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Law of Universal Gravitation

  1. Mar 24, 2006 #1
    Hello all,

    I was wondering if someone could please help me with a problem. It is regarding the law of universal gravitation, this time using conservation of mechanical energy.

    We know that F = GMm/R^2

    **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 27% of the escape speed.**

    I approached this problem two ways and have stopped in the middle, because I don't know how to continue. I was wondering if someone can help me find h or the height in meters.

    First thing I tried:
    .5mv^2 - GmM/Radius of earth = .5mv^2 - GmM/h
    0 = GmM/Radius of earth - GmM/h goes to h = Radius of earth
    I stopped here

    Second thing I tried:
    .5mv^2- GmM/Radius of earth = .5m(.27v)^2- GmM/Radius of earth + h
    I do not know how to manipulate this problem to get radius of earth + h


    Thank you for your assistance!:smile:
     
  2. jcsd
  3. Mar 24, 2006 #2

    Hootenanny

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    As we are considering energy, it is more useful to use the equation for gravitation potential;

    [tex]U = -\frac{GMm}{r}[/tex]

    This gives the potential energy of an object on the surface. Now if the kinetic energy of an object is greater than or equal to the potential energy of an object the object will 'escape' from the gravitational field. Knowing this can you formulate an equation to find the escape velocity of an object?
     
  4. Mar 24, 2006 #3
    I do not know what escape velocity is. Does it happen when the object is not on the Earth's surface or when the object leaves the ozone layer/atmosphere and is in space?

    Thank you!

    **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 27% of the escape speed.**
    I am thinking that this means that when the object is launched from Earth it leaves the Earth's atmosphere and enters space. To do this its speed must equal the escape speed.
    So, if its speed is only 27% of the escape speed, then the object has not entered space yet. It is still in the Earth's atmosphere/realm. I don't know what you call it. So, I guess I want to find the height of the object because it has not reached space yet. So, it hasn't "escaped Earth" yet. I feel really really stupid to have had to write all of this out. I feel like Physics brings the dumbness out of me b/c it takes a long time for me to digest the math/concepts.

    Anyway back on track, so I have the KE and PE of the object initially and after written above. I don't know why there is conservation of ME. Does anyone know why?
    I do not know if the object has any PE or KE on in the Earth's realm. It technically is not on the surface anymore b/c there is some height we have to find, so the PE and KE initial must be nonzero. There should also be some PE and KE final because the object is moving higher and faster I suppose.

    So, I do not think we can get rid of PE nor KE in our equation of conservation of ME.
     
    Last edited: Mar 24, 2006
  5. Mar 24, 2006 #4

    Hootenanny

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    Escape velocity is the minimum initial velocity required to 'overcome' the earth's gravitational pull. I'll start you off with an equation. As I said above "if the kinetic energy of an object is greater than or equal to the potential energy of an object the object will 'escape' from the gravitational field".

    [tex]\frac{1}{2}mv^2 = \frac{GMm}{r}[/tex]

    Do you follow?
    Can you find an equation for the escape velocity from here?

    -Hoot:smile:
     
  6. Mar 24, 2006 #5

    nrqed

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    No. It is a theoretical concept. Imagine that the Earth was at rest, imagine that there is no atmosphere. The escape speed is the initial speed such that an object will barely make it to an infinite distance (barely means that it will reach an infinite distance with a speed going toward zero....strictly speaking, it would take an infinite amount of time to reach infinity).

    So, by its definition,

    [itex] { 1 \over 2} m v_{escape}^2 - {G m M \over R_{Earth}} = {1 \over 2} m v_{infinity}^2 - {G m M \over r = \infty} = 0 [/itex]

    So you can solve for the escape speed (by the way, if you set the escape speed to the speed of light, let R be an unknown and solve R, you get the radius of the event horizon of a Schwarschild black hole!)

    Patrick
     
  7. Mar 24, 2006 #6

    Hootenanny

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    Cool, I didn't know that :biggrin:
     
  8. Mar 24, 2006 #7
    Why is the initial PE of the object 0?

    Is the final KE of the object a nonzero number?

    Thank you for your help.

    I also edited my message up there.:smile:
     
  9. Mar 24, 2006 #8
    Hmmm.. I am really dumb, this I know.

    So, I am inferring that there is 0 final PE and 0 final KE from the posts above. I don't know why.

    So initial KE and inital PE are set to equal each other. And, I guess I will solve for (.27v) because there is 27% of the velocity.

    If I am right, thank you all for your help.
     
  10. Mar 24, 2006 #9

    Hootenanny

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    The initial PE is not zero it is given by;

    [tex]U = \frac{GMm}{r}[/tex]

    What do you mean by the final KE?
     
  11. Mar 24, 2006 #10

    Hootenanny

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    Yes you are correct :smile:
     
  12. Mar 24, 2006 #11
    What do you mean by the final KE?[/QUOTE]

    The .5mv^2 on the second side of the ME eqn. I guess the final velocity of the object when it exits the field of the Earth and enters space. I guess I am wrong. The object must have infinite KE not final KE I suppose or a 0 KE. I don't know.
     
  13. Mar 24, 2006 #12
    Can I get height from this equation? Will it be the r?

    Thank you for your assistance.
     
  14. Mar 24, 2006 #13

    Hootenanny

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    When the object is an infinite distance away from the earth it will have a KE of zero if it was launched at the escape velocity.

    More information is available here, you can peruse at your leisure:
    http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html#gpt

    Remeber to find the distance when v is 27% you will first have to find the escape velocity, then find 27% of that velocity, then solve for r. Remember r is the distance from the centre of mass of the earth so you need to subtract the radius of the earth to find the height.

    -Hoot :smile:
     
    Last edited: Mar 24, 2006
  15. Mar 24, 2006 #14
    Wow thank you.
    You are very kind and helpful. I am going to look at my problem now. I have about 3 different sketches/equations to look at and see where I went wrong. I have to log off now b/c the lab is closing.
     
  16. Mar 24, 2006 #15

    Hootenanny

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    No problem. If you have any further problems I hope you'll come back to PF!

    -Hoot:smile:
     
  17. Mar 24, 2006 #16

    nrqed

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    Of course not!
    I think you understand by now...but by the definition of the escape velocity, the object will reach an infinite distance with a kinetic energy equal to zero. And at an infinite distance the potential energy is zero.

    The starting point is at the surface of the Earth. So you can find the escape velocity in terms of the radius of the Earth, its mass and G.

    Pat
     
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