This series is intended to revisit some of the deeper, more thoughtful questions directed at me during my time serving on All Experts. Each question will be given, followed by the answer.

*Question: I am puzzled by Einstein’s conclusion that Mercury’s advance of perihelion is*

__43 arc seconds per year__. How can that be obtained from the equation he used in The Principle of Relativity?**Diagram showing (exaggerated) advance of Mercury's perihelion.**

Ans. What
Einstein is really concerned with here is the amount of rotation of 'the
planetary ellipse' due to the effects of gravitation in general relativity. In this case the ellipse is that of the planet
Mercury.

The equation used and purported to show the amount of this rotation (i.e. advance of perihelion) is given as:

e = [24 (p)

where e is the advance (or rotation) in

Einstein, on page 164, states that for Mercury e= 43 seconds per century

The equation used and purported to show the amount of this rotation (i.e. advance of perihelion) is given as:

e = [24 (p)

^{3}a^{2}]/ T^{2}c^{2}[1 - e^{2}]where e is the advance (or rotation) in

*seconds of arc,*T is the period of revolution in seconds, c the velocity of light and e the eccentricity of the orbit.Einstein, on page 164, states that for Mercury e= 43 seconds per century

The
speed of light c = 3 x 10

^{10}cm/s in the text.In this context, 1 astronomical unit (AU) = 1.5 x 10

^{13}cm. But from a Table of distances, Mercury's semi-major axis = 0.387 AU or:

a = (1.5 x 10

^{13}cm) (0.387) = 5.8 x 10

^{12}cm

The period, T (in seconds) is just the length of Mercury's year (in days = 87.96, again from a Table) multiplied by the seconds-length of an Earth day, or 86,400 s, hence::

T = 7.6 x 10

^{6}s

The eccentricity, e from a similar Table is e = 0.205.

Substituting all these values into the given equation yields:

e = 5.036 x 10

^{-7}radian

To get the equivalent seconds of arc (or arcsec) we use 1 rad (radian) = 57.3 degree where one degree has 3600 seconds. Thus, 1 radian will have:

2.063 x 10

^{5}arcsec

So, the associated arcsec for e will be:

(5.036 x 10

^{- 7}rad) x (2.063 x 10

^{5}arcsec/ rad) =

0.104 arcsec

We are still not finished because the quantity is defined per CENTURY

At this point, you need to recall the PERIOD of Mercury is 0.2405 YRS.

So, the number of arcsec of perihelion advance per Earth years is:

0.104 arcsec/ 0.2405 years = 0.432

and over 100 years:

eta = 100 yr x (0.432 arcsec/ yr) ~ 43.2 arcsec or near to what Einstein noted on the page.

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