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Third cosmic velocity

  1. Mar 7, 2009 #1
    ... is the minimal speed (relative to the Earth) which must be imparted on a body resting on the Earth's surface to escape the Solar (and Terrestrial) gravity. So, if one uses the conservation of energy (neglecting the Earth's rotation about its axis), one has

    1/2 m (v_3 + v_o)^2 = G m (M_s / R_s + M_e / R_e), where

    m - mass of the body
    M_s/e, R_s/e - mass of the Sun (Earth) and distance from its center to the body
    v_3 - third cosmic velocity
    v_o - the speed at which the Earth orbits the Sun
    G - gravitational constant

    Now, this equation gives v3 = 13 km/s, while all the sources cite v3 = 16 km/s. Am I missing something?

  2. jcsd
  3. Mar 8, 2009 #2


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    Shouldn't your R_s be the distance of the Earth from the Sun?
  4. Mar 8, 2009 #3
    Yes, R_s is the distance form the Sun to the Earth. Anyhow, all other refrences give completely different expression than mine (e.g. Irodov - problem circa 1.230). I think that my energy calculation is ok. Or am I missing something?
  5. Mar 8, 2009 #4


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    The sum v_o+v_3 should be a vector sum. Your answer is for the two velocities parallel.
    There is a range of answers for v_3, depending on the angle between them.
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