Question #55052

Giant pulses from energetic pulsars have been observed with S5GHz = 10 kJy, but with durations of only 10 ns. The region from where a pulse of radiation originates must be no larger than the distance that light can travel during the duration of a pulse. If the pulsars are at a distance of 1 kpc, estimate the brightness temperature of these sources.

Expert's answer

Answer on Question 55052, Physics / Astronomy | Astrophysics

Question:

Giant pulses from energetic pulsars have been observed with S5GHz = 10 kJy, but with durations of only 10 ns. The region from where a pulse of radiation originates must be no larger than the distance that light can travel during the duration of a pulse. If the pulsars are at a distance of 1 kpc, estimate the brightness temperature of these sources.

Solution:


S5GHz=B5GHzΩ=10kJy=1019ergscm2×s×HzS_{5GHz} = B_{5GHz} \Omega = 10kJy = 10^{-19} \frac{\text{ergs}}{\text{cm}^2 \times s \times Hz}Ω=π(rR)2ergscm2×s×Hz=2×1038sr\Omega = \pi \left(\frac{r}{R}\right)^2 \frac{\text{ergs}}{\text{cm}^2 \times s \times Hz} = 2 \times 10^{-38} \text{sr}B5GHz=2kTv2c2B_{5GHz} = \frac{2kTv^2}{c^2}Tb=1019×(3×1010)22×(1.38×1016)×(5×109)2×(2.9×1038)=4.5×1035K\therefore T_b = \frac{10^{-19} \times (3 \times 10^{10})^2}{2 \times (1.38 \times 10^{-16}) \times (5 \times 10^9)^2 \times (2.9 \times 10^{-38})} = 4.5 \times 10^{35} \text{K}


Answer: Tb=4.5×1035KT_b = 4.5 \times 10^{35} \text{K}

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