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Projectile and gravitation problem

  1. Sep 10, 2008 #1


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    1. The problem statement, all variables and given/known data
    A body is fired vertically from the ground of the Earth with a velocity [tex]v_0=1km/s[/tex]. If the gravitational force is not considered as constant and supposing that the Earth is a sphere of radius [tex]6371km[/tex],
    a)Find the maximum height reached by the body.
    b)Compare the previous result with the same experience if g is a constant. (That is [tex]9.8 m/s^2[/tex].)

    2. Relevant equations [tex]r_{max}=\frac{1}{\frac{1}{r_0}-\frac{v_0^2}{2GM_E}}[/tex]

    3. The attempt at a solution I found the formula above in my class notes. If I understand it well, it gives "[tex]r_{max}[/tex]", which I interpret as the distance of a body from the center of the Earth. [tex]r_0[/tex] is the radius of the Earth and [tex]M_E[/tex] is the mass of the Earth.
    I don't know how to find this formula so I think I will have to learn it by heart (sadly...). If you know how to reach to it, please let me know.
    Now using the formula, I found that the projectile will reach [tex]6690.36km[/tex]! Oh wait.... this is the height from the center of the Earth, not the ground as I thought... So it's probably right then.
    Because for the b) I get that it reach only [tex]51.020km[/tex], but it is from the ground of the Earth. Now that I think it makes about [tex]319 km[/tex] (if g is not a constant), I think it's too much, isn't it?
  2. jcsd
  3. Sep 10, 2008 #2


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    Re: Projectile/gravitation

    You are right, forget the formula. It's too specific to a particular problem. The concept you want is that the sum of the gravitational potential energy (PE) and kinetic energy (KE) is a constant. If you consider the gravitation force to be constant PE=mgr. If you use Newton PE=-G*M*m/r. 319km does seem like a bit much. I agree with your 51.02km for the constant g and g doesn't vary so much near the Earth.
  4. Sep 10, 2008 #3


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    Re: Projectile/gravitation

    Thank you very much. I'll try it tomorrow, if I have any problem I'll ask for further help, but I think all will be all right.
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