I have one problem:

On picture below we can see Erigone family asteroids in plane semimajor axis - absolute magnitude. Based on V-type as a consequence of Yarkovsky effect, estimate age of this family bearing in mind that members of this family have spectral of C-type. For average value of change of semimajor axis which belongs to this family take $da=5 \times 10^{-4} \text{AU}$ (for object with diameter $D=1 \,\text{km}$ and time $t=1$ million years). Ignore initial size of family.

enter image description here

I really don't know how to solve this problem (I am studying math, not astronomy). Professor won't tell us how to do it, and I tried to find the answer in his textbook and in two more books in my language, but there is no similar problem at all.

Could you point me in the right direction?

  • $\begingroup$ @called2voyage I'm sorry but I really don't see anything relevant in what you re-added. The purpose of the site is to make the question relevant to a broad audience, it doesn't matter if the person who originally asked has a teacher or searched for an answer on some books in his language. $\endgroup$ Commented Jan 15, 2014 at 17:07
  • $\begingroup$ @EduardoSerra You are incorrect: the data is relevant to the particular incident that the user needs solved and doesn't obstruct the overall goal of the site. This is how Stack Exchange was designed to operate. $\endgroup$
    – called2voyage
    Commented Jan 15, 2014 at 17:08
  • $\begingroup$ It also gives record of prior research on the OP's part and the level of detail he needs in an answer. $\endgroup$
    – called2voyage
    Commented Jan 15, 2014 at 17:09

1 Answer 1


I'm not a professional astronomer, but I'm going to take a stab at this. If anyone sees any problems, feel free to correct them.

The Yarkovsky effect basically means that spinning asteroids over time will be pushed into a further orbit (more precisely, it increases the semimajor axis of the orbit) due to uneven heating and cooling.

Apparently to estimate the age of an asteroid family, using only the Yarkovsky effect, we would start out with this equation:

$$0.2H = \log(\Delta a / C)$$

This equation assumes a fixed geometric albedo for the whole family. $H$ here is the absolute magnitude. $\Delta a$ is equal to $a - a_c$ where $a$ is the semimajor axis.

$$C = \sqrt{p_v} (da/dt)_0 T \cos \epsilon$$

$p_v$ is the geometric albedo. $(da/dt)_0$ is the maximum Yarkovsky drift rate for a body of size $D_0$, where $D_0$ is an arbitrary reference size (we could use your provided $D$ value above). $T$ is the age of the family that we are trying to calculate. $\epsilon$ is the spin axis obliquity.

Note: I've had to remove my example solution because I could not figure out how to correct my errors from accidentally putting $C$ outside the logarithm.

You can check out more details in my source below.


"Yarkovsky/YORP chronology of asteroid families" - Vokrouhlickýa, Broža, Bottke, Nesvorný, Morbidelli

  • $\begingroup$ Thank you for this! I will study that paper. One note: I think your $C$ is wrong, becaue $0.2H=\log\left(\dfrac{\Delta a}{C}\right)$ ($C$ is in logarithm), so $C=\dfrac{0.06}{e^3}$. $\endgroup$
    – user110822
    Commented Jan 15, 2014 at 14:11
  • $\begingroup$ @user110822 You are correct, I missed that $C$ was in the log. Thanks. $\endgroup$
    – called2voyage
    Commented Jan 15, 2014 at 14:14
  • $\begingroup$ And should we use $H=9$, because on page 125 of your source it says (Fig 4.): (163) Erigone is shown as a large filled square. In my picture, that is star. And I will see why did you choose other parameters like that (probably from that paper, I will see). If I have problems I'll ask you :-) I will accept this answer later, maybe somebody want to say something more on this question. $\endgroup$
    – user110822
    Commented Jan 15, 2014 at 14:22
  • $\begingroup$ @user110822 I had several errors in my example solution. I have removed it for now. Perhaps someone can help me work out how to fix it later. $\endgroup$
    – called2voyage
    Commented Jan 15, 2014 at 14:25
  • $\begingroup$ All right, but I can't see what is problem with finding $C$? $$0.2H=\log\left(\dfrac{\Delta a}{C}\right) \implies e^{0.2H}=\dfrac{\Delta a}{C} \implies C=\frac{\Delta a}{e^{0.2H}} $$ (if $\log$ is for base $e$) $\endgroup$
    – user110822
    Commented Jan 15, 2014 at 15:20

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