Question

Galaxy NGC2300 has an flat disk dominating the visible light image. Surface brightness is given by
the exponential law
I(R) = I0 exp(−R/Rd)
where I0 is the central surface brightness, R distance from the center, and Rd = 4 kpc the exponential
radial scale of the disk (e-folding distance). The total luminosity of the galaxy equals L = (2.5·10^10)L(sun).
Considering that the total luminosity is surface brightness I(R) integrated over the area of the
whole disk from R = 0 to R = ∞ (not over the radial distance, a mistake some people make!), compute
I0 (in units of L(sun)/pc^2)

Answer 1

Answer on Question #71096, Physics / Astronomy | Astrophysics

Question: Galaxy NGC2300 has an flat disk dominating the visible light image. Surface brightness is given by the exponential law $I(R) = 10 \exp(-R / Rd)$ where $I0$ is the central surface brightness, $R$ distance from the center, and $Rd = 4$ kpc the exponential radial scale of the disk (e-folding distance). The total luminosity of the galaxy equals $L = (2.5 \cdot 10^{\wedge}10)L(\text{sun})$. Considering that the total luminosity is surface brightness $I(R)$ integrated over the area of the whole disk from $R = 0$ to $R = \infty$ (not over the radial distance, a mistake some people make!), compute $I0$ (in units of $L(\text{sun}) / \text{pc}^{\wedge}2$)

Solution:

$$L = \int_{0}^{\infty} I_{o} e^{-\frac{r}{r_{d}}} 2\pi r dr = 2\pi I_{0} \int_{0}^{\infty} e^{-\frac{r}{r_{d}}} r dr = 2\pi I_{0} r_{d}^{2}$$ $$I_{0} = \frac{L}{2\pi r_{d}^{2}} = 2.5 * \frac{10^{10} L_{sun}}{3.14 * 2 * 16 * 10^{6} pc^{2}} = 248 \frac{L_{sun}}{pc^{2}}$$

Answer:

$$I_{0} = 248 \frac{L_{sun}}{pc^{2}}$$

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