How Nuclear Fusion Directly Affects The Earth’s Orbit

We all know (or maybe not!) that the Sun produces its light via the process of nuclear fusion (turning hydrogen into helium), and loses a little of its mass in doing so.

As a result, an interesting question arises. How much of an effect would this loss of mass for the Sun affect the Earth’s orbit? Since the Earth’s orbit depends on the gravity provided by the Sun, which is in turn related to the Sun’s mass, then there definitely would be a change in Earth’s orbit around our local Star. What we really want to know is how much our orbit will change – will it be small so as not to worry about, or will the change be dramatic enough to change the climate here on Earth and in turn reshape the world in which we will?

To start off with, lets take a look at the fusion of Hydrogen into Helium. Lets look at nuclear fusion. Proton-Proton fusion (the type that occurs in the Sun) is the main process responsible for the electromagnetic radiation radiation that is emitted from it [1]. This is extremely important for providing energy to the rest of the Solar System as radiation is the only form of energy which can propagate through the vacuum (as conduction and convection both require mediums to pass through) [2].

Essentially, Proton-Proton fusion (as the name implies) is the fusing of multiple protons into larger and heavier elements (Note: when referring to a proton, we are really talking about the most common type of isotope* (see definitions) of Hydrogen, which is a single proton in the nucleus) [1]. There are a number of steps involved in describing the fusion of hydrogen into Helium, and involves more than just two Hydrogen’s joining together. Rather than explaining this process in writing, I shall explain the picture below (as it is a little easier to grasp with a diagram).

Image from Hyperphysics [1]

Step 1: What this is showing is the fusion of two 1Hydrogen atoms (which is really just two protons).

Step 2: The masses for a proton and a neutron are very similar; 1.6726 x 10^(-27) kg and 1.6749 x 10^(-27) kg respectively [3]. The close resemblance of their masses is no coincidence. The Standard Model was developed in the early 1970’s, and described what all matter (including a set of particles referred to as Baryons – which contain the protons and neutrons) in the universe was made of – the fundamental particles of quarks* and leptons [4]. And as it turns out, protons and neutrons aren’t actually that different from each other. The single large difference between them is their charge; protons are positively charged particles and neutrons have no charge. But more importantly, their internal structures are almost identical [5]. The proton consists of three quarks; two up quarks and one down quark. The neutron also consists of three quarks, however it only has one up quark but two down quarks [5]. The image [5] below will help you visualise this:

The squiggly lines between the quarks simply describe the strong nuclear force (messenger particles called gluons) which is responsible for holding the nucleus together, but this is not important for our response [5]. Just another quick addition, the quarks aren’t actually coloured (well a quark has never been seen to verify this), they are just given a colour by us to describe how a baryon (proton/neutron) can exist – that is with all three colours colours mixing together to create a colourless/white combination. It is merely an analogy known as colour charge to explain why individual quarks cannot exist by themselves outside the particle (which is why we have never seen them) [6]. This again is out of the scope of the question, but is interesting nonetheless.

Anyway now this allows us to explain step 2 properly. It describes how a proton can transmute* into a neutron by emitting a positron* and an electron neutrino*. This is known as the process of beta-plus decay [8]. This is possible as the proton and neutron are fundamentally the same (as described by quarks above). Thus we have now created an isotope of Hydrogen (as the atomic number which is based on the number of protons has remained the same – only the number of neutrons have changed). This is called deuterium.

Step 3: This is simple, where the deuterium isotope now combines with another proton (Hydrogen) to form a new element; Helium-3 (the 3 standing for the atomic mass, being 2 protons + 1 neutron). It is important to understand that this is not the most common type of isotope of Helium that exists, as a more stable isotope exists (where the stability of an isotope is roughly ruled by the ratio of ~ 1 proton:1 neutron) [7].

Step 4: Two He-3 isotopes combine to form an alpha particle (He-4, which is the most abundant isotope of Helium) and the release of two protons.

Step 5: The release of these two protons also releases a large amount of energy in the form of gamma rays. Note that smaller doses of gamma radiation were also given off in the previous processes [1].

Just to finish up on the process of nuclear fusion, lets see why this process even occurs in the first place. For nucleons to remain stable in the nucleus, energy must be converted from their mass to hold them together. This is known as the binding energy. Essentially it is the amount of energy that has to be put in to separate the nucleons* from each other [9]. So why would energy be wasted trying to achieve this process in nuclear fusion? Well, we see that there is a trend which dictates that each subsequent element has a lower potential energy in the nucleus – or in other words, the nucleus becomes more stable. Considering that reactions that take something to a lower energy state are favourable, nuclear fusion occurs under the right conditions. Note that the most stable nucleus is that of Iron – from there on in, the stability of the nucleus decreases so fusion does not occur past iron in any star [10]. Here is an image [10] of the stability trend:

So now that we have a good understanding of what occurs during nuclear fusion, as well as why it occurs, lets return to the later part of the question.

