This is only an example from Kraus Antenna 3rd edition page 404. The question is really a math problem involves calculation of ratio of solid angles. Just ignore the antenna part. this is directly from the book:(adsbygoogle = window.adsbygoogle || []).push({});

Example 12-1.1 Mars temperature

The incremental antenna temperature for the planet Mars measured with the U.S. Naval Research Lab 15-m radio telescope antenna at 31.5mm wavelength was found to be 0.24K(Mayer-1). Mars subtended an angle of 0.005 deg at the time of the measurement. The antenna HPBW=0.116 deg. Find the average temperature of the Mars at 31.5mm wavelength.

The answer from the book:

[tex]T_s=\frac {\Omega_A}{\Omega_S}\Delta{T_A}\;\;\approx\;\frac {0.116^2}{\frac{\pi}{4}0.005^2}(0.24)=164K[/tex]

I don't understand where the [itex]\frac {\pi}{4}[/itex] in the denominator comes from. This is just a simple problem where the ratio of the area of two disk one with Θ=0.116/2 and the other with Θ=0.005/2. I try to use the following to calculate:

[tex]\Omega=\int_0^{2\pi} d\phi\;\int_0^{\theta/2}\sin {\theta} d \theta[/tex]

So

[tex]\frac{\Omega_A}{\Omega_S}\;=\;\frac{\int_0^{2\pi} d\phi\;\int_0^{0.116/2}\sin {\theta} d \theta}{\int_0^{2\pi} d\phi\;\int_0^{0.005/2}\sin {\theta} d \theta}\;=\;\frac{2\pi\int_0^{0.116/2}\sin {\theta} d \theta}{2\pi\int_0^{0.005/2}\sin {\theta} d \theta}[/tex]

[tex] \frac{\Omega_A}{\Omega_S}\;=\;\frac{(1- 0.999999487)}{(1-0.999999999)}=538.2[/tex]

But using the answer from the book:

[tex]\frac{0.116^2}{\frac{\pi}{4}0.005^2}\;=\;685.3[/tex]

I just don't get the answer from the book. Can it be the limitation of my calculator to get cos 0.0025 deg? Can anyone use their calculator to verify the number? please help.

Thanks

Alan

**Physics Forums - The Fusion of Science and Community**

# Please help with calculation involve solid angle.

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

- Similar discussions for: Please help with calculation involve solid angle.

Loading...

**Physics Forums - The Fusion of Science and Community**