Temperature of gas in a cluster

  • #1
physgeek2033
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TL;DR Summary
I am just trying to parse the following together:
The following link states that: "since kT ~ mpv2. A typical velocity dispersion 700 km/s implies T ~ 6 x 107 K from this source alone."

How did they get 6*10^7K ?

When I try this, using mp = 1.67*10^-27 , k =1.38*10^-23 I end up getting 84 as a final answer, nowhere near 6*10^7. Can anyone clarify this?


https://pages.astronomy.ua.edu/keel...o the temperature associated with the cluster
 

Answers and Replies

  • #2
Drakkith
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Check your work. Plugging in ##\frac{m_p v^2}{k}## gives me about ##6*10^7##.
Remember to convert 700 km/s to m/s.
 

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