1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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=\frac{GMm}{r_2}-\frac{GMm}{r_1}+\frac{p}{r_2}-\frac{p}{r_1}[/tex]
    factor..
    [tex]\frac{1}{2}v^2=\bigg( \frac{p}{m}-GM\bigg) \bigg( \frac{1}{r_1}-\frac{1}{r_2} \bigg)[/tex]
    so
    [tex]v=\sqrt{2\bigg( \frac{p}{m}-GM\bigg) \bigg( \frac{1}{r_1}-\frac{1}{r_2} \bigg)}[/tex]
    Let
    [tex]2\bigg( \frac{p}{m}-GM\bigg)=b[/tex]
    so
    [tex]v=\sqrt{b \bigg( \frac{1}{r_1}-\frac{1}{r_2} \bigg)}[/tex]
    or
    [tex]v=\sqrt{b \bigg( \frac{r_2-r_1}{r_1r_2} \bigg)}[/tex]
    then
    [tex]\frac{1}{v}=\sqrt{\frac{r_1r_2}{b(r_2-r_1)}}[/tex]

    [tex]v=\frac{dr}{dt}[/tex]
    and
    [tex]dt=\frac{dr}{v}[/tex]
    then
    [tex]t=\int \frac{1}{v}dr[/tex]
    and.....
    [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]
    and
    [tex]\tan \theta = \sqrt{x-a}[/tex]
    so
    [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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: I need a fresh critique of this 4 hour proof from another post
Loading...