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Homework Help: I need a fresh critique of this 4 hour proof from another post

  1. May 27, 2012 #1
    1. The problem statement, all variables and given/known data

    1. The problem statement, all variables and given/known data
    A motionless spacecraft powered by a “solar sail” at a distance of 1.20 AU
    from the sun has a perfectly reflective sheet of lightweight material that allows the spacecraft to accelerate by using the radiation pressure from the sun. If a certain sail is in the shape of a square of side 800 m, its velocity is 38.5 km/s when it reached a distance of 3.50 AU from the sun. The Force at 1.2 AU is 4.18 N and at 3.54 AU is 0.492.
    Answer range: 3.0 to 3.4 years

    3. The attempt at a solution
    Okay this is the function I have for velocity and I think I have the integration....

    Change in energy=Work due to gravity (GMm/radius) + work due to radiation pressure(I'll call it p/radius)

    [tex]\frac{1}{2}v^2=\bigg( \frac{p}{m}-GM\bigg) \bigg( \frac{1}{r_1}-\frac{1}{r_2} \bigg)[/tex]
    [tex]v=\sqrt{2\bigg( \frac{p}{m}-GM\bigg) \bigg( \frac{1}{r_1}-\frac{1}{r_2} \bigg)}[/tex]
    [tex]2\bigg( \frac{p}{m}-GM\bigg)=b[/tex]
    [tex]v=\sqrt{b \bigg( \frac{1}{r_1}-\frac{1}{r_2} \bigg)}[/tex]
    [tex]v=\sqrt{b \bigg( \frac{r_2-r_1}{r_1r_2} \bigg)}[/tex]

    [tex]t=\int \frac{1}{v}dr[/tex]
    [tex]t=\int \sqrt{ \bigg( \frac{r_1r_2} {b(r_2-r_1)}\bigg)}dr[/tex]

    I think so.....then for easier typing let $r_1=a$ and $r_2=x$
    [tex]t=\int \sqrt{\bigg( \frac{ax} {b(x-a)}\bigg)}dx[/tex]
    now if we let
    [tex]x=a\sec^2 \theta[/tex]then[tex]dx=2a\sec^2 \theta \tan \theta d \theta[/tex]
    and[tex]t=\frac{1}{\sqrt{b}}\int \sqrt{\frac{a^2 \sec^2 \theta}{a \sec^2 \theta - a}}2a \sec^2 \theta \tan \theta d \theta[/tex]
    [tex]t=\frac{1}{\sqrt{b}}\int \sqrt{\frac{a^2 \sec^2 \theta}{a \tan^2 \theta}}2a \sec^2 \theta \tan \theta d \theta[/tex]
    [tex]t=\frac{1}{\sqrt{b}}\int \frac{a\sec \theta}{\sqrt{a} \tan \theta}2a \sec^2 \theta \tan \theta d \theta[/tex]
    [tex]t=\frac{2a^2}{\sqrt{ab}}\int \sec^3 \theta d \theta[/tex]
    By integral table
    [tex]\int sec^3 \theta d \theta = \frac{1}{2} \bigg( \sec \theta \tan \theta +\ln |\sec \theta + \tan \theta| \bigg)+C[/tex]
    If [tex]x=a \sec^2 \theta[/tex]
    then[tex]\sec \theta = \sqrt{\frac{x}{a}}[/tex]
    [tex]\tan \theta = \sqrt{x-a}[/tex]
    [tex]t=\frac{a^2}{\sqrt{ab}}\bigg( \sqrt{\frac{x(x-a)}{a}}+\ln \bigg|\sqrt{\frac{x}{a}}+\sqrt{x-a}\bigg| \bigg) +C[/tex]

    ...that's it. But I'm not sure about the limits though. AU or meters....?
  2. jcsd
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