How much hydrogen does the sun use?

[Gazabone]

Sorry if this seems a silly question, I was never much good at science at school and, in any case, that was a lot of years ago.

However, here's the question; there seems to be a consencious that the sun burns (or fuses) 600 million tons of hydrogen each second. I'm having difficulty understanding quite what this means. As hydrogen on earth is lighter than air, the only way to measure 600 million tons of the stuff would be to measure its upwards pressure presumably (I have this ridiculous image in my mind of a set of scales bolted to the ceiling lol). Or does it mean 600 million tons as weighed on the sun because the gravity of the sun is clearly massively greater than the earth's. If it's this latter definition, then I have no idea what this equates to (eg 1 cubic meter of hydrogen, or 100,000,000 cubic miles etc).

[lukebl]

A damn good question and I look forward to hearing the answers. Also, I'd like to know how it's kept going for billions of years. I just don't get it.

[Vega2]

Rather than think of it as hydrogen gas try and think of it as hydrogen nuclei, or simply put protons. The sun is a big ball of plasma comprised of protons and the number of proton-proton fusion reactions to helium occurring in the suns core is estimated to be 3.6×10^38 per second. Now the mass of one single proton is ~1.67x10^-24 grams and doing the multiplication of these two numbers gives the 600 million tonnes per second you quote.

What I find astonishing is that 'on average' it takes a billion years for a single proton-proton fusion reaction to initiate within the suns core, yes that is right a Billion years. And that the energy density in the sun is less than a back yard compost heap - the sun is just really really big which is why it's so hot ... Oh and it takes hundreds of thousands of years for the high energy photons to travel from the suns core to its surface losing enough energy to turn from gamma rays to visible light. Pretty amazing.

It's also amazing that here on Earth we are trying to do similar feats using both inertial & spatial confinement fusion research in places like ITER in France (and before at JET in the UK) and NIF in the states.

[Vega2]

Here wiki to the rescue I meant to quote it in the text above - http://en.m.wikipedia.org/wiki/Solar_core

Note when I wrote energy density I should have written 'fusion power production' - see article. Can we edit posts here?

[gkec]

First of all don't confuse mass and weight.  An object will always have mass but will only have weight when subject to the gravitational field of another object with mass*.  To make things easy a 1kg mass weighs 1kg in a 1G gravitational field.  

So how do you weigh hydrogen.  Well thought experiments are good for this.  If you had a tank of water and weighed it, and then added oil the oil would float on top and you could measure the weight of the oil.  If you removed the water you could still measure the weight of the oil.

So you could measure a tank containing a vacuum and then add hydrogen and reweigh the tank and measure how much the hydrogen weighs. 

*There is obviously a caveat as a free falling object doesn't weigh anything until it hits the ground.  But then how do things weigh in aircraft.  Obviously it's to do with gravitational attraction v the frame of reference of the weighing machine but how is this formally described?  I suppose it has to be stationary wrt to the gravitational field with the weighing machine providing an equal and opposite force keeping it stationary. Thus weighing it. 

[Gazabone]

Thanks all, that helps a lot. I won't pretend I understand every aspect but it's dispelled my wrong thinking.

Cheers :-)

[spartan45]

Interesting question. I hope this helps, I've included all I've done on the effect of the Sun's annual loss of mass as well.

Objective is to calculate and show how mass loss of Sun effects Earth’s orbit.

E = 3.846 x 10²⁶ Joules per sec (Sun’s luminosity NASA 2013 figure)

m = kg

c = 2.99792458 x 10⁸ m/s  (speed of light in metres per sec)

m = E

       c² 

m = 3.846 10²⁶ J/sec        

       (2.99792458 x 10⁸)² m/s

m = 3.846 x 10²⁶ J/sec   

        8.987551787 x 10¹⁶ m/s

m = 4,279,252,116 kg/sec

Sidereal year in seconds = 31,558,149.8 seconds

Sun’s mass loss sidereal yr = 4,279,252,116 kg x 31,558,149.8s = 1.350452793 x 10¹⁷ kg

Sun’s mass loss per sidereal year = 1.350452793 x 10¹⁷ kg.    

