be in advance of the pace of the course. Then, some point in the textbook that will be fully dealt with in the course in due time has been a source of misunderstanding for OP
     
  13. Oct 2, 2013 #12

    Simon Bridge

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    Oh that makes sense ... I've been assuming current coursework.
     
  14. Oct 2, 2013 #13

    D H

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    Minor problems:
    • Use a superscript (250×106 km), the caret symbol "^" (250×10^6 km), or scientific "e" notation (250e6 km). We could tell from context that you did not mean 250×106=26500 km.
    • G is approximately 6.67×10-11 m3/s2/kg. You stated 6.67 × 10−11 m2/s2 · kg, in the opening post. That's just a typo, as as you later had it as m^3.

    Those are just minor problems. The answer you obtained is correct.
     
  15. Oct 2, 2013 #14

    Simon Bridge

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    Which is what I'm thinking ... there is a problem with the question.
    Either there was something in the context that changes the outcome, the question was not written out verbatim, or the choices are just plain wrong. It happens.

    Puzzling that the prof wouldn't discuss it - but that happens too.
     
  16. Oct 2, 2013 #15

    VU2

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    Simon Bridge and D H,

    Sorry for the typo and late posts, been busy with school, but what D H corrected is right. My professor told me to come back tomorrow to look over it. So I will have an answer by tomorrow night and will see what went wrong.
     
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