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Received power by a radiotelescope

  • Thread starter RHK
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RHK
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Homework Statement



The specific power received by a radiotelescope is given by:

[itex]P_\nu = A \int {I_\nu (θ,\phi) f (θ,\phi) dθ d\phi}[/itex]

where:
θ, [itex]\phi [/itex] are the angular coordinates on the celestial sphere around the pointing direction;
[itex]I_\nu [/itex] is the angular distribution of the specific intensity of the source (in units of W/m2 Hz std);
A f(θ, [itex]\phi [/itex]) is the effective area of the antenna in the direction (θ, [itex]\phi [/itex]).

The radiotelescope, operating at a frequency of 3 GHz, of area A= 1000 m2 and with f(θ, [itex]\phi [/itex])= 1 for θ0>θ and f(θ, [itex]\phi [/itex])= 0 for θ>θ0, with θ0=1.9 arcminutes, is pointed in the direction of Mars (distance from the Earth d= 56 x 106 km, diameter D= 6794 km), which has an emission approximated with a black body at T=210 K.
Calculate the detected power by the antenna in a band with Δ[itex]\nu[/itex]=30 GHz around the working frequency.

[The specific intensity of a black body is: [itex]\frac{2h}{c^2}\frac{\nu^3}{e^{h\nu/KT}-1}[/itex]
K=1.381 x 10-23 J/K
h = 6.6 x 10-34 J s ]

Homework Equations



The intensiy of a black body

The Attempt at a Solution



I can not go so far.
Just I have calculated the angular diameter of Mars: δ=arctan D/d = 25"=0.417'.

Any suggestion please?
I can not how to carry out the result, but I think that this is a particular educational exercise.
Thanks in advance.
 

Answers and Replies

  • #2
RHK
64
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Can I have just a start please?
 

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