---
title: ASTR 597 W23 HW 6
author: Andy Tzanidakis and friends
tags: LSST Class
---
[TOC]
## Problem 1 - Geometry Absolute Magnitude
An asteroid's absolute magnitude, commonly denoted as H, is the visual magnitude an observer would record if the asteroid were placed 1 Astronomical Unit (au) away, and 1 au from the Sun and at a zero phase angle (⍺; see the included figure). Sketch out the Observer-Sun-Asteroid configuration satisfying the above requirement (assuming the observer is on the Observer). What is the value of d (the Observer-Sun distance) in such configuration?
In reality, all this is telling us is that the absolute magnitude ($H$) at phase-angle ($\alpha$) of 0 degrees is that we're viewing the asteroid from the perspective of the Sun (i.e d=0 AU). Schematically, this is what it would look like:

## Problem 2 - Asteroid Distance Modulus
The relationship of an asteroids’s magnitude to flux, at phase angle ⍺=0 (also known as “being in the opposition”), follows the familiar definition:
$$\begin{equation}
𝑚 = 𝐻 − 2.5 log_{10} \bigg{(} \frac{f}{f_0}\bigg{)}
\end{equation}$$
Starting with the above, derive the expression for m(r, 𝚫), the magnitude as a function of the Sun-Asteroid distance r, and the Observer-Asteroid distance 𝚫, at phase angle ⍺=0. When deriving this expression, keep in mind that the asteroid shines in reflected light.
We can express the ratio of the fluxes such that:
$$\begin{equation}
\frac{f}{f_0} = \frac{L_{\odot}}{4 \pi d^2} \times 4\pi r^2 \Delta^2 \frac{1}{L_{\odot} d^2 q\times\Phi(\alpha)} = \frac{d^4 a_0 \times \Phi(\alpha)}{r^2 \Delta^2}
\end{equation}$$
In the above equation we have the phase function (i.e what phase of the asteroid we are seeing), and $a_0$ the albedo of the asteroid. We can assume if the surface brightness is isotropic across the entire asteroid (meaning albedo $a_0$=1), generally we can assume that the phase will be approximately:
$$\begin{equation}
\Phi(\alpha) \approx cos(\alpha) = cos(\alpha=0) = 1
\end{equation}$$
Then we can re-write the expression:
$$\begin{equation}
𝑚 (r, \Delta) = 𝐻 − 2.5 log_{10} \bigg{(} \frac{d^4}{\Delta^2 r^2}\bigg{)} = 𝐻 − 2.5 log_{10} \bigg{(} \frac{1 AU^4}{\Delta^2 r^2}\bigg{)}
\end{equation}$$
## Problem 3 - Asymptotic Behavior of Apparent Magnitude
Assuming, r = 𝚫 + 1au (i.e., the asteroid is at opposition) show the asymptotic behavior of m(𝚫) as 𝚫 → ∞ .
How does it compare to the behavior of flux as a function of distance for stars (which falls off as 1/distance 2 )? Why the difference? We will begin with the expression we derived from the previous problem:
$$\begin{equation}
𝑚 (r=\Delta + 1AU, \Delta) = 𝐻 − 2.5 log_{10} \bigg{(} \frac{1 AU^4}{\Delta^2 (\Delta + 1AU)^2}\bigg{)}
\end{equation}$$
We continue to expand the following expression:
$$\begin{equation}
𝑚 (r=\Delta + 1AU, \Delta) = 𝐻 − 2.5 log_{10} \bigg{(} \frac{1 AU^4}{\Delta^2 (\Delta^2 + 2\Delta AU + 1 AU^2)}\bigg{)}
\end{equation}$$
$$\begin{equation}
𝑚 (r=\Delta + 1AU, \Delta) = 𝐻 − 2.5 log_{10} \bigg{(} \frac{1 AU^4}{\Delta^4 + 2\Delta^3 AU +\Delta^2 AU^2}\bigg{)}
\end{equation}$$
In the limit of increasing $\Delta$, we can disregard the floating AU units and see that the apparent magnitude limit will be:
$$\begin{equation}
lim_{\Delta\rightarrow \infty} -2.5 log_{10} \bigg{(}\frac{1}{\Delta^4 + 2\Delta^3 +\Delta^2} \bigg{)}\approx \infty
\end{equation}$$
From this exercise we can see that as the asteroid separation from the observer increases the magnitude also approaches infinity - meaning that it becomes very faint. From this exercise, we can also see that asteroid fluxes drop of as 1/d$^4$ instead of the typical d$^{-2}$ relation we see for stars. This is due to the fact we're observing a reflected surface that takes both into account the incident flux and reflected flux. In the even more complex case, it would also be interesting to explore how this relationship would change if the phase angle was not zero.
## Acknowledgements
Many concepts and ideas for the workflow of this project were inspired by conversations and code I shared with Tom Wagg, David Wang, Tobin Wainer, and Jake Kurlander, and John Franklin Crenshaw.