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

[tempneff]

Homework Statement

1. Homework Statement
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

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...?

 

[anathema]

Thank you for posting this interesting problem! I would like to offer some suggestions and corrections to your solution attempt.

Firstly, it is important to note that the force of radiation pressure on the spacecraft is dependent on the shape and size of the solar sail, as well as the intensity of sunlight at the given distance from the sun. Therefore, the value of p in your equation should be specified in terms of these parameters.

Secondly, it seems that you have mixed up some of the variables in your solution. The distance from the sun at which the force is given (1.2 AU and 3.54 AU) should correspond to the values of r1 and r2 in your equation. Additionally, the velocity at a given distance should correspond to the value of v in your equation, not the change in energy.

Furthermore, your integration should be done in terms of time (t), not distance. This is because the problem asks for the time taken for the spacecraft to travel from 1.2 AU to 3.54 AU, not the distance traveled. Therefore, your limits of integration should be in terms of time (t), and the integral should be evaluated with respect to time, not distance.

Finally, the units of your final answer should be years, as specified in the problem. Therefore, you should make sure to use consistent units throughout your solution, whether it be meters, AU, or other units.

I hope this helps in your solution attempt. Good luck!