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Satellite Velocity

  1. Dec 18, 2016 #1
    1. The problem statement, all variables and given/known data
    On the surface of a distant planet, objects experience an acceleration due to gravity of 20 m/s2. The distance from the center of the planet to its surface is 8.0 * 106 m. Find the velocity of a moon that is circular orbit about the planet at a height of 300,000 m above the planet's surface.

    2. Relevant equations
    a = v2/r

    3. The attempt at a solution

    20 = v2/(8e6 + 300,000) = 12,884.1

    I know this isn't right because my teacher said the acceleration due to gravity should not be 20. But I don't know how to solve it another way. Could anyone shed some insight?
     
  2. jcsd
  3. Dec 18, 2016 #2

    Bandersnatch

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    The acceleration is caused by the gravity of the planet. Can you write the equation for gravitational acceleration at the specified height?
     
  4. Dec 18, 2016 #3
    The only formula I can think to use would be a = v2/r which would leave me with a = v2/ (8e6 + 300,000) and then I'm still completely lost
     
  5. Dec 18, 2016 #4

    Bandersnatch

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    What equations were introduced when you talked about gravity in class? Look in your notes, or use the internet. Here's a hint: it's all Newton's idea.
     
  6. Dec 18, 2016 #5
    I suppose you mean the F = GMm/r2 equation? My teacher doesn't really teach so it's mostly us students trying to guess at things. So it would become:
    GM/r = v2 leaving me with (6.67e-11)(M)/(8e6+300,000) = v2?

    I'm sorry thank you for your patience
     
  7. Dec 18, 2016 #6
    Actually could you solve it so that:

    a= v2/r
    20= v2/8e6
    v2= 160,000,000
    v= 12,649.1 m/s

    GM/r= v2
    (6.67e-11)(M)/8e6 = 12649.12
    6.67e-11M= 1.28e15
    M = 1.92e25

    GM/r = v2
    (6.67e-11)(1.92e25)/(8e6+300,000)= v2
    154293975.9 = v2
    12421.5 m/s = v

    Was this what you were getting at? I was staring at it a little more and the idea came to mind but I'm not really sure if it works out.
     
  8. Dec 18, 2016 #7

    Bandersnatch

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    That's pretty much what I had in mind, yes. :)
    Looks about right. I'll go over the numbers later, though, I have to go now. But should be fine.
     
  9. Dec 18, 2016 #8
    No worries! Thank you so much for your help!!
     
  10. Dec 18, 2016 #9

    Bandersnatch

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    Yup, numbers look good too.
    Myself, I'd probably write ##ma=\frac{GMm}{R^2}##, get the mass of the planet from there, and then plug it into ##V=\sqrt{\frac{GM}{R+r}}##. Saves you an equation, and is more straightforward reasoning-wise.

    Do you understand why you used those equations?
     
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