The distance to the Large Magellanic Cloud

Adapted from IOAA 2017 (Phuket), data-analysis problem 1, part 3. We have restated it in our own words and kept the original data. The relations to use are the ones found in Sections D2 and D3, which take the place of the relations found in the earlier parts of the original problem. The solution was last checked on 20 October 2026.

The table gives the periods $P$, the mean apparent magnitudes $⟨V⟩$, already corrected for dust, and the mean Wesenheit magnitudes $⟨W⟩$ of five Cepheids in the Large Magellanic Cloud.

Use the relations $M_V=-2.48\log (P/\text{day})-1.44$ and $M_W=-3.43\log (P/\text{day})-2.34$ to find the distance modulus and the distance of each star in both bands. Then find the mean distance to the Large Magellanic Cloud and its standard deviation for each band, say whether the two mean distances differ by more than their scatter and whether the two standard deviations differ, and decide which band gives the better distance. The distance in parsecs is $d={10}^{(\mu +5)/5}$. In the original paper this part was worth 24 marks.

AnswerThe Large Magellanic Cloud is 46 600 ± 4 300 pc away from the V magnitudes and 47 400 ± 900 pc from the Wesenheit magnitudes. The two distances agree within their scatter, but the Wesenheit distances scatter about five times less, because they are free of the uneven dimming by dust, so the Wesenheit band gives the better distance. The official solution, with slightly different relations, finds 46 200 ± 4 100 pc and 48 100 ± 900 pc and the same conclusions.