Processing math: 100%
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Physics LibreTexts

2.10: S11. The Friedmann Equation - SOLUTIONS

( \newcommand{\kernel}{\mathrm{null}\,}\)

Exercise 11.1.1

Answer

K.E. is 12mv2, where v=˙a, so the test particle's kinetic energy is

12m˙a22

Exercise 11.1.2

Answer

P.E. is GM(<d)md, where by M(<d) we mean the mass contained in the sphere of radius d, so M(<d)=43πd3ρ.

sphere_11_1_2.jpg

Therefore the test particle's potential energy is

G43πρd2m

Exercise 11.1.3

Answer

12m˙a22G43πρd2m=κ

Exercise 11.1.4

Answer

Recall that d=a=da, substituting this in and rearranging our equation we get

12md2˙a2a212md28πGρ3=κ

dividing through by 12md2 gives

˙a2a28πGρ3=2κm1d2

Then we substitute back in d=a and solve for (˙aa)2:

(˙aa)2=8πGρ3+2κml2×1a2


This page titled 2.10: S11. The Friedmann Equation - SOLUTIONS is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Lloyd Knox.

  • Was this article helpful?

Support Center

How can we help?