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The Institute of Space and Astronautical Science in Japan proposes to place a radio telescope into an even higher orbit than the HALCA telescope. Using this telescope in concert with a ground-based radio telescope, baselines as long as 25,000 km may be obtainable. Astronomers want to use this combination to study radio emission at a frequency of 43 GHz from the molecule silicon monoxide, which is found in the interstellar clouds from which stars form. (1 GHz = 1 gigahertz = 109 Hz.) (a) What is the wavelength of this emission? (b) Taking the baseline to be the effective diameter of this radio-telescope array, what angular resolution can be achieved?

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Answer:

0.00697 m

[tex]3.40465\times 10^{-10}\ rad[/tex]

Explanation:

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

f = Frequency of the wave = 43 GHz = [tex]43\times 10^9\ Hz[/tex]

[tex]c=f\lambda\\\Rightarrow \lambda=\frac{c}{f}\\\Rightarrow \lambda=\frac{3\times 10^8}{43\times 10^9}\\\Rightarrow \lambda=0.00697\ m[/tex]

Wavelength is 0.00697 m

Angular resolution

D = Diameter of the telescope = 25000000 m

[tex]\theta=1.22\frac{\lambda}{D}\\\Rightarrow \theta=1.22\frac{\frac{3\times 10^8}{43\times 10^9}}{25000000}\\\Rightarrow \theta=3.40465\times 10^{-10}\ rad[/tex]

The angular resolution of the telescope is [tex]3.40465\times 10^{-10}\ rad[/tex]

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