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Muzzle velocity

  1. Dec 4, 2014 #1
    < Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template is shown >

    Suppose you are on the lunar surface and want to find out the speed a bullet has when it is fired from the revolver. You shoot a projectile vertically from the surface of the moon, and notes that it rises to 330km above the lunar surface. Calculate the projectile muzzle velocity.

    Is this possible to calculate, and if so, what would the answer be?
     
    Last edited by a moderator: Dec 4, 2014
  2. jcsd
  3. Dec 4, 2014 #2

    Danger

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    This sounds suspiciously like a homework question, and if so should be posted in that section. You are required to submit your best thoughts about the matter before being told any answers.
    To start with, what factors do you think might be involved in the situation?
     
  4. Dec 4, 2014 #3

    HallsofIvy

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    I presume, though perhaps you should check, that the height will not be sufficient that the "[tex]1/r^2[/tex]" factor does not come into play. If that is the case, the acceleration is constant- you can look up the "acceleration due to gravity" on the surface of the moon. Using that value, use exactly the same equations you would for motion on the earth.
     
  5. Dec 4, 2014 #4
    Hallsoflvy, if I was to use the 1/r2, how would I do it?
     
  6. Dec 4, 2014 #5
    Danger, it's not homework, it's a booklet of physics belonging to the students two grades above me, this question made me super confused, but I promise you, I'm not trying to cheat! I actually thought of posting it in the homework area but as I'm in no hurry, I didn't bother. Regarding what I know, not much, I can barely understand the question.
     
  7. Dec 4, 2014 #6

    Danger

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    Oh, okay. Sorry, but I had to check.
    I don't do math, so I'll leave that up to Halls and others. I will point out that you do not have to factor in air resistance as you would on Earth.
     
  8. Dec 4, 2014 #7

    berkeman

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    Welcome to the PF.

    All schoolwork-type questions go in the Homework Help forums, even if they are for self-study. I've moved your thread to the appropriate forum. :-)
     
  9. Dec 5, 2014 #8

    HallsofIvy

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    i
    For an object on a planet of mass M, the acceleration at height r is [tex]\frac{GM}{r^2}[/tex]. That was what I was referring to. If r does not change much then the acceleration can be treated as a constant, just as on the surface of the earth the acceleration is the constant -9.81 m/s2.

    For a greater range in r, you can [tex]a= -\frac{GM}{r^2}[/tex] and, because acceleration is the derivative of velocity, find the velocity by integrating:
    [tex]v= \int -\frac{GM}{r^2} dt[/tex]. That's a bit complicated because we want to integrate with respect to t but have the acceleration written as a function of r. Fortunately there is a standard method called "quadrature" for this case. By the chain rule we have [tex]\frac{dv}{dt}= \frac{dv}{dr}\frac{dr}{dt}[/tex] but v, the velocity, is itself dr/dt so that is [tex]\frac{dv}{dt}= v\frac{dv}{dr}= -\frac{GM}{r^2}[/tex] so that [tex]vdv= -\frac{GM}{r^2}dr[/tex]. Integrating that, [itex]\frac{1}{2}v^2= \frac{GM}{r}+ C[/itex]. You can then solve for v= dr/dt and integrate again to find r as a function of t. That's why it is so much simpler to assume that r does not really change much and you can treat a as a constant.

    (But notice that [itex]\frac{1}{2}v^2= \frac{GM}{r}+ C[/itex] is the same as [itex]\frac{1}{2}v^2- \frac{GM}{r}= C[/itex]. That's "conservation of energy"- [itex]\frac{1}{2}v^2[/itex] is the kinetic energy and [itex]-\frac{GM}{r}[/itex] is the potential energy- their sum is the total energy.)
     
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