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Math Help - Differential equation for velocity

  1. #1
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    Differential equation for velocity

    i was working on a differential equation to find the initial velocity of a rocket to get to height h.

    so i have, which i got and its correct cause its the solution in the book, v^2 = 2g(R/(R+h))-2gR+V(0)^2

    v = velocity
    g = acceleration of gravity at earths surface
    R= radius of earth
    h = height
    and i assume v(0) is the velocity v at height=0 so v(0)=0?

    Problem is when i keep getting a negative number. so i assume I'm making an amateur mistake.

    What am i doing wrong? ive tried converting etc. im using km
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  2. #2
    Senior Member BAdhi's Avatar
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    Re: Differential equation for velocity

    The initial velocity cannot be 0. Unless how are you gonna send the rocket up? It is velocity at the maximum height that should be zero.
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    Re: Differential equation for velocity

    so should i be trying to solve for v(0) instead of v? i guess you're saying the velocity goes to 0 when we reach the height we want to achieve right?
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  4. #4
    Senior Member BAdhi's Avatar
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    Re: Differential equation for velocity

    That's right. In your question it self has asked to find the initial velocity hasn't it?
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    Re: Differential equation for velocity

    Yes, to get to a certain height. i will work on it, then post what i have.
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  6. #6
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    Re: Differential equation for velocity

    v^2 = 2g((R^2)/(R+h))-2gR+V(0)^2

    v0^2 = 2gR - 2g((R^2)/(R+h)) + v^2

    so v^2 is 0 cause at the velocity of h the rocket stops. and now solve for V0

    v0 = √(2gR) - √(2g((R^2)/(R+h)))

    R= 6378.1 km
    h= 450 km
    g= 9.80665 m/s2 = 127094.184 km/hr2

    v0 = √((2)(127094.184)(6378.1)) - √(2(127094.184)((6378.1^2)/(6378.1+450)))

    v0 = 1328.1287 km/hr = 368.9246 m/s


    Did i solve this correctly?

    when i try to do this in metres instead of kiometers using the variables
    R= 6378100 m
    h= 450000 m
    g= 9.80665 m/s2

    i get v0 = 374.7107 m/s

    is this more or less accurate?
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  7. #7
    Senior Member BAdhi's Avatar
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    Re: Differential equation for velocity

    Now your main equation is,

    v^2=2g\left(\frac{R^2}{R+h}\right)-2gR+v_0^2 right...

    with v=0, we have,

    v_0^2=2g\left[R-\left(\frac{R^2}{R+h}\right)\right]

    v_0=\sqrt{2g\left[R-\left(\frac{R^2}{R+h}\right)\right] not v_0=\sqrt{2gR}-\sqrt{2g\left(\frac{R^2}{R+h}\right)}
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  8. #8
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    Re: Differential equation for velocity

    OMG you are right! i feel a fool. im getting matching answers now. of 2.8713 km/s and 2871.2917 m/s
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