Sun’s mass = 1.9885 x 10³⁰ kg (NASA 2013 figure)

Earth Sun distance = 1AU = 1.495978707 x 10¹¹ metres (adopted 2014 figure)

 Annual AU increase =    Sun’s mass loss per sidereal year kg   x   AU distance metres

                                         Sun’s mass kg

Annual AU increase  =   1.350452793 x 10¹⁷ kg   x   1.495978707 x10¹¹m

                                        1.9885 x 10³⁰ kg

Annual AU increase  =        6.791314021 x 10^-14   x   1.495978707 x 10¹¹m     

Annual AU increase =  0.01016 metres = 1 cm.

Kepler’s third law modified : Orbital separation proportional to 1 

                                                                                                      M

Orbital period proportional to  1  x2

                                                 M

For every 1% decrease of M orbital separation will increase by 1% and orbital period by 2%.

Annual Earth orbit period increase = (6.791314021 x 10^-14)  x 31,558,149.8 secs  x  2

Sidereal yearly increase due to Sun’s mass loss = 4.286426104 x 10^-6 seconds.

[a10ken]

Interesting question. I hope this helps, I've included all I've done on the effect of the Sun's annual loss of mass as well.

Objective is to calculate and show how mass loss of Sun effects Earth’s orbit.

E = 3.846 x 10²⁶ Joules per sec (Sun’s luminosity NASA 2013 figure)

m = kg

c = 2.99792458 x 10⁸ m/s  (speed of light in metres per sec)

m = E

       c² 

m = 3.846 10²⁶ J/sec        

       (2.99792458 x 10⁸)² m/s

m = 3.846 x 10²⁶ J/sec   

        8.987551787 x 10¹⁶ m/s

m = 4,279,252,116 kg/sec

Sidereal year in seconds = 31,558,149.8 seconds

Sun’s mass loss sidereal yr = 4,279,252,116 kg x 31,558,149.8s = 1.350452793 x 10¹⁷ kg

Sun’s mass loss per sidereal year = 1.350452793 x 10¹⁷ kg.    

Sun’s mass = 1.9885 x 10³⁰ kg (NASA 2013 figure)

Earth Sun distance = 1AU = 1.495978707 x 10¹¹ metres (adopted 2014 figure)

 Annual AU increase =    Sun’s mass loss per sidereal year kg   x   AU distance metres

                                         Sun’s mass kg

Annual AU increase  =   1.350452793 x 10¹⁷ kg   x   1.495978707 x10¹¹m

                                        1.9885 x 10³⁰ kg

Annual AU increase  =        6.791314021 x 10^-14   x   1.495978707 x 10¹¹m     

Annual AU increase =  0.01016 metres = 1 cm.

Kepler’s third law modified : Orbital separation proportional to 1 

                                                                                                      M

Orbital period proportional to  1  x2

                                                 M

For every 1% decrease of M orbital separation will increase by 1% and orbital period by 2%.

Annual Earth orbit period increase = (6.791314021 x 10^-14)  x 31,558,149.8 secs  x  2

Sidereal yearly increase due to Sun’s mass loss = 4.286426104 x 10^-6 seconds.

my brain just popped!!

[Cath]

Another possible way to think about the 'weight' of what an area of hydrogen would be, would be to assume no other atmosphere, and so the hydrogen would no longer have any 'buoyancy', just the usual 'downward' force to the planets surface (due to the gravitational well/attraction).

[ollypenrice]

A damn good question and I look forward to hearing the answers. Also, I'd like to know how it's kept going for billions of years. I just don't get it.

It has a huge fuel consumption and an even more huge fuel tank. It really is that simple. Units like 'tonnes' in solar terms are rather like units such as 'miles' in astronomy. They are too small to be make much sense to us.

Olly

[Maximidius]

I read today that the sun converts approx 4 million tons of hydrogen into energy every second. Which is the equivalent of burning its fuel at the rate of a very big skyscraper every second. When ya think of it like that, it's nearly as much as my old landy lol.