Estimating aphelion and perihelion for Mercury and Venus

In an earlier post, I estimated the orbital radii for Mercury and Venus, by taking the sine of the average greatest elongation angles for each, using data for 2014 to 2029 from here.

In fact, I've been treating the orbits as being near-circular, but recently I've been wondering how to find their true shapes (which we know from Kepler and Newton are ellipses). Specifically, can I use the extremes of the greatest elongation angles to find accurate figures for the furthest (aphelion) and closest (perihelion) approaches of these planets to the Sun?

So, here is a summary for the inner planets. I've separated greatest elongation observations by East and West. The sines of the minimum and maximum angles within each set of observations provide estimates of perihelion and aphelion respectively and as can be seen are nearly exactly the same as the true figures, with exceptions to the exact figures highlighted with an *. This surprises me, because I'd assumed that I would have to find and make an allowance for the Earth's orbital eccentricity (loosely, non-circularity) for greater accuracy (which I have not done yet), especially so in the case of Venus, because the Earth's orbit is actually the more eccentric of the two planets.

Notice how eccentric Mercury's orbit is!

Mercury

• Greatest Elongation West
• 50 observations
• min. 17.9 (sine: 0.31)
• max. 27.8 (sine: 0.47)
• Greatest Elongation East 
• 51 observations
• min. 18.1 (sine: 0.31)
• max. 27.4 (sine: 0.46*)
• wikipedia: perihelion 0.31 AUs, aphelion 0.47 AUs

Venus

• Greatest Elongation West
• 10 observations
• min: 45.8 (sine: 0.72)
• max 47 (sine: 0.73)
• Greatest Elongation East
• 10 observations
• min. 45.4 (sine: 0.71*)
• max. 47.2 (sine: 0.73)
• wikipedia: perihelion 0.72 AUs, aphelion 0.73 AUs

The Sun's gravitational pull at the Earth's surface

We know from an earlier post that the acceleration due to a massive object is:

i. 4π²k/d²

where k is the Kepler constant for the massive object (yes, I'm avoiding G and gravitational mass) and d is the distance from that object.

Hence at the Earth's surface the ratio of the gravitational acceleration due to the Sun to that due to the Earth is:

ii. (kₛ.dₑ²)/(kₑ.dₛ²).

Where kₛ and kₑ are the Kepler constants for the Sun and Earth respectively and dₑ is the distance from centre of the Earth, i.e. one Earth radius, and dₛ is the distance from the Sun, i.e. 1 AU.

We know also that:

iii. kₛ/kₑ ~ 10⁶/3 &

iv. dₑ/dₛ ~ 1/24000

Hence, at the surface of the Earth, the ratio of the gravitational acceleration due to the Sun to that due to the Earth is about

10⁶/(3 x 24000²) ~ 0.0006.

This tiny, tiny acceleration, less than one-thousandth that we experience due to the Earth, is enough to keep the Earth in its orbit about the Sun, year in, year out!

At Saturn, ~ 9.6 AU from the Sun, the acceleration due to the Sun is about 100 times smaller than at the Earth and yet, it's sufficient to keep Saturn orbiting too!

Gravity is such a puny force and yet it permeates the cosmos!

Earth 'weighed' against the Sun

Having discovered that the size of the AU can be estimated in Earth radii using the telescope, there remains one task: to convert Earth's Kepler's constant to AU³/year², so that we can it with those of the Sun, Jupiter and Saturn:

• 1 AU ~ 24000 Earth radii, i.e. the radius of the Earth is ~ 4.2 x 10⁻⁵ AUs
• we cube this: ~ 7.4 x 10⁻¹⁴ and multiply our Kepler constant (297) by it
• 1 day is ~ 1/365 years
• we square this: ~ 7.5 x 10⁻⁶ and divide the above result by it 

In Solar System units, the Earth's Kepler constant is therefore ~ 3 x 10⁻⁶, i.e. about one 300,000th the value of that for the Sun. Hence, we suspect (pre-Cavendish) that the Earth's mass is the same proportion of the Sun's, one 300th the mass of Jupiter and one 100th the mass of Saturn.

Further, given the Sun's angular size is about one half a degree of arc, we can deduce that its radius is about one hundred Earth radii and its volume about that of one million Earths.

Hence the density of the Earth is about 3 times that of the Sun (and of Jupiter).

Earth's Keplerian system & towards universal gravitation

Updated 31 May 2026

Having investigated the Solar, Jovian and Saturnian Keplerian Systems, let's take a closer look at home and start with trying to determine the value of its Kepler 'constant' (should it exist) by looking at the Moon's orbit about the Earth.

It's at a distance (measured with reasonable accuracy by Ptolemy nearly 2000 years ago) of about 60 Earth radii and takes about 27 days to complete one orbit.

From this we can say the value of the Kepler 'constant' for the Earth is ~ 60³/27² ~ 296.

In the case of the other systems, there are many orbiting bodies and we can calculate an average, but in this case, if we go back many decades at least, there is only one such body orbiting the Earth.

However, let's take inspiration from Newton, who realized that an apple falling to the ground also experiences an acceleration towards the centre of the Earth. This acceleration is commonly referred to as g and can be measured accurately by means of a pendulum, length l, the period of which is 2π√(l/g).

We can generalize the formula from the previous post and say that the acceleration at distance d, regardless of whether the object is orbiting, is:

4π²k/d².

At the Earth's surface, we state in this case that d = 1 (the same as if all the Earth's mass were concentrated at its centre) and g = 4π²k, where k is the Kepler 'constant' for the Earth. If we know g, we can therefore determine that:

k = g/(4π²).

In metric units g ~ 9.8 m/s² and we need to convert this to the units used when calculating the value from observation of the Moon, i.e. Earth radii/day²:

• the radius of the Earth (also measured, albeit in different units, with reasonable accuracy by the Ancient Greeks) is ~ 6.4 x 10⁶ m and we divide our value of g by this number
• there are 86400 seconds in a day, squared is ~ 7.5 x 10⁹ and we multiply the above result by this number.

In these new units g ~ 11719 and dividing by 4π² gives ~ 297, which is very close indeed to the value we derived earlier.

We can conclude therefore that the Earth, as with the Sun, Jupiter and Saturn also has its own Keplerian system, where the central body exerts a gravitational force that follows an inverse-square law.

From earlier considerations, that force appears to be related to the mass of the central object (Sun, Jupiter etc), possibly simply proportional to it, in which case the Kepler 'constant' for the system is also proportional to the mass of the central object.

We can surmise therefore that all massive bodies, including those bodies without satellites of their own, exert such a gravitational force on surrounding matter. (Surface features on the Moon, for example, and the near spherical shape of all planets, including those without moons, provide strong evidence for this).

This was to be confirmed experimentally by Henry Cavendish in the laboratory at the end of the 18th century - an incredible achievement for the time given the weakness of the gravitational force. 

From Kepler's third law to the inverse-square law

When a small object orbits a massive one, such as a moon going about its planet, say, it experiences an acceleration towards it. In perfect circular motion (an approximation), that acceleration is: r⍵²,where r is the radius of the orbit and ⍵ is its angular speed, which is 2π/t, where t is its orbital period. We know this from Newton.

We know also from Kepler's third law that

r³/t² = k,

where k is the Kepler 'constant' for that system. 

Hence the acceleration 

r⍵² = r4π²k/r³ = 4π²k/r²

This is the famous inverse-square law for the gravitational force (also from Newton of course).