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An object is launched from the Earth to escape the Sun

  1. Dec 2, 2017 #1
    1. The problem statement, all variables and given/known data
    (a) What is the escape speed from the sun for an object in the Earth's orbit (of orbital radius R) but
    far from the Earth? (b) If an object already has a speed equal to the Earth's orbital speed, what
    additional speed must it be given to escape as in (a)? (c) Suppose an object is launched from Earth
    in the direction of the Earth's orbital motion. What speed must it be given during the launch so
    that when it is far from Earth, but still at a distance of about R from the sun, it has that additional
    speed calculated in (b) and thus can escape from the sun?
    upload_2017-12-2_18-32-22.png
    2. Relevant equations
    upload_2017-12-2_18-33-8.png
    upload_2017-12-2_18-35-22.png

    F = ma

    3. The attempt at a solution
    I was able to find the answers for parts (a) and (b), but am completely unsure of where to go for part (c). At first, I thought that I needed to find the distance away from the Earth at which the object has a speed of 12.3 km/s (obtained from part (b)), and then use that distance to calculate the velocity of launch. When I did that, I just ended up with the Earth's orbital speed, as calculated in part (b).

    1. F = ma = mv^2/R = G * M(sun) * m/R^2
    2. v = (G*M(sun)/R)^1/2 = 29.8 km/s

    Once I did that, I realized I must not be on the right track. I'm completely unsure of where to begin!
     
  2. jcsd
  3. Dec 2, 2017 #2

    mfb

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    Staff: Mentor

    The Sun is irrelevant for part (c) (apart from using the result from (b)) as you only consider the motion close to Earth. If you launch an object with speed v from the surface of Earth, what is its speed far away from Earth?
     
  4. Dec 2, 2017 #3
    Would it be v = sqrt(G*M(earth)/R) where R is the distance between the center of the Earth and the object?
     
  5. Dec 2, 2017 #4

    mfb

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    That is related to (but not equal to) the escape velocity, the minimal velocity to get it away from Earth. But then it has "zero velocity" far away from Earth, not the 12.3 km/s it needs to escape from the Sun.

    It is also not an answer to my question, which should be a velocity as function of the unknown v.
     
  6. Dec 2, 2017 #5
    Then I'm just not sure of what its speed far away from Earth would be.
     
  7. Dec 2, 2017 #6

    mfb

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    Staff: Mentor

    Think of how the escape velocity is derived. Which conservation law is used?
     
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