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Grade 12 Gravitational Force Fields Questions.

  1. Jan 21, 2012 #1
    1.Nasa fires a 1 tonne space probe, from a stable orbit of altitude 13,620km using a giant rail gun. The rail gun provides the probe with 9.97165*10^10 J of additional kinetic energy. Determine if the probe has enough velocity to escape the earth's gravity well. The Earth's mass and radius are 5.98*10^24kg and 6380km respectively.
     
  2. jcsd
  3. Jan 21, 2012 #2
    Hi TheSerpent,
    The formula for escape velocity is v=√(2GM/r) where v= velocity, G=universal gravitational constant (G=6.67*10-11 Nm2kg-2), M is the mass of the body, and r is the separation distance.

    v=√(2GM/r)
    Ekinetic=0.5mv2

    m (rocket)=1000 kg
    r (orbit)=13620 km
    Ekinetic=9.97165*1010 J
    m (earth)=5.98*1024 kg
    r (earth)=6380 km

    First, find the velocity that the rocket has...
    Ekinetic=0.5mv2
    9.97165*1010 J = 0.5(1000 kg)v2
    v=√((2*(9.97165*1010 J))/1000 kg)
    v=1.412*104 m/s

    Then, find the velocity that the rocket would need to escape...
    v=√(2GM/r)
    v=√(2GM/rorbit+rearth)
    v=√((2*(6.67*10-11 Nm2kg-2)*(5.98*1024 kg))/(13620 km + 6380 km))
    v=1.997*105 m/s

    Compare the two -- the rocket will not have sufficient velocity to escape.
    Good luck! I hope this helps.
     
  4. Jan 21, 2012 #3

    gneill

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    Staff: Mentor

    First determine the initial velocity of the space probe in its original orbit (prior to receiving a boost from the rail gun). Remember that you're given its altitude, not its orbit radius.

    From that velocity determine what the escape velocity would be; escape velocity has a particular relationship to the circular orbit velocity. Then determine what the new velocity would be given the rail gun boost of energy. Compare to the escape velocity.
     
  5. Jan 21, 2012 #4

    gneill

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    Staff: Mentor

    The rocket begins in a circular orbit at the given altitude. Therefore it will have an initial velocity and kinetic energy appropriate for a circular orbit at that location. Next additional energy is supplied via a rail gun boost. The probe will gain that energy as KE and thus have a new velocity. Compare to the escape velocity for that orbital radius.
     
  6. Jan 21, 2012 #5
    Thanks for the correction!
     
  7. Jan 21, 2012 #6
    So what's going to be changed in the calculation process (how do you find the new velocity?)
     
  8. Jan 21, 2012 #7

    gneill

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    Staff: Mentor

    The rail gun adds kinetic energy. Velocity depends upon kinetic energy.
     
  9. Jan 21, 2012 #8
    Find orbital velocity:
    v = √(GM/r)
    v = √((6.67*10^-11)(5.98*10^24)/(13620000 + 6380000)
    v = 4.466 * 10^3 m/s

    Find orbital kinetic energy:
    K = mv^2/2
    k = (1000)(4.466*10^3)^2/2
    K = 9972578000 = 9.973 * 10^9 J

    Find escape velocity:
    v esc = √2 v
    v esc = √2 (4.466 * 10^3)
    v esc = 6.316 * 10^3 m/s

    How would you include the energy boost from the rail gun into the answer, that is what I am not sure of.
     
  10. Jan 21, 2012 #9

    gneill

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    Staff: Mentor

    Okay! Good to here!
    Kinetic energy adds. Add the boost energy to the energy k that you found above. Find the new velocity from that.
     
  11. Jan 21, 2012 #10
    Find new velocity:

    K = mv^2 / 2
    (9.973 * 10^9)+(9.97165 * 10^10) = (1000)v^2 / 2
    v = √(2)(1.096895*10^11)/1000
    v = 14811.44827 = 1.481 * 10^4 m/s

    This is the actual velocity so therefore the rocket will have sufficient velocity to escape.
     
  12. Jan 21, 2012 #11

    gneill

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    Yup. That looks good.
     
  13. Jan 21, 2012 #12
    Thank you for your help!!!! Can you also help us in our other post you commented on? We posted an idea but unsure!
     
  14. Jan 21, 2012 #13

    gneill

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    Staff: Mentor

    You're welcome.

    I see that you have two threads, this one and one other. I've made a suggestion for you to work on in that other thread.
     
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