The kinetic energy of any moving particle is KE = ½mv 2 , where m is the particl

Question

The kinetic energy of any moving particle is KE = ½mv2, where m is the particle mass, and v is its velocity. Maxwell and Boltzmann made a conversion for ideal gases, which is, KE = 3kT/2, where K is the Boltzmann Constant (1.38 × 10-23 Joules per Kelvin). Accordingly, the average velocity of any hydrogen gases (H = 1 amu) in the intergalactic medium will be found to be v = (3 kT / m), which (for Hydrogen) reduces to v = (3 kT) . Modified from Chapter 18, we have the escape velocity from a mass M, is v2 = GM/r. Manipulating the math, we find GM/r = 3 kT. Further manipulation gives us M = 3kTr/G. With a temperature of 1,000,000 kelvin, Newton's Constant (G) 6.673×10-11 N m2 kg-2, and by further manipulation of the math, giving = kT/G = 2.07x10-7, which produces M = 3r, where r = 15 million (1.5 x 107) light years, and 1 light year = 9.461x 1012 km. Determine the minimum total mass (both visible and dark matter) in Kg that is binding these gases across the above galactic cluster distance.

   Answer = ____________________________ Kg.

Explanation / Answer

Since K=kT/G

where temperature T=106K

Gravitatitional constant G=6.673*10-11Nm2kg-2

boltzman constant k=1.38*10-23 J/K

K=1.38*10-23*106/6.673*10-11

=2.07*10-7

Now M=3Kr

K=2.07*10-7

r=1.5*107light years=1.5*107*9.461*1012 km

=14.19*1019km

Now M=3Kr

=3*2.07*10-7*14.19*1019

=88.11*1012 Kg

Therefore minimum total mass of visible and dark energy is 88.11*1012kg

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