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The precision in measurements of radial velocities by DE?

  1. Sep 23, 2016 #1
    Hello, I have an exercise here that I need help with.

    The precision in measurements of radial velocities by the Doppler effect is currently 1 m/s. Can a Jupiter like planet orbiting a star similar to the Sun at a distance from the mother star equal to the Sun-Jupiter distance be detected? (Use www or other sources to find the mass of Jupiter, the Sun and the distance between the two which are the only data you are allowed to use).

    If found the following variables:

    mJup = 1.9*1027kg
    d = 7.78*108 m
    mSun = 1.99 * 1030 kg

    Relevant equations:

    γ - Gravitational constant
    ms - Mass of Star
    mp - Mass of planet
    P - period
    vsr - Radial velocity of star

    mp*sin i = ((ms) *vsr*p1/3)/((2*γ*)1/3)

    P = √((r^3*4*π^2)/(γ*mS))


    I found:

    P = 1.52*10-5 s

    Assumes that i = 90 °:

    vsr = ((mp*(2*π*γ)1/3)/((ms 2/3) * (P1/3))

    vsr = 3.62 * 105 m/s

    This is clearly not the correct answer. The correct answer is vsr ≈ vs = 12.2 m/s
  2. jcsd
  3. Sep 23, 2016 #2


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    1.5 * 10-5 s would be 15 microseconds for an orbit...
  4. Sep 23, 2016 #3
    I also meant r = d (distance between sun and planet)

    Yeah, I forgot r3

    Here is equation (1) and (2)written in LATEX

    $$ P_{star} = \sqrt{ \frac{r^{3}4 \pi^{2} }{ \gamma m_{s} } } $$

    $$ v_{rs} = \frac{m_{p}(2 \pi \gamma )^{ \frac{1}{3} } }{m_{s}^{ \frac{2}{3} }p^{ \frac{1}{3} } } $$



    So the period was P ≈ 11835 seconds ≈ 3.3 hours
    Now I got vsr ≈ 395 m/s
    Still far from 12 m/s
    Last edited: Sep 23, 2016
  5. Sep 23, 2016 #4


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    3.3 hours for the Jupiter orbit is still way too short.
    You might also want to check the value you used as distance. Especially the unit.
  6. Sep 24, 2016 #5
    I believe it's the period of the sun around the sun- jupiter center of mass system( a little outside the sun). Vrs is the radial velocity of the sun around this center of mass.

    I only used SI - units (m/s, s, kg)
  7. Sep 24, 2016 #6


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    Quick consistency check: Earth needs one year for the orbit. Jupiter can be seen during the night, so it has to be more distant than Earth, therefore its orbital period has to be longer than a year. And certainly longer than 3.3 hours!

    You can also look up its orbital period - it is about 12 years.
    Or look up its distance: 7.78*108 km.
  8. Sep 27, 2016 #7
    I found vs by using eq (3):

    $$ v_{p} = \sqrt{\gamma m_{s}} \approx 13061 m/s $$

    then using (4)

    $$ v_{s} = \frac{m_{p} v_{p}}{m_{s}}\approx 12 m/s $$

    But vs is ≈ vrs, but still not vrs

    But couldn't I use a different method with eq(1) and eq(2) to find vrs?
  9. Sep 27, 2016 #8


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    The difference between Jupiter's speed and the relative speed is just 0.1%. You can use the reduced mass to take this small difference into account, or take into account that the radius of the Jupiter orbit is smaller than the distance between Jupiter and sun. But you would not really gain precision with that, because the other planets get relevant at that level, they influence the position of the sun as well.
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