PS09_lensing_solutions_modified

3. We observe a galaxy, and measure a velocity dispersion of σ = 400 km/s. The galaxy has a radius of 20 kpc. [20 pts] (a) Assuming that the galaxy is an isothermal sphere, and has isotropic motion of the stars, what is the mass of the galaxy (in solar masses)? Using the virial theorem, we can estimate the mass of a galaxy as M = 3 β η Rσ 2 G . Assuming an isothermal, isotropic sphere, β = η = 1. Then M = 3(2 × 10 4 pc)(3 . 086 × 10 16 m/pc)(4 × 10 5 m/s) 2 6 . 67 × 10 − 11 m 3 kg − 1 s − 2 = 4 . 44 × 10 42 kg = 2 . 23 × 10 12 M ⊙ (b) When observing this galaxy, we see two images of a more distant galaxy. These two images appear at 0.840’ and 0.540’ from the nearer galaxy (’ = arcminutes). Using your answer from part (a), and assuming the nearer galaxy is 10 Mpc away, how far away is the more distant galaxy? (You may ignore any effects from the expansion of the Universe in this problem.) We must first convert the image locations to radians: | θ + | = (0 . 840 ′ )(1 ◦ / 60 ′ )( π/ 180 ◦ ) = 2 . 44 × 10 − 4 | θ − | = (0 . 540 ′ )(1 ◦ / 60 ′ )( π/ 180 ◦ ) = 1 . 57 × 10 − 4 . Then, we can relate the distance to the lens galaxy D ℓ = 10 Mpc to the mass of the lens galaxy M , the distance to the source D s , and the distance between the source and the lens D ℓs = D s − D ℓ as M = c 2 4 G D ℓ D s D ℓs | θ + θ − | = c 2 4 G D ℓ D s D s − D ℓ | θ + θ − | D s − D ℓ D s = c 2 D ℓ 4 GM | θ + θ − | 1 − D ℓ D s = c 2 D ℓ 4 GM | θ + θ − | D ℓ D s = 1 − c 2 D ℓ 4 GM | θ + θ − | = 1 − (3 × 10 8 m/s) 2 (10 Mpc)(3 . 086 × 10 16 m/pc) 4(6 . 67 × 10 − 11 m 3 kg − 1 s − 2 )(4 . 44 × 10 42 kg) (2 . 44 × 1 . 57) × 10 − 8 = 1 − 0 . 898 = 0 . 102 D s = (0 . 102) − 1 D ℓ = 9 . 82(10 Mpc) = 98 . 2 Mpc . 3