Weighing the spiral galaxy NGC 7083

Adapted from IOAA 2011 (Katowice), data-analysis problem 2. We have restated it in our own words and kept the original data. The original problem gave a photograph and a spectrum to measure, and we give the measurements that the official solution made on them. The solution was last checked on 20 October 2026.

The spiral galaxy NGC 7083 lies 40 Mpc away. The original problem gave its photograph and part of its spectrum, taken with the slit of the spectrograph along the long axis of the galaxy's image. Along the slit one pixel covers 0.82″. Two bright emission lines, with rest wavelengths 6564 Å and 6584 Å, run across the spectrum, tilted because one side of the galaxy approaches us and the other recedes. The official solution measured the following on the images. The two lines are 22 ± 2 pixels apart along the wavelength axis. Measured 10 pixels from the core on either side, the two arms of the 6564 Å line are 7 ± 1 pixels apart along the wavelength axis. On the photograph the image of the galaxy is 11.5 ± 0.5 cm long and 6.5 ± 0.5 cm wide.

(a) Find the speed of rotation of the galaxy 10 pixels from its core, taking the disc to be round. (b) Taking the central bulge to be a sphere, estimate its mass within that distance. The whole problem, which also asked for the rotation curve to be plotted, was worth 25 marks.

AnswerThe galaxy rotates at about 176 km/s, and its bulge holds $(1.1\pm 0.4)\times {10}^{10}$ solar masses within 1.6 kpc of the centre. The official solution follows the same method, but in the step for the tilt it writes 216.7 km/s where its own numbers give 177.5 km/s, and so it finds $1.7\times {10}^{10}$ solar masses.