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Projectile problem

  1. May 27, 2007 #1
    1. The problem statement, all variables and given/known data

    A projectile of mass m is fired from the surface of the earth of radius R at an angle a from the vertical with initial speed equal to (1/[sqrt 2]) the escape velocity.How high does the projectile rise?What is its range on earth's surface?Neglect air resistance and earth's rotation.

    2. Relevant equations
    3. The attempt at a solution

    I solved the first part easily by using work-energy theorem.The answer is 2R/3.I hope it is correct.

    But I am stuck to solve the 2nd part.Will the normal formulas for the projectile problem give the answer?
    Last edited: May 27, 2007
  2. jcsd
  3. May 27, 2007 #2


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    If you can find the equation of the projectile in terms of [itex]r[/itex] and [itex]/theta[/itex], you can find when it intersects the earth again, and the range on the earth's surface from that.
  4. May 27, 2007 #3
    Should we consider the earth as sphere or a flat space?
    What equation are you referring to?Is it the Energy equation in central force motion?
  5. May 27, 2007 #4


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    If you take the Earth as a point mass, then this problem will correspond to a motion of a small mass (ie, the projectile) in a central force field.

    So, you can get a differential equation for the shape of the orbit for
    [tex] \frac{d \theta}{dr} [/tex], which you can solve by a change of variable of u=1/r. The soln will be of the form

    [tex] r= \frac{A_1}{1 + \epsilon \cos \theta} [/tex]

    Then from the initial condition of [itex] v, a ,R_{earth} [/tex], I think you can calculate the values of the constants, and then find what theta is when R=R_{earth} again, from which you should be able to get the range on earth's surface.
    Last edited: May 27, 2007
  6. May 27, 2007 #5
    Yes,I am familiar with Central force notations.
    I did the firdt part last night.Now I am trying the 2nd part.

    e can be found directly from another formula.The A_1 you have written is (L^2/mk).So,it can also be found easily without solving the equation.

    the problem is theta which is the polar angle.
    Initially,we have theta=0,but the direction of projection is a(=alpha)
  7. May 28, 2007 #6


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    Since [tex]\vec{L} = \vec{r} \times \vec{p} [/tex], won't you need the angle to calculate L, and hence e and A_1? And how do you get [tex]\theta[/tex] as 0? What coordinate system are you working with?
  8. May 28, 2007 #7
    When we applied the angular momentum conservation for the first part,we did it from a frame whose origin is at the centre of the earth.

    I said that when you are taking L=r x p,you have to take the angle between the vextors r and p and hence, a(=alpha).

    But if you are writing the orbit formula,there theta corresponds to polar angle same as in a spherical polar co-ordinate system.So,initially, theta=0(the particle is assumed to be in the North pole),but a is not equal to 0.

    The A_1 you have written is semi-latus rectum=(L^2)/mk
    and eccentricity e is given by

    e^2=1+(L^2/k^2)((v_0)^2 - (v_e)^2 ) where v_0 is the initial speed and v_e is the escape speed.

    These two give the value of the constants...
    I cannot understand the next step.
  9. May 28, 2007 #8


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    Yeah, but remember that if you imagine that the projectile is launched at the north pole, then the ellipse will be "inclined". In that case, if you use the polar angle as [itex] \theta [/itex], the equation should look like,

    [tex] r= \frac{h^2/ \mu}{1 + \epsilon \cos (\theta - \theta_0)} [/tex]

    Next, have you set the value of the [itex]r[/itex] as the radius of the Earth (ie, when the projectile will hit the earth again) and solved for the value of [itex] \theta[/itex]?
    Last edited: May 28, 2007
  10. May 28, 2007 #9
    Latest Attempt:

    Draw an ellipse with principal axes aligned to a rectangle coordinate system. This ellipse is the trajectory of the projectile. Now, draw a circle with the same center as the ellipse such that the ellipse and the circle intersect at four points. The circle is the Earth's surface.

    I put the major axis of the ellipse along the z axis and the North pole of the earth is lying on this axis.The minor axis is along the x axis.

    So,the ellipse is symmetrical about z axis.
    Say,the particle was thrown from the intersection point of the 1st quadrant. And it reaches the 2nd quadrant.Then the orbit is symmetrical about z axis.

    The two thetas are equal...

    Applying the orbit equation the projection point,we can find theta in terms of R.Then we are to multiply it with 2 to get the total angular range.
    So,range is 2R*theta

  11. May 28, 2007 #10


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    No, becuase for central force motion, the origin (ie, the COM of the earth) is at a focal point, not the center.
    Besides, I don't understand how you can arbitarily say that

    You've negelected the information of the angle at which the projectile is thrown.
    Last edited: May 28, 2007
  12. May 28, 2007 #11
    But this diagram should yield the result.
    I am trying.
  13. May 28, 2007 #12
    Kolahal,are you sure that this diagram is working?

    I did not get anything so far.
    Can you please show your work?
  14. May 29, 2007 #13
    OK,it's done...

    The semimajor axis equals the earth radius R.You can see this from
    2*semi-major axis=(-GMm)/E.Also note that the Perigee point is at the centre of the earth.(The ellipse cuts the circle at two points).

    So,in launching (and landing) point the projectile is at the minor axis, and the straight line (tunneling) distance would be it's length, 2R.sin a.

    Measuring on earth's surface, it is the arc length 2.a.R
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