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Velocity of an object some distance from the surface of the earth. Watch

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    Okay so I'm looking at some mechanics work and stumbled across this formula.

    \displaystyle v^2=\frac{2gR^2}{s}+u^2-2gR

    Apparently this is the velocity of an object travelling some distance s away from the earth of radius R. This is more accurate than using constant acceleration formulas and assuming constant acceleration. My question is how is this formula derived, I've had a mess about with Newton's law of gravitation, circular motion formulas and energy conservation and I just can't see where this comes from and how it even makes sense.

    It then goes on to do a binomial expansion to get the formula



    What's the point of doing so? (It looks maybe like you can show for small h that the formula is almost like constant acceleration which would seem to make some sense but I'm just getting confused).

    Could anyone clear this up for me?

    Thanks.
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    (Original post by poorform)
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    From a bit of playing around, I think "s" has to be the distance from the centre of the Earth.

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    Using suffix "1" for our object and "E" for the Earth.

    F=\displaystyle\frac{Gm_1m_E}{x^  2}=-m_1a

    From which acceleration \displaystyle a=-\frac{Gm_E}{x^2}...........(1)

    When d=R acceleration is -g, so:

    \displaystyle -g=-\frac{Gm_E}{R^2}..............(2)

    Then (1) divided by (2) gives:

    \displaystyle -\frac{a}{g}=\frac{R^2}{x^2}

    and:

    \displaystyle a=-g\frac{R^2}{x^2}

    Now: a= v\frac{dv}{dx}, so

    \displaystyle v\frac{dv}{dx}=-g\frac{R^2}{x^2}

    This is a separable differential equation and with a minor abuse of notation on one limit:

    \displaystyle \int_u^v v\; dv=\int_R^s -g\frac{R^2}{x^2}\;dx

    :holmes: I seem to have made a sign slip somewhere, but other than that it gives the desired result.

    Edit: Resolved. Using the positive value for g requires a minus sign as used above.


    For the binomial they've set s=R+h, divided numerator and denominator by R, and are thus expanding (1+\frac{h}{R})^{-1}
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    (Original post by ghostwalker)
    From a bit of playing around, I think "s" has to be the distance from the centre of the Earth.
    Spoiler:
    Show

    Using suffix "1" for our object and "E" for the Earth.

    F=\displaystyle\frac{Gm_1m_E}{x^  2}=-m_1a

    From which acceleration \displaystyle a=-\frac{Gm_E}{x^2}...........(1)

    When d=R acceleration is -g, so:

    \displaystyle -g=-\frac{Gm_E}{R^2}..............(2)

    Then (1) divided by (2) gives:

    \displaystyle -\frac{a}{g}=\frac{R^2}{x^2}

    and:

    \displaystyle a=-g\frac{R^2}{x^2}

    Now: a= v\frac{dv}{dx}, so

    \displaystyle v\frac{dv}{dx}=-g\frac{R^2}{x^2}

    This is a separable differential equation and with a minor abuse of notation on one limit:

    \displaystyle \int_u^v v\; dv=\int_R^s -g\frac{R^2}{x^2}\;dx

    :holmes: I seem to have made a sign slip somewhere, but other than that it gives the desired result.

    Edit: Resolved. Using the positive value for g requires a minus sign as used above.

    For the binomial they've set s=R+h, divided numerator and denominator by R, and are thus expanding (1+\frac{h}{R})^{-1}
    Yeah that the definition of s, sorry I forgot to type that up. I don't currently have time to work through it myself but I imagine there will be no problems now (I will let you know if there are). Thanks, much appreciated.
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    (Original post by ghostwalker)
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    Actually would you mind just explaining the integral bit in a bit more detail (I'm not sure exactly what it means with the limits (what is the aim))? I've never seen differential equations solved with limits of integration like that my lecturer always integrates and then uses conditions to solve for the constant of integration.
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    (Original post by poorform)
    Actually would you mind just explaining the integral bit in a bit more detail (I'm not sure exactly what it means with the limits (what is the aim))? I've never seen differential equations solved with limits of integration like that my lecturer always integrates and then uses conditions to solve for the constant of integration.
    You can do it the way you usually do.

    Initial conditions are x=R and v=u (the lower limits) going to x=s and v=v (the upper limits).

    It boils down to the same thing whether you use limits or whether you have a constant of integration which you resolve via initial conditions.

    If you work it through both ways you'll see how they tie up.
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    (Original post by poorform)
    Actually would you mind just explaining the integral bit in a bit more detail (I'm not sure exactly what it means with the limits (what is the aim))? I've never seen differential equations solved with limits of integration like that my lecturer always integrates and then uses conditions to solve for the constant of integration.
    This is incredibly surprising (you're a uni student correct?).
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    (Original post by Phichi)
    This is incredibly surprising (you're a uni student correct?).
    Yeah.

    It's a 1st year maths module and is quite short, not sure if that has anything to do with it.

    I agree though it does seem weird as I see that elsewhere in lots of places. (I think it was even used in a-level), I'm guessing it's just the lecturers preference maybe?

    Thanks for the help guys anyway.
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    (Original post by poorform)
    Yeah.

    It's a 1st year maths module and is quite short, not sure if that has anything to do with it.

    I agree though it does seem weird as I see that elsewhere in lots of places. (I think it was even used in a-level), I'm guessing it's just the lecturers preference maybe?

    Thanks for the help guys anyway.
    I've always seen it sloppy practice to not stick the limits in straight away if available, saves time and looks tidier.
 
 
 
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