Expert: Philip Stahl - 3/13/2010

Question
The mass of the Sun is 1.99 x 10^30 kg. Every second, it radiates 3.8 x 1026 J of energy.
(a) How much mass does the Sun lose each day due to this radiation of energy. (b) What percentage of its mass will the Sun lose in 100 years?; in 100 million years?

Answer
Hello,

The key here (actually, the luminosity is closer to 3.9 x 10^26 J/s) is using this figure in conjuntion with the Einstein mass-energy equivalence equation:

E = mc^2

to obtain m, which will be the mass (in kg) converted to energy per second.

Doing this:

m = E/c^2  =  (3.9 x 10^26 J)/ (3 x 10^8 m/s)^2

where c is the velocity of light,

so m = 4.3 x 10^9 kg

or just over 4 billion kilograms of mass converted to energy each second.

To get the amount lost in a day (86400s) just multiply that result by the seconds in a day:

m(86400) =  (4.3 x 10^9 kg/s) (86400s) =  3.7 x 10^14 kg/day


To get the amount of mass lost in 100 years, you need to multiply m by the total seconds in 100 years.


The previous result gave the mass lost in a day, so you just need to multiply that by the number of days in 100 years.


T_100y = (100y)(365.25 d/y) = 3.65 x 10^4 d

So;

(3.7 x 10^14 kg/ day)x (3.65 x 10^4 day)

= 1.36 x 10^19 kg

The percentage of mass lost will be:

% (m) = (1.36 x 10^19 kg)/ (1.99 x 10^30 kg) x 100%  = 6.8 x 10^-10 %

or a minuscule amount!

e.g.

0.00000000068%



I will let you do the % for a 100 million years because it simply follows on from the previous computations. No biggie!

You just need to calculate the days in 1 million years. (You already have it for 100 years)

Then multiply that by what you have already for 100 yrs.

Take the % in similar fashion to what I did above.