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Rocket propulsion - Differential equations

  1. Oct 8, 2012 #1
    1. The problem statement, all variables and given/known data


    Suppose a rocket is launched from the surface of the earth with initial velocity
    [itex] v_0 = \sqrt(2gR) [/itex], the escape velocity.

    a) Find an expression for the velocity in terms of the distance x from the surface of the earth (ignore air resistance)

    b) Find the time required for the rocket to go 240,000 miles. Assume R = 4000 miles, and g = 78,545 miles/h2


    3. The attempt at a solution

    So I figure we use [itex] \frac{dv}{dt} = -\frac{mG}{(R+x)^2} [/itex] and multiply by [itex]\frac{dt}{dx} = \frac{1}{v}[/itex] to get

    [itex] \frac{dt}{dx} \frac{dv}{dt} = \frac{dv}{dx} = -\frac{mG}{v (R+x)^2} [/itex]

    which gives [itex] v(x) = \sqrt(\frac{2mG}{R+x}) + C [/itex]

    Using [itex] v(0) = \sqrt(2gR) [/itex] gives

    [itex] C = \sqrt(2gR) - \sqrt(\frac{2mG}{R}) [/itex]


    I don't know where to go from here....
     
  2. jcsd
  3. Oct 8, 2012 #2
    g, G, m and R are related. Use the relationship to eliminate m and G from your formula for v(x). That should answer (a).

    For (b), you just need to integrate (a).
     
  4. Oct 9, 2012 #3

    HallsofIvy

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    Problems like this really annoy me! The whole point of a rocket is that, unlike simply throwing a rock or ball in the air, is that a rocket engine stays on during its flight. In this problem we are apparently to assume that is not true.
     
  5. Oct 9, 2012 #4
    Lol. But what if we assume the rocket is very big ball thrown by a really strong dude? o_O


    As for the actual math...

    What? How do I do that??? It seems that whenever I integrate of differentiate they will always remain since they multiply y? The only way I can see they are related is that they appear in the same formula for y''... How can I cancel them?
     
  6. Oct 9, 2012 #5
    mg equals the force of gravity on the surface of the Earth. What is the the force of gravity on the surface of the Earth?

    edit: "m" here is not the mass of the Earth as you used it above. It is the mass of some small body, such as the rocket. I suggest you denote the mass of the Earth as M to eliminate confusion.
     
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