# I need a fresh critique of this 4 hour proof from another post

1. May 27, 2012

### tempneff

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....

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

$$v=\frac{dr}{dt}$$
and
$$dt=\frac{dr}{v}$$
then
$$t=\int \frac{1}{v}dr$$
and.....
$$t=\int \sqrt{ \bigg( \frac{r_1r_2} {b(r_2-r_1)}\bigg)}dr$$

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

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