How much mass is lost, and how does this effect the Earth’s orbit?

All that is necessary to find how much mass is being lost is to determine the energy being expended from the binding energies being broken (and we are only going to take into account the proton-proton fusion as fusion between larger elements is microscopically small for the Sun at the moment; this is because it still has a large amount of Hydrogen fuel to burn with the composition* being 91% Hydrogen and 8.7% Helium; also amongst that are a total of 0.3% various metals [11]). To find a value for this we simply need to find the difference in atomic mass* for 4 1H atoms and 1 4He atom. this equates to:

amu* 1H = 1.00794
amu 4He = 4.0026202

Mass change (lost) in percentage = { [(4 * 1H) – (1 * 4He)] / [4He] } x 100

{ [4*(1.00794) – (4.002602)] / [4.0026202] } x 100 = 0.728476%

This figure falls closely in line with what was measured from other educational sources; The Cornell University astrophysics department measured a rounded value of 0.7% [14]. This mass difference is the binding energy discussed above. This means that at most, if all the hydrogen was converted to Helium, the Sun would end up weighing only 0.7% less in total. “In actuality, not all of the Sun’s mass is hydrogen to start with, and only the mass in the inner core of the Sun gets hot enough to undergo fusion reactions, so we really only expect around 0.07% of the mass to get converted“.

To obtain a deeper understanding, lets look at the rate at which mass is lost per second. Thanks to Einstein, we can convert the binding energy that has been lost into mass from the equation E = Mc^2 ; where E is the energy (in the form of electromagnetic radiation) emitted from the Sun, M is the mass of the Sun, and c is the speed of light [14]. Since we want to find the mass, we can do a little rearranging to obtain: M = E/c^2 . We know that c^2 is a constant which is (3 x 10^8)^2, so all we need to do is measure the luminosity* of the Sun to determine the energy. We do this by measuring the intensity of sunlight reaching the top of the Earth’s atmosphere (1360 watts per square metre) and the distance that light has travelled in all directions. This last part is the surface area of a sphere with radius equalling the distance from the Sun to the Earth (2.8 x 10^23 m^2) [15]. By multiplying the two together we obtain:

Energy = (1360) * ((2.8 x 10^23) = 3.808 x 10^26 Watts

which is in agreement from with a value for Cornell once again (3.8 x 10^26 W) [14].

Now simply find the mass lost per second through M = E/c^2, so

M = (3.8 x 10^26) /  (3.0 x 10^8)^2 = 4.2 x 10^9 kg

which is in line with an approximation of 4,200,000,000 kg from Cornell. (Note*: that is 4.2 billion kg per second!)

However don’t be fooled by the large number. The mass of the Sun is approximated to 2.0 x 10^30 kg. Conceptually, it helps to remember that a power is an exponent, so increasing the power produces an exponential graph rather than a linear one. One example that I refer to produced by Cornell is the mass loss over 100 years – it equates to 1.3 x 10^19 kg of loss, but still remains only a fraction of the Sun’s total mass (being 6.6 x 10^-12 as a ratio, or equivalently to 6.6 parts in a trillion) [14].

Moving onto the effect on the orbit of Earth, we must take a look at orbital energies. The total mechanical energy of a system can be mostly be divided up into 2 main energies; that of kinetic energy* and potential energy*. The addition of these 2 energies gives (the equations below were derived with reference to [16])

E(mechanical) = (mv^2/2) – (GmM/r) = some constant (as the kinetic and potential energies change at equal rates in opposite magnitudes)

dividing by the mass of Earth gives:

E(mechanical)/m = (v^2/2) – (GM/r)

Now looking at Kepler’s Third Law (to find orbital period, T):

T^2 = 4 * Pi^2 * a^3 / GM  where a is the semi-major axis (average distance to the Sun); G = Newtons Universal Law of Gravitation; M is the mass of the Sun

Just like an ellipse/circle, the circumference (orbit) is 2 * Pi * a. Divide this by the time to obtain velocity (distance/time)

v = 2 * Pi * a / T

To save the tedious mathematics, squaring this and replacing the period with Kepler’s 3rd Law gives:

v^2 = GM / a

Another equation (which does not need to be derived here) proves that the mechanical energy is equal to the negative of the kinetic energy.
Now plugging the above equation into the kinetic energy and solving gives:

E(kinetic) = GM / 2a

This also works out to be half the potential energy. Thus, as the mass M (of the Sun) in the numerator becomes smaller, so too does the potential energy. As potential energy is defined to be zero at some point at an infinite distance away, this means that is it always negative. Therefore a smaller magnitude means that it must become less negative, and is approaching 0. Using the same logic, it must then be approaching a distance an infinite distance away. Hence, the orbital radius will increase as the Sun’s mass decreases [16]. But by how much, that is the real question.

