0.07%

0.07%


That's a small number.


7 parts in 10,000.


But this small fraction is a significant number in terms of the evolution of our universe. This is the mass fraction difference between one helium nucleus and four hydrogen nuclei. During the proton-proton chain, four protons (essentially, the nucleus of a hydrogen atom is just a proton) are fused into one helium nucleus (two protons and two neutrons). The helium nucleus is smaller than four hydrogen nuclei by only 0.07%. This extra 0.07% mass is converted to energy via E=mc².


This fraction may seem insignificant, but in reality, it is very important. If the mass fraction was only 0.06%, then stars would take too long to evolve and may even be unstable. The energy released by the proton-proton chain would not be enough to hold back gravitational forces from the outer layers of the star, and the star would collapse.*


*The balance between the energy from fusion in the core of a star and the gravity of the star pushing in is called hydrostatic equilibrium. This is why stars are stable. It is also why stars tend to expand and contract. As the star evolves, the fusion in the core increases as it begins to use up its fuel and the star expands as it evolves on the main sequence. When fusion stages end (at the end of the Main Sequence, for example), gravity will overcome the fusion energy and push in. See more about the evolution of a star here.


If the mass fraction was just a tad higher, at 0.08%, the fusion of hydrogen into helium would occur too fast and stars would use up their fuel too quickly (and possible even dissipate the outer layers of the star as the energy would overcome gravity and push out the gases in the outer layers). If the stars were stable, the stars would use up their mass too quickly for planets to form and in turn, life would probably not evolve.

Sidereal Day Vs. Synodic Day

I’ve talked here before about sidereal days and synodic days. What exactly is the difference between the two?

It all comes down to one thing: the object that you are using as a reference. For example, the sidereal day of the Earth is equal to 23 hours, 56 minutes, and 4 seconds. The sidereal day uses a distant star (which is basically any star that is not the Sun) to measure against. You would determine how long a star would take to reach the same location in the sky on consecutive days. Another way to look at it, is the length of time it takes the Earth to rotate 360°.

However, the synodic day is 24 hours. The best way to think of the synodic day is to just determine the length of time between two consecutive noons. A synodic day means the Earth rotates more than 360° to get the Sun in the same location in the sky. The reason this happens is because as the Earth rotates on its axis, it is also moving around the Sun. Granted, it is only moving a little less than a degree in its orbit (since the Earth year is about 365.25 days), but that is long enough that the Earth has to rotate just a tad more to get the Sun back to noon.

 

Geosynchronous and Geostationary Orbits

Objects can orbit the Earth in different ways. Most orbits look the same from above, a sine curve (or if you like different phasing, a cosine curve) with the Earth’s equator as the x-axis. The difference is how fast the satellite or spacecraft or space station takes to orbit the Earth.

However, there is a special orbit which does not orbit the entire Earth, but stays above a particular longitude. These orbits are called geosynchronous orbits. These orbits have a period that is just equal to the Earth’s sidereal day (23 h, 56 min, and 4 seconds). Because of this orbit, they tend to remain around the same longitude on Earth and if you were to look down on this orbit, the satellite would trace out something called an analemma, which is just a fancy term for the figure 8. Depending on the inclination of the orbit, these satellites are not visible from all parts of the Earth. These orbits are used mostly for communications and weather satellites. This is why you do not have to move your satellite dish if you have satellite television, as a non-geosynchronous orbit would be a pain if you are watching your favorite TV shows.

There is a special geosynchronous orbit called a geostationary orbit. Not only does this have a period of one sidereal day, but a satellite in this orbit does not move at all. It is always above the same place on Earth, and by definition, the location in the sky must be above the equator. If we were to build a space elevator (more on this concept later), the receiving station for the elevator must in a geostationary orbit. All geostationary orbits are geosynchronous, but not all geosynchronous orbits are geostationary.

How far up is an object in a geosynchronous/geostationary orbit? Just using some basic concepts from Newtonian mechanics, the calculation is relatively simple.

First, to be in a stationary orbit, the force of gravity on the satellite must be counteracted by the centripetal force, i.e.:

Where:

• G is the gravitational constant, 6.67x10^-11 m³/kg·s²
• M_E is the mass of the Earth, 5.972x10²⁴ kg
• m is the mass of the satellite
• v is the orbital velocity, m/s
• R is the radius of the orbit (assuming circular orbit), in m

Equating these two and we get:

We know the period of the orbit (P) has to be one sidereal day, 23h56m4s, which in seconds is 86,164 seconds (60 seconds in a min, 60 min in an hour) and the orbital velocity is just the length of the orbit (the circumference of the orbit, 2πR) divided by the period, P.

Plug v=2πR/P into the above equation and simplifying, we get:

and plugging in all the constants, we find that the orbital radius is 42,164 km. (If you like, you can solve this yourself and see if I’m right.) Note, that this is the radius of the orbit from the center of the Earth. If we take into account the Earth’s radius, the orbital altitude is 35,786 km (R_E = 6378 km at the equator).