Surface Density of a Galaxy

Classical Mechanics Level 2

The surface density of luminous matter in a galaxy has units of mass per area and obeys the equation

D ( r ) = D 0 e − r / r 0 , D(r) = D_0 e^{-r/r_0},

where D 0 D_0 is some constant with units of mass per area. Suppose we have some galaxy where D 0 = 0.25 kg / m D_0 = 0.25 \text{ kg}/\text{m} . What is D ( 2 r 0 ) D(2r_0) in units of kg / m \text{ kg}/\text{m} , to the nearest thousandth?


The answer is 0.034.

Matt DeCross by Matt DeCross , 27, USA

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1 solution

Matt DeCross
Jan 26, 2016

Evaluate directly: D ( 2 r 0 ) = . 25 kg / m ∗ e − 2 ≈ . 034 kg / m . D(2r_0) = .25 \text{ kg}/\text{m} * e^{-2} \approx .034 \text{ kg}/\text{m}.

3 Helpful 2 Interesting 0 Brilliant 2 Confused

I’m confused

Kerry White - 1 year, 11 months ago

The problem asks you to plug in for the constant D 0 and evaluate the function at r=2r 0, giving e^{-2} as the exponential. Can you say where you are confused?

Matt DeCross - 1 year, 10 months ago

Very interesting.🧐 very very interesting. Ah I love dark matter.

Emily Fleming - 11 months, 1 week ago

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