The lifetime of the sun Essay

The lifetime of the sun

1 - The lifetime of the sun Essay introduction. Why can we say that the rate at which the sun gives off energy at the surface( the luminosity) must be equal to the rate at which it produces energy deep down in the core?

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Temperature affects pressure, that is temperature and pressure are directly proportional. If the core is giving less energy than the sun’s surface, there will automatically be a pressure difference that will make the sun collapse in itself, and the results will be devastating. On the other hand, if the temperature within the core is more than that on the sun’s surface, the pressure difference may cause the sun to explode in a matter of speaking.

 

2.                     4H atoms———————————— 1 He atom + energy

This nuclear reaction has an input ( 4 H atoms) and two outputs, 1He and energy. Since the He atom has less mass than the 4H atoms, that difference in mass must have been converted to the energy. If energy produced is via E=mc2, how much energy is produced by producing 1 He atom from 4H atoms?

E=mc2

m= 0.048 * 10 ^ -27kg= 4.8 * 10^-26 g

c2= (3 * 10^10 cm/ sec)2 = 9* 10^20 cm2/sec2

 

therefore : E= (4.8 *10^26)g * ( 9* 10^20)cm2/sec2

 

=4.32 *10 ^47 gcm2/sec2

 

If 1gm cm2/sec2= 10 ^-7 J

then 4.32 *10 ^47g cm2/sec2= 4.32 *10^40J

 

3. If the sun gives off 3.89*10^33 ergs every second, how many H atoms are being destroyed every second?

I erg = 1g cm2/sec2

 

I H atom = 1.67352* 10^-24g

4 H atoms therefore: 4 x 1.67352*10^24g

= 6.69408*10^24g

 

Energy produced every time one of these four atoms is destroyed

E=mc2

E=  1.67352* 10^-24gx 9* 10^20 cm2/sec2

= 1.506168*10^45cm2/sec2( ergs)

If 1 atom is destroyed by 1.506168*10^45 ergs

then 3.89*10^33 ergs destroys:

3.89*10^33 x 1 atom

1.506168*10^45

= 2.5827* 10^ -12 atoms are destroyed every second

 

4. Why are these reactions confined to the core? The core acts as the nucleus of the sun. In every chemical or physical reaction, reactions always occur in the nucleus. In this case, the Hydrogen protons are only found in the core, hence their reaction taking place there.

 

5. The total mass of the sun is 2*10^33gm. How long will it take until the sun has used up all hydrogen atoms in the core?

 

E=mc2

= 2*10^33g x 9*10^20 cm2/sec2

= 1.8 *10^54gcm2/sec2

 

 

If 1.5606168*10^45 ergs is produced every to destroy 1 hydrogen atom,

then 1.8*10^54 ergs destroy:

 

=   1.8*10^54 x 1 H atom

1.5606168*10^45

=1153390121 atoms are to be destroyed.

 

If it takes 1 second to destroy 2.5827*10 ^ -12 hydrogen atoms, then

time remaining to exhaust the hydrogen atoms will be:

1153390121atoms x 1 Sec

2.5827*10^-12 atoms

 

= 4.46583*10^20 seconds

 

1 year= 3.15*10^7 secs

Therefore: 4.46583*10^20

3.15*10^7

= 1.417724*10^13 years.

 

6. Compare the lifetime of the sun to the current age. How soon will the sun running out of fuel be a problem?

 

Current age———-4.5 billion years

 

The sun has a lifetime of about 11.5 billion years.

 

About 3 billion years to come the sun will run out of fuel.

 

 

 

 

 

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