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).

The Saturnian Keplerian system

Decades after the discovery of Jupiter's moons, Saturn was observed to have its own satellites: the first being Titan, the largest, in 1655.

Let's try to compute the Kepler value for the three largest, using the same method as for Jupiter, noting however, that accuracy and precision would likely have been low in the early days, so the figures below (especially its density) may not have been achievable until much later.

There is no orbital resonance with these, so the table shows the period in days.

(Rhea's orbit is actually about 4.5 Saturnian diameters. Were we able to estimate to this precision, then we should expect the Kepler value to be ~4.5).

We see that the values in the last column suggest a Kepler 'constant' for Saturn of ~ 4 and hence that Saturn, as well as the Sun and Jupiter, also has its own gravitational system obeying the same laws.

Converting this to Solar system units:

• as Saturn's angular diameter at opposition is ~ 1/3 minute of arc and it's 8.4 AUs away, it's diameter is ~ 0.0008 AUs
• we cube this:  ~ 5 x 10^(-10) and multiply our Kepler constant by it
• 1 day is 1/365 years, so we square this ~  7.5 x 10^(-6) and divide the above result by it

Hence, the average Kepler 'constant' for Saturn is ~ 0.0003.

We deduce that its gravitational pull is ~ 0.0003 times that of the Sun and ~ 0.3 times that of Jupiter. 

We suspect its mass is also in the same proportions. That being the case, as we know its diameter is ~ 0.8 of Jupiter's, its volume is therefore around half of Jupiter's and its density ~ 0.6 times Jupiter's, indicating a different composition.

Moon	Orbital radius (r) in Saturnian diameters	r^3	Period (t) days	t^2	r^3/t^2
I. Rhea	4	64	4.5	20	3.2
II. Titan	10	1000	16	256	4
III. Iapetus	30	27000	79	6241	4.3

Jupiter vs. the Sun

To compare the Jovian Kepler constant with the Solar one, we must convert it to the same units (AUs cubed over years squared):

• we determine the diameter of Jupiter is ~ 0.001 AUs by estimating* its angular diameter is about 3/4 of one minute of arc at opposition when we know the planet is 4.1 AUs away (its orbital radius minus that of the Earth's)
• we cube this: ~ 10⁻⁹ and multiply our Jovian Kepler constant by it
• we determine the period of moon I (Io) is ~1.8/365.25 or 0.0049 years
• we square this: ~ 2.4 x 10⁻⁵ and divide the above result by it

(* If there were two stars of known angular separation visible in the eyepiece of the telescope at the same time as Jupiter, their presence could provide a useful yardstick against which to make a reasonably accurate estimate).  

Hence, the average Kepler constant for the Jovian system is ~  0.001, i.e. 1/1000th of the value (1) for the Solar System. This means that for two different objects orbiting at the same distance from the Sun and from Jupiter, in the former case the object's speed is 10√10 times that of the latter. (Similarly, if they have the same orbital period, then the object orbiting the Sun does so at 10 times the distance of that orbiting Jupiter). 

We conclude that the Sun's gravitational pull is much stronger than Jupiter's and that the ratio of their Kepler constants is a measure of this.

Given its angular diameter at opposition is as stated earlier, then we conclude that at the distance of the Sun (1 AU), it would be 4.1 times larger, i.e. ~ 0.05°, compared with the Sun's  ~ 0.5°. That is, Jupiter's diameter is about 10% that of the Sun and its volume is about one-thousandth, i.e. about the same as the ratio of its Kepler constant to the Sun's.

Here then is a first inkling about what may be behind the strength of a body's gravitational pull. Volume seems an unlikely candidate as we shouldn't expect a diffuse cloud of gas to have the same effect as a solid object of the same size. Perhaps the gravitational pull is simply proportional to its mass then? If so, then Jupiter's density is about the same as the Sun's. And yet, what different bodies they seem to be!

The Jovian Keplerian system

Updated 11 May 26

The discovery of Jupiter's moons through the newly-invented telescope back in the early 1600s must have caused great excitement at the time and a wish to investigate further to see if the Jovian system follows Kepler's laws.

However, imagine trying to track these moons individually over many days and weeks in an attempt to measure the orbital radius and orbital period for each!

