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Kepler's 3rd Law vs. Newton's second for Satellite

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data

    We are to place 1000 kg satellite in circular orbit 300 km above earths surface...find speed and period...

    2. Relevant equations

    [tex]F_g=ma_c=\frac{GMm}{R^2}[/tex]

    [tex]a_c=\frac{v^2}{R}[/tex]

    Kepler's Third Law and Constant for Earth:

    [tex]K_E=2.97*10^{-19}s^2/m^3[/tex]

    [tex]T^2=K_Ea^3[/tex]

    3. The attempt at a solution

    This is all very straight forward, except the values I obtain are not consistent using both methods. I can use newtons second law for centripetal motion to solve for v obtaining:

    [tex]v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{6.67*10^{6}}}≈7.7*10^3 m/s[/tex]

    R is the earth radius PLUS the height above the earth's surface. The period:

    [tex]T=\frac{2\pi r}{v}≈5.4*10^3 s[/tex]

    If I apply Keplers 3rd law, I have:

    [tex]T=\sqrt{(2.97*10^{-19})(6.67*10^6)^3}≈9.4 s[/tex]

    I can't tell where I'm going wrong here. The fast orbit is preposterous. My instructor accepted the value on my homework I obtained using second law; but I'm just reviewing the homework and tried the Kepler equation and it doesn't work...it should work though because Kepler's third law can be derived from the inverse square law of orbits.
     
    Last edited: Jan 30, 2012
  2. jcsd
  3. Jan 30, 2012 #2
    How did u get the acceleration? Did you include the earths radius plus 300km?
     
    Last edited: Jan 30, 2012
  4. Jan 30, 2012 #3
    Yes, the radius plus the altitude is [tex]6.67*10^6[/tex]. I didn't calculate the acceleration, just the velocity and then the period. The acceleration is just centripetal.
     
  5. Jan 30, 2012 #4
    My best guess is that Kepler's 3rd law applies to an elipse (http://en.wikipedia.org/wiki/Semi-major_axis), where as what you're doing is plucking the value of a circular radius directly into Kepler's 3rd law.

    Whether I'm right or wrong, I'm equally interested to hear from someone who can explain why.
     
  6. Jan 30, 2012 #5

    D H

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    That is the Kepler constant for the Sun, not the Earth.
     
  7. Jan 30, 2012 #6
    [tex]\text{That shouldn't be } 2.97*10^{-19}. \text{ Kepler's 3rd law is }\frac{r^3}{t^2} = \frac{Gm}{4\pi^2}[/tex]

    [tex]\text{Rearranging } T = \sqrt{\frac{4\pi^2((6.67*10^6)^3}{(6.67*10^{-11})(6*10^{24})}}[/tex]

    [tex]T = \sqrt{4\pi^2((6.67*10^6)^3(4.002*10^{-14})}[/tex]

    [tex]T = \sqrt{(6.67*10^6)^3(1.58*10^{-12})}[/tex]

    I think that should cover your mistake
     
    Last edited: Jan 30, 2012
  8. Jan 30, 2012 #7
    Ok; in my book there is a table "useful planetary data" and I just grabbed it from there. I should have tried calculating it.

    Looking at the data though, I see the values are nearly identical for all the planets so I guess the author just took the square of the orbital period and divided it by the cube of the mean distance from the sun to obtain slightly different values of K FROM each planet. I guess the table was just trying to show that you get the same value of K from the periods and mean distances for all planets. I mistook these values as K values for the planets themselves.

    If I use 9.91 * 10^{-14} it works out, about 5400 seconds....

    It's valid for circular and elliptical orbits. The equation is the same infact, you just replace the semi major axis "a" with the radius "r."
     
  9. Jan 30, 2012 #8
    I think there's an error in there, although you're right about the K value being wrong in my original calculation, I calculated the real value to be something like 9.91*10^{-14}
     
  10. Jan 30, 2012 #9
    It is likely I made a mistake with it, just did it quickly
     
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