- #1
Barbequeman
- 7
- 1
- Homework Statement
- The surface density of stars in a galaxy at a radius of 1 kpc from the centre is 100 Msun/pc^2. The surface density at 10 kpc is 10 Msun/pc^2. Assuming that the surface density of stars is given by the exponential law, calculate
(a) The scale length of the disk.
(b) The surface density at the centre of the galaxy.
- Relevant Equations
- σ(r) = σ0e−r/h
a.)
The scale length of the disk is the length over which the surface density of stars decreases by a factor of e. In this case, the surface density decreases by a factor of 10 over a distance of 9 kpc, so the scale length is 9 kpc. The surface density of stars at a radius of r from the center of the disk is given by: σ(r) = σ0e−r/h where σ0 is the central surface density and h is the scale length of the disk. We can rearrange this equation to solve for h: h = −r/ln(σ(r)/σ0) plugging in the values from the question, we get: h = −9 kpc/ln(10/100)
h = 9 kpc
The scale length of the disk is the length over which the surface density of stars decreases by a factor of e. In this case, the surface density decreases by a factor of 10 over a distance of 9 kpc, so the scale length is 9 kpc. The surface density of stars at a radius of r from the center of the disk is given by: σ(r) = σ0e−r/h where σ0 is the central surface density and h is the scale length of the disk. We can rearrange this equation to solve for h: h = −r/ln(σ(r)/σ0) plugging in the values from the question, we get: h = −9 kpc/ln(10/100)
h = 9 kpc
