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Satellite bullet problem

  1. Mar 28, 2014 #1
    1. The problem statement, all variables and given/known data

    An earth satellite is revolving in a circular orbit of radius 'a' with velocity 'v0'. A gun is in the satellite and is aimed directly towards the earth.A bullet is fired from the gun with muzzle velocity v0/2.Neglecting resistance offered by cosmic dust and recoil of gun,calculate maximum and minimum distance of bullet from the center of earth during its subsequent motion.

    2. Relevant equations



    3. The attempt at a solution

    Orbital speed of satellite is [itex]\sqrt{\frac{GM}{a}}[/itex]

    Initial velocity of the bullet [itex]v_{i} = \sqrt{{v_o}^2+(\frac{v_0}{2})^2} = \frac{\sqrt{5}v_{0}}{2}[/itex]

    Let P be the point at which bullet is fired and Q be point where distance is maximum/minimum.

    Applying conservation of angular momentum at P and Q

    [itex]mv_{i}a=mvr[/itex]

    or , [itex]v = \frac{v_{i}a}{r} = \frac{\sqrt{5}}{2}\frac{av_0}{r}[/itex]

    Applying conservation of mechanical energy at P and Q

    [itex]\frac{1}{2}m{v_i}^2 - \frac{GMm}{a} = \frac{1}{2}m{v}^2 - \frac{GMm}{r}[/itex]

    Solving the equations , I get [itex]3r^2-8ar+5a^2 = 0 [/itex] which gives r =5/3a and a .

    The answer i am getting is incorrect .

    The correct answer given is 2a and 2a/3 .

    I would be grateful if somebody could help me with the problem.
     
    Last edited: Mar 28, 2014
  2. jcsd
  3. Mar 28, 2014 #2

    gneill

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    Staff: Mentor

    Check your algebra for that last term in your quadratic. Otherwise you've done fine up to that point.
     
  4. Mar 28, 2014 #3
    Hi gneill...

    Sorry...I couldn't find any algebraic error .I redid the calculations . Maybe I am committing the same mistake again .

    I keep on getting r=5a/3 and a .
     
  5. Mar 28, 2014 #4

    gneill

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    Staff: Mentor

    Can't fix what we can't see...

    Can you elaborate your derivation of the quadratic a bit?
     
  6. Mar 28, 2014 #5
    [itex]\frac{1}{2}m{v_i}^2 - \frac{GMm}{a} = \frac{1}{2}m{v}^2 - \frac{GMm}{r}[/itex]

    [itex]\frac{1}{2}m\frac{5}{4}\frac{GM}{a} - \frac{GMm}{a} = \frac{1}{2}m\frac{5}{4}\frac{a^2}{r^2}\frac{GM}{a} - \frac{GMm}{r}[/itex]

    [itex]\frac{5}{8a}-\frac{1}{a} = \frac{5}{8}\frac{a}{r^2}-\frac{1}{r}[/itex]

    [itex]\frac{-3}{8a} = \frac{1}{8r^2}(5a-8r)[/itex]

    [itex]3r^2-8ar+5a^2 = 0[/itex] which gives [itex]r=5a/3 ,a[/itex]
     
    Last edited: Mar 28, 2014
  7. Mar 28, 2014 #6

    gneill

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    Staff: Mentor

    The LHS looks fine. But the velocity used on the RHS should be the velocity as obtained via the conservation of angular momentum:

    ##r \cdot v = a \cdot v_o## {angular momentum depends on the velocity component perpendicular to the radius vector}

    ##v = \frac{a}{r} v_o##

    ## v^2 = \left( \frac{a}{r} \right)^2 v_o^2## where: ##~~~v_o^2 = \frac{GM}{a}##
     
  8. Mar 28, 2014 #7
    Thanks gneill :smile:
     
  9. Mar 28, 2014 #8

    D H

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    Staff Emeritus
    Science Advisor

    Here's the source of your error:

    The radial component of velocity does not contribute to angular momentum. This means that firing the gun doesn't change the bullet's orbital angular momentum.
     
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