Question

The mass of the sun is 2*1030 kg. It currently consists to about 70% (mass) of hydrogen. The dominant source of energy for the sun is the pp-chain (type I) fusion process in which four protons fuse into a 4He core consisting of two protons and two neutrons. This process releases 26.22 MeV ( = 4.2*10-12 J) of energy. Assume that to first order the luminosity of the sun remains constant at 1.1 times its current-day value. How long will it take for the sun until the fraction of hydrogen is reduced from 70% to 30%?

Answer 1

Mass of sun = 2*10³⁰ kg.

70 % of this mass is hydrogen. This is to be reduced to 30%.

Assuming that the mass loss due to the nuclear fusion process is negligible(this actually reduces the total mass of sun)

The mass loss required = 40%

In kg, this mass is 2*10³⁰ *40/100 = 0.8*10³⁰ kg.

The mass of a single hydrogen atom is 1.67355*10^-27 kg.

So, the above mass constitutes 0.8*10³⁰ /(1.67355*10^-27 ) = 0.478*10⁵⁷ atoms.

4 hydrogen atoms can produce one helium atom with the release of 4.2*10^-12 J of energy.

So, energy produced by 0.478*10⁵⁷ atoms. = 0.478*10⁵⁷ *4.2*10^-12 /4 = 0.5019*10⁴⁵ J.

The current solar luminosity as defined by the International Astronomical Union is 3.828*10²⁶ W (W=joules/second)

1.1 times this is 4.2108*10²⁶ W

The time taken for the radiance of energy produced above = total power/luminosity = 0.5019*10⁴⁵/(3.828*10²⁶ ) =

1.31113*10¹⁸ seconds.

This time in years = 1.31113*10¹⁸ /(3.154*10⁷) = 4.157*10¹⁰ years.