Dark Matter — Question 1

Section 44-12 · Physics pages 7512–7513

Question 1 A flat rotation curve implies increasing enclosed mass Assume a spherical mass distribution and circular speed 200 km/s at r=20 kpc. Use G = 4.30 × 10−6 kpc (km/s)2 per solar mass. r vc Schematic; labels, not drawing lengths, specify the data. (a) Derive and calculate enclosed mass. (b) If speed remains 200 km/s at 40 kpc, find the added mass between the two radii.
Show solutionHide solution
Question 1 - Solution A flat rotation curve implies increasing enclosed mass r vc (a) Method and calculation v2 c/r = GM(< r)/r2 ⇒M(< r) = v2 cr/G . M(< 20) = 2002(20)/(4.30 × 10−6) = 1.86 × 1011M⊙. (b) Interpretation and further calculation M(< 40) = 3.72 × 1011M⊙. ∆M = 1.86 × 1011M⊙. Physical check A flat speed is inconsistent with a fixed enclosed point mass, which predicts speed proportional to the inverse square root of radius. Disk geometry changes detailed inference.

Question 1

A flat rotation curve implies increasing enclosed mass

Assume a spherical mass distribution and circular speed 200 km/s at r=20 kpc. Use G=4.30×10−6G=4.30\times10^{-6} kpc (km/s)2^2 per solar mass.

Diagram shown in the original SVG worksheet.

  1. Derive and calculate enclosed mass.

  2. If speed remains 200 km/s at 40 kpc, find the added mass between the two radii.

Question 1 - Solution

A flat rotation curve implies increasing enclosed mass

Diagram shown in the original SVG worksheet.

(a) Method and calculation

vc2/r=GM(<r)/r2⇒M(<r)=vc2r/G.v_c^2/r=GM(<r)/r^2\Rightarrow\boxed{M(<r)=v_c^2r/G}.M(<20)=2002(20)/(4.30×10−6)=1.86×1011M⊙.\boxed{M(<20)=200^2(20)/(4.30\times10^{-6})=1.86\times10^{11}M_\odot}.

(b) Interpretation and further calculation

M(<40)=3.72×1011M⊙.\boxed{M(<40)=3.72\times10^{11}M_\odot}.ΔM=1.86×1011M⊙.\boxed{\Delta M=1.86\times10^{11}M_\odot}.

Physical check

A flat speed is inconsistent with a fixed enclosed point mass, which predicts speed proportional to the inverse square root of radius. Disk geometry changes detailed inference.