The final equation to work this out is given by [16]:

a(new) / a(original) = ( M(new) / M(original) ) / (2 M(new) / M(original) -1)

However, luckily for us we can eliminate the mathematics.For small percentage changes in the Sun’s mass, the formula can reduce to a very simple approximation [14]. “For every percentage decrease in the Sun’s mass, the orbital separation of the planet will increase by the same percentage, and the orbital period of the planet will increase by twice the percentage“. This means that if we return to the values used in the above example of the mass loss for over 100 years (where the Sun’s mass decreased by about 6.6 parts per trillion), then according to the above statement , Earth’s orbit will also increase by 6.6 parts per trillion! So what does this equate to in terms of its current distance (of 1.5 x 10^11 m)? A single metre. Just 1 metre over 100 years. Also of note is the increase in the orbital period (which is related to orbital distance by Kepler’s Law), is also a mere 0.4 milliseconds. So at the end of every century, a year has grown by 0.4 milliseconds [14]. Due to this result, the effect is pretty much immeasurable over any reasonable period of time.

So there is little to worry about in terms of variation. The Earth’s orbit will not deviate into a large elliptical orbit, and no, we wont fling off into the galaxy and/or spiral into the Sun! Rejoice!


Proton is used in reference to 1H.

isotope: variations in the number of neutrons contained within a specific element. There exists multiple isotopes for each element.

quark: fundamental particles that make up all matter including protons and neutrons

transmute: change in form; i.e. a proton turning into a neutron

positron: the anti-particle to the electron (identical but opposite in charge)

electron neutrino: a possibly massless neutral particle

nucleons: protons and neutrons

composition: measured in % of atoms rather than total mass

atomic mass: measurement of mass for nucleons in atomic mass units (1/12 the mass of an unbound neutral atom of carbon-12)

amu: atomic mass unit

luminosity: the brightness of a celestial object

kinetic energy: energy which a body possesses by virtue of being in motion (planet moving along its orbit

potential energy: energy available for an object to use; possessed by virtue of a bodies position relative to others (energy of Earth due to its position in the Sun’s gravitational field)

[1] – Proton-proton fusion. 2014. Proton-proton fusion. [ONLINE] Available at: [Accessed 31 August 2014].

[2] – Halliday, Resnick, Walker,  2014, Fundamentals of Physics: 10th edition, WILEY, Chapter 18

[3] – Neutron vs. Proton Mass . 2014. Neutron vs. Proton Mass . [ONLINE] Available at: [Accessed 31 August 2014].

[4] – The Standard Model | CERN. 2014. The Standard Model | CERN. [ONLINE] Available at: [Accessed 31 August 2014].

[5] – Protons and Neutrons: The Massive Pandemonium in Matter | Of Particular Significance. 2014. Protons and Neutrons: The Massive Pandemonium in Matter | Of Particular Significance. [ONLINE] Available at: [Accessed 31 August 2014].

[6] – The Color Force. 2014. The Color Force. [ONLINE] Available at: [Accessed 31 August 2014].

[7] – ESRL Global Monitoring Division – Education and Outreach – Isotopes of CO2 . 2014. ESRL Global Monitoring Division – Education and Outreach – Isotopes of CO2 . [ONLINE] Available at: [Accessed 31 August 2014].

[8] – Beta Decay. 2014. Beta Decay. [ONLINE] Available at: [Accessed 31 August 2014].

[9] – E=mc2 and binding energy. From Einstein Light. 2014. E=mc2 and binding energy. From Einstein Light. [ONLINE] Available at: [Accessed 31 August 2014].

[10] – Nuclear Energy – The Theory. 2014. Nuclear Energy – The Theory. [ONLINE] Available at: [Accessed 31 August 2014].

[11] – What is the Element Composition of the Sun? 2014 Solar Scientific Composition. [ONLINE] Available at: [Accessed 31 August 2014].

[12] – Chemical – Hydrogen (H). 2014. Chemical – Hydrogen (H). [ONLINE] Available at: [Accessed 31 August 2014].

[13] – Chemical – Helium (He). 2014. Chemical – Helium (He). [ONLINE] Available at: [Accessed 31 August 2014].

[14] – Curious About Astronomy: When the Sun converts mass to energy, do the orbits of the planets change?. 2014. Curious About Astronomy: When the Sun converts mass to energy, do the orbits of the planets change?. [ONLINE] Available at: [Accessed 31 August 2014].

[15] – Climate and Earth’s Energy Budget : Feature Articles. 2014. Climate and Earth’s Energy Budget : Feature Articles. [ONLINE] Available at: [Accessed 31 August 2014].

[16] – The Effect of Stellar Mass Loss on Planetary Orbits. 2014. The Effect of Stellar Mass Loss on Planetary Orbits. [ONLINE] Available at: [Accessed 31 August 2014].

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