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Homework Help: Escape Speeds

  1. Nov 22, 2005 #1
    Question 1: If a projectile is launched from Earth with a speed equal to the escape speed, how high above the Earth's surface is it when its speed is 96% of the escape speed?
    On this question i understand i am not finding how high it would go if it were launched at 96% of the escape velocity, but rather that when it gets to 96% of the escape speed. I am unsure of how I am to find this... and I would like to atleast get a concept for this one... which will help me with question 2 as well. I believe it will deal with energy conservation, but i cannot think of a way that i can involve velocity with this. Can I get atleast an intuitive way to help lead me to get somewhere with this? :confused: If I can get this question I can surely get question 2.
    Question 2: A projectile is launched vertically from the surface of the Moon with an initial speed of 2300 m/s. At what altitude is the projectile's speed 55% of its initial value?
     
  2. jcsd
  3. Nov 22, 2005 #2
    Gravity will act on the projectile. So what is the formula for its final velocity in terms of t, its initial velocity and the acceleration?
     
  4. Nov 22, 2005 #3
    if a projectile is launched from earth at escape speed, then its total energy is equal to 0 ...

    K + U = 0

    from this relationship, escape speed is calculated as v = sqrt(2GM/R)
    where G = grav. constant, M = earth mass, and R = earth radius.

    I'm assuming that you already know all the above, but the point is this ...
    during its entire trip, total energy remains a constant equal to zero.

    let v = escape velocity, k = .96, and r = variable distance from earth's center

    (1/2)m(kv)^2 - GMm/r = 0

    since escape velocity v = sqrt(2GM/R)

    (1/2)m*k^2*2GM/R - GMm/r = 0

    solving the above equation for r yields ...

    r = R/k^2

    where r = distance from the earth's center when it is at 96% of escape speed
     
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