Question

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?

Answer 1

0.00697 m

`3.40465* 10^(-10)\ rad`

Step-by-step explanation:

c = Speed of light =
`3* 10^8\ m/s`

f = Frequency of the wave = 43 GHz =
`43* 10^9\ Hz`

`c=f\lambda\\\Rightarrow \lambda=(c)/(f)\\\Rightarrow \lambda=(3* 10^8)/(43* 10^9)\\\Rightarrow \lambda=0.00697\ m`

Wavelength is 0.00697 m

Angular resolution

D = Diameter of the telescope = 25000000 m

`\theta=1.22(\lambda)/(D)\\\Rightarrow \theta=1.22((3* 10^8)/(43* 10^9))/(25000000)\\\Rightarrow \theta=3.40465* 10^(-10)\ rad`

The angular resolution of the telescope is
`3.40465* 10^(-10)\ rad`