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Escape velocity

  1. Jan 2, 2005 #1

    DB

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    I can't seem to obtain the right answer using the escape velocity formula. I think my problem is with the gravitational constant. I would like to find the escape velocity of the sun.

    [tex]v_e=\sqrt{\frac{2GM}{r}}[/tex] or [tex]v_e=\sqrt{\frac{2\mu}{r}}[/tex]

    [tex]v_e_\odot=617.54(km/s)[/tex]

    [tex]v_e_\odot=\sqrt{\frac{2*6.673e-11*1.9891e30(kg)}{696000(km)}}[/tex]

    [tex]v_e_\odot=\sqrt{~3.8141e14}\neq 617.54(km/s)[/tex]

    What am I doing wrong?
    Thanks
     
  2. jcsd
  3. Jan 2, 2005 #2

    dextercioby

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    Okay,i'll make some approximations,but i'll get the result and the order of magnitude.
    [tex]v_{esc.,Sun}\sim\sqrt{\frac{4\cdot6.673}{696}}\sqrt{\frac{10^{30}10^{-11}}{10^{6}}}\frac{m}{s}\sim \sqrt{0.0383}\sqrt{10}\cdot 10^{6} \frac{m}{s}\sim 619278\frac{m}{s}\sim 619.3\frac{Km}{s} [/tex]

    Using more precise values for radius ad Sun's mass will give about [itex] 617.54\frac{Km}{s} [/itex].

    Daniel.

    PS.I don't know how u got that figure under the radical.
     
  4. Jan 2, 2005 #3

    DB

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    Thnx. I see what youve done but don't understand why. How come you broke it down to 2 factors:

    [tex]v_{esc.,Sun}\sim\sqrt{\frac{4\cdot6.673}{696}}\sqrt{\frac{10^{30}10^{-11}}{10^{6}}}\frac{m}{s}[/tex]

    And how come you multiplied by 4 instead of 2?
    Thnx
     
  5. Jan 2, 2005 #4

    pervect

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    Here's the easy way to calculate it using google calculator

    The mistake in your calculation is that the value of G you use is in MKS units
    G = 6.67300 × 10^-11 m^3 kg^-1 s^-2

    and you put in the radius of the sun in km, not meters.
     
  6. Jan 2, 2005 #5

    dextercioby

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    Since
    [tex] \sqrt{ab}=\sqrt{a}\sqrt{b} [/tex]
    ,i could break it into 2 square roots,putting the powers of 10 aside.
    That '4' appears from the product between 2 (initially in the formula) and 2 (from the Sun's mass).

    Daniel.
     
  7. Jan 2, 2005 #6

    DB

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    Ahh, ok 1.9891
     
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