天文学代写|astronomy代考|高质保分天文学作业

Problem 1
A star in the Andromeda galaxy yields a bolometric flux at the Earth of $F=1.0 \times 10^{-13} \mathrm{ergs}$ $\mathrm{cm}^{-2} \mathrm{sec}^{-1}\left(1.0 \times 10^{-16} \mathrm{Watts} \mathrm{m}^{-2}\right)$. It has a B-V color index of -0.24 . Take the distance to Andromeda to be $1 \mathrm{Mpc}$. Make use of the table below, where appropriate, to answer the following questions (interpolate between entries using any interpolation scheme that is reasonable). The reference flux for a bolometric magnitude of 0.0 is $F_0=2.5 \times 10^{-5} \mathrm{ergs}$ $\mathrm{cm}^{-2} \mathrm{sec}^{-1}\left(2.5 \times 10^{-8}\right.$ Watts m$\left.{ }^{-2}\right)$.
a. Find the bolometric magnitude, $M_{\mathrm{bol}}$ of the star.
$$
M_{\mathrm{bol}}=-2.5 \log \left(\frac{1 \times 10^{-13}}{2.5 \times 10^{-5}}\right) \simeq 21
$$
b. What is the approximate effective temperature, $T_{\text {effective }}$, of the star.
$T_e \simeq 17250 \mathrm{~K}$, based on a linear interpolation from the table; or $\sim 16850 \mathrm{~K}$, based on a logarithmic interpolation. Let’s adopt $T_e \simeq 17,000 \mathrm{~K}$.
c. Calculate the approximate radius of the star.
$$
\begin{gathered}
F=\frac{L}{4 \pi d^2}=\frac{4 \pi \sigma R^2 T^4}{4 \pi d^2} \simeq 1 \times 10^{-13} \
R=\sqrt{\frac{F d^2}{\sigma T^4}} \simeq 6.4 R_{\odot}
\end{gathered}
$$

Problem 5
The disk of a galaxy can be modeled as a uniform slab of material of mass density, $\rho$, that is of (full) thickness, $2 H$, in the $\hat{z}$ direction, and is effectively infinite in the $\hat{x}$ and $\hat{y}$ directions. Assume that the mass density for $z>H$ is zero.
a. Compute the effective gravity, $\vec{g}$, at an arbitrary distance, $z$, inside and above the disk. Sketch $\vec{g}(z)$ for all $z$ (i.e., for + and – values of $z$ ). Hint: make use of Gauss’ law for gravity $\int \vec{g} \cdot \overrightarrow{d A}=-4 \pi G M$
$$
\int \vec{g} \cdot \overrightarrow{d A}=-4 \pi G M
$$
For a cylindrical “pillbox” of end area, $A$ :
$$
\begin{gathered}
-g(z) 2 A=-4 \pi G \rho 2 A z \quad \text { (inside the disk) } \
-g(z) 2 A=-4 \pi G \rho 2 A H \quad \text { (outside the disk) }
\end{gathered}
$$
or
$$
\begin{gathered}
g(z)=4 \pi G \rho z \quad \text { (inside) } \
g(z)=4 \pi G \rho H \quad \text { (outside) }
\end{gathered}
$$
pointing toward the midplane
b. Find the speed, $v_z$ that a star must have, starting at the middle of the disk, to get above height $H$, i.e., just outside of the mass distribution. Express your answer in terms of $\rho, G$, and $H$.
$$
\phi=\int_0^z g(z) d z=2 \pi G \rho z^2 \quad \text { (inside) }
$$
In order for a star to reach $z=H$ from the midplane, its kinetic energy must exceed the potential energy which is equal to the result found in part (a) evaluated at $z=H$, and multiplied by the mass of the star:
$$
\frac{1}{2} m v^2 \gtrsim 2 \pi G \rho H^2 m
$$
or,
$$
v^2 \geq 4 \pi G \rho H^2
$$