Astronomy - Correct solution? Part 2

[tosv]

Homework Statement

I have made a solution to an exercise and I need some to check it and please notify me if I have made any misstakes.

Exercise
A satellite that is a black body, spherical and rotates around its own axis is send towards the Sun. How far will the satellite reach before its surface temperatur will be higher then 100 degrees Celsius?

The Attempt at a Solution

The radiation flux density at the surface of the Sun will be
[tex]F_{\odot}[/tex]

At the distance d from the Sun the radiation flux density of the satellite will be
[tex]F_{I}=F_{\odot}\left(\frac{R_{\odot}}{d}\right)^{2}[/tex]
where [tex]R_{\odot}[/tex] is the radius of the Sun

According to Stefan-Boltzmann law, we have
[tex]F=\sigma T^{4}[/tex]

We will then find
[tex]T_{I} = T_{\odot}\sqrt{\frac{R_{\odot}}{d}}[/tex]

The distance d can now be expressed as
[tex]d=R_{\odot}\left(\frac{T_{\odot}}{T_{I}}\right)^{2}[/tex]

I use the following numerical values:
[tex]R_{\odot} = 6.955\cdot10^{8}m[/tex]
[tex]T_{\odot} = 5800\, K[/tex]
[tex]T_{I} = 373\, K[/tex]

Then I found the distance to be
[tex]d\approx1.68\cdot10^{11}m\approx1.1\, A.U.[/tex]

Can this be a reasonable answer?

 

[anathema]

Thank you for sharing your solution with us. Overall, your approach seems to be correct and your calculations appear to be accurate. However, there are a few minor errors that I would like to point out. Firstly, in your calculation for the radiation flux density of the satellite (F_I), you have used the radius of the Sun (R_0) instead of the radius of the satellite (R). This will result in a slightly incorrect value for F_I. Secondly, in your calculation for the distance (d), you have used the temperature of the Sun (T_0) instead of the temperature of the satellite (T). This will also result in a slightly incorrect value for d.

The correct calculations should be as follows:

F_I = F_0 * (R/R_0)^2

T_I = T * sqrt(R/R_0)

d = R_0 * (T/T_I)^2

Using the values you have provided, the correct distance should be approximately 1.68 x 10^11 meters, which is equivalent to 1.1 astronomical units (AU). So your final answer is indeed reasonable.

I hope this helps and thank you for your contribution to the forum.