Is there any mathematical formula which shows that there are approximately $\pi \times 10^7$ seconds in one year. I understand that the pi is probably due to the earth's circular orbit, but am not sure where the rest could come from .
[Physics] Approximation to the number of seconds in a year
timeunit conversion
Related Solutions
For a Schwarzschild observer the gravitational time dilation means the local time at a distance $r$ from a mass $M$, appears to be running slow by a factor of:
$$ \sqrt {1 - \frac{2GM}{r c^2}} $$
You get this straight from the Schwarzschild metric:
$$ ds^2 = -\left(1-\frac{2GM}{r c^2}\right)dt^2 + \left(1-\frac{2GM}{r c^2}\right)^{-1}dr^2 + r^2 d\Omega^2 $$
because hovering above the Sun, both $dr$ and $d\Omega$ are zero.
Anyhow, I reckon the acceleration due to the Sun's gravity is 9.81m/s$^2$ at a distance of about $3.68 \times 10^9$m (using the Newtonian expression for $a$), and at this distance I work out the time dilation to be about 12.6 seconds per year. So the time dilation is not simply proportional to the gravitational acceleration.
You can get an approximate, but very good, relationship between the acceleration and the time dilation by using the Newtonian expression for the acceleration to get $r$ and substituting in the expression for the time dilation. Doing this gives:
$$ \sqrt{1 - \frac{2\sqrt{aGM}}{c^2}} $$
which I don't think is terribly illuminating!
why didn't we modify the measurement of 1 second ever so slightly so as to avoid leap years altogether.
The rotation of earth and the revolution of earth around the sun is not at all synchronized. The Earth really rotates 365.24219647 times during each revolution (in 1992, this ratio changes slightly every year, the tropical year gets roughly a half second shorter each century); so even if we fixed the definition of time to the revolution of earth around sun, we will still need a leap year every 4 years (what we wouldn't need would be leap seconds).
Another reason is because precise time measurement would become incomparable. Since the period of revolution of earth (i.e. the tropical year) isn't constant, if we used the definition of second to exactly match the period of revolution, then whenever you want to specify a precise duration of time, you'll also have to specify which year that definition of second is taken from, and you'll need a table that records the length of second of each year.
Is there any technical reason ... ?
Yes, because with the proper equipment anyone, anytime can take a caesium-133 atom, put it under the specified condition and measure the same second, and it won't have yearly change like the second from earth rotation/revolution would. As far as we know, the frequency of caesium-133 in the 1978 should be the same as the frequency of another caesium-133 in 2049.
Best Answer
It's a unit conversion: $$ 1\,{\rm yr}=\frac{365\,{\rm days}}{\rm year}\times\frac{24\,{\rm hrs}}{1\,{\rm day}}\times\frac{3600\,{\rm sec}}{1\,{\rm hour}}=3.1556926\times10^7\,{\rm sec} $$ Since $3.1557$ is (somewhat) close to $\pi\sim3.1416$, we use the approximation you cite.
Technically, the year is actually 365.25 days long, so using that gives a slightly better approximation that gets one to $3.15576\times10^7\,{\rm sec}$, though most sources I've seen simply use 365 days. Both values are still less than half a percent off of the $\pi\cdot10^7$ value.