1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Escape Speeds
  1. Escape Speed (Replies: 1)

  2. Escape speed (Replies: 2)

  3. Escape speed (Replies: 2)

  4. Escape speed (Replies: 3)

  5. Escape Speed (Replies: 2)

Loading...