How do you calculate the lifetime of the Sun?
How can I calculate the age of the sun using classical physics? I know it's estimated to have a life of about 10 billion years, but was this calculated?
It requires a bit more than classical physics, but still, you can estimate the sun's lifetime from a very simple calculation.
First of all, if you want the current age of the sun (around 5 billion years) this number is determined from radioactive dating of objects in the solar system which are known to have formed around the same time as the sun, as stated in the answer to a previous question.
The total lifetime of the sun before it becomes a red giant is, as you say, around 10 billion years (meaning that the transition will occur around 5 billion years from now). This can be estimated by assuming that the sun will "die" when it runs out of energy to keep it shining. The time for this to occur is roughly the total energy the sun has that can be turned into light, divided by the rate at which the sun is giving off energy, or:
lifetime = (energy) / (rate [energy/time] at which sun emits energy)
The rate at which the sun emits energy (its luminosity) is around 3.8 x 1026 Watts (that's the number 38 followed by 25 zeroes - quite a lot of lightbulbs!). This number can be determined from measurements of how bright the sun appears from Earth as well as its distance from us.
The total energy that the sun has to burn requires a little extra knowledge (for example, some nuclear physics) to understand. We know that the sun shines via nuclear reactions in the core that transform four hyrogen atoms into one helium atom. If you look at a periodic table, you will see that one helium atom has a little less mass than four hydrogen atoms combined; about 0.7% of the original mass has "disappeared". This "missing mass" gets transformed into energy, and this is the energy that causes the sun to shine. Therefore, using Einstein's famous formula E=mc2 for the conversion between mass and energy we have that the available energy in the sun is:
E = 0.007 x M c2
where c is the speed of light and M is the amount of mass in the sun that is capable of undergoing the above nuclear reactions.
Now, it turns out that only the central part of the sun is at a high enough temperature to actually undergo these reactions. You would need to use a detailed model of the sun's structure to figure out exactly how much of the sun is at a high enough temperature, but if we're just estimating things we can say that on the order of 10% of the sun's mass is in the central part of the sun where it is hot enough to undergo nuclear reactions. We then have:
E = 0.007 x 0.1 x Msun c2
where Msun is the total mass of the sun, 2 x 1030 kilograms. We therefore can calculate that the total energy the sun has to burn is around 1.3 x 1044 Joules. Dividing 3.8 x 1026 Watts (the rate at which the sun is giving off energy) into this number gives an approximate value of 10 billion years for the sun's lifetime.
Get More 'Curious?' with Our New PODCAST:
- Podcast? Subscribe? Tell me about the Ask an Astronomer Podcast
- Subscribe to our Podcast | Listen to our current Episode
- Cool! But I can't now. Send me a quick reminder now for later.
- Is it true that the Sun burns off a billion tons of gas every 5 seconds?
- How do we know the age of the Universe and the Earth?
- Will the sun go supernova in six years and destroy Earth (as seen on Yahoo)?
- When the Sun converts mass to energy, do the orbits of the planets change?
How to ask a question:
If you have a follow-up question concerning the above subject, submit it here. If you have a question about another area of astronomy, find the topic you're interested in from the archive on our site menu, or go here for help.Table 'curious.Referrers' doesn't existTable 'curious.Referrers' doesn't exist
This page has been accessed 92356 times since November 11, 2002.
Last modified: November 11, 2002 2:16:46 PM
Ask an Astronomer is hosted by the Astronomy Department at Cornell University and is produced with PHP and MySQL.
Warning: Your browser is misbehaving! This page might look ugly. (Details)