The orbital radius could be estimated crudely (being imprecise and prone to human error) using the diameter of the planet as the base unit, e.g. the innermost moon being found to orbit at around 3 Jovian diameters from the centre of the planet.

I've tried this myself on just a few occasions with binoculars and it was difficult! Doing this to an acceptable level of accuracy and precision in the early 17th century would have demanded the best optics of the time, a solid mount, good skies over many weeks and an experienced and patient observer. And then no doubt repeating the exercise again and again. 

The orbital period could be measured with more precision by timing the intervals between successive identical locations in a moon's orbit (e.g. maximum elongation), although we should bear in mind that this was before the invention of the pendulum clock.

(Would we have to allow for the movement of the Earth introducing errors due to parallax?)

See the table at the bottom of the page for an investigation. (The table had to be imported and in this case, if not placed there, messes up the font of the text following it!). The moons are numbered according to their proximity to Jupiter.

It turns out that the orbital periods of moons I, II and III are in the ratio 1:2:4, so I'm using the orbital period of moon I as the unit of time here.

The average estimate of the Jovian Kepler 'constant' (last column) is 26, but with a large range of values, no doubt due to the lack of precision in measuring orbital radii in whole Jovian diameters, which are then cubed. (In practice, there would be even greater variation with this method, because, unlike the case here, estimates would likely be inaccurate, as well as imprecise). However, these values are of a similar order of magnitude and a more precise and more accurate method should yield results that are more consistent*. Therefore, we tentatively assert that these moons do indeed obey Kepler's third law.

(* If instead we were to estimate accurately using the radius of Jupiter as our measuring unit, then we should find that the Jovian Kepler constant has an average of 202 with a proportionately smaller range of 182 to 216. However, I expect this advantage would have been easily removed by increased human error, in estimating, say, that Callisto orbits at 26 units from Jupiter! I imagine that the later invention of the reticule would have been revolutionary for making accurate and precise measurements). 

Moon	Orbital radius (r) in Jovian diameters	r^3	Period (t) as a multiple of that of moon I	t^2	r^3/t^2
I. Io	3	27	1	1	27
II. Europa	5	125	2	4	31
III.Ganymede	7	343	4	16	21
IV. Callisto	13	2197	9.4	89	25

Determining the orbital radius for Saturn

Updated 5 May 26

To complete the orbital radii of the classical planets, let's now find Saturn's:

rₛ ~ (dₑ + sin θ)/ ((dₑ/pₛ) + sin θ)

where pₛ is its orbital period, for which we'll give the value of 29.45 that we found earlier.

This time, let's double Δt to 4 days, because we know θ may be too small to measure otherwise. As a result, distance dₑ will now be 8π/365.25 AUs. However, as this is an approximation (chord vs. arc of a circle), we will have increased the error compared with a time of 2 days.

Considering the opposition of Saturn on October 4 2026 and using Stellarium to determine its position 48 hours before and 48 hours after, and this tool, θ, the angular distance Saturn is seen to have moved in those four days is approximately 0.32° of arc, slightly more than for Jupiter over 2 days, again hopefully measurable.

Using these values, rₛ ~  9.4 AUs, while the accepted figure is around 9.6 AUs, which perhaps we could have achieved by averaging the results from many opposition events.

Again, checking Kepler's third law:

• the cube of its orbital radius is 9.4³ ~ 831
• the square of its orbital period is 29.45² ~ 867
• the former divided by the latter is ~ 0.96

Note: in no case has the estimate of the Solar Kepler constant ever exceeded 1!

Determining the orbital radius for Jupiter

Updated 27 April 26

Adapting the formula that we used for Mars, the radius of Jupiter's orbit, rⱼ, is given by:

rⱼ ~ (dₑ + sin θ)/ ((dₑ/pⱼ) + sin θ)

where pⱼ is its orbital period, for which we'll give the value of 11.85 that we found earlier.

Having said earlier that for Jupiter etc, it would be better to consider a larger Δt, for now, I'm going to use exactly the same value as for Mars, i.e. 2 days, and see how it goes, so as before, distance dₑ is 4π/365.25 AUs.

Considering the opposition of Jupiter on January 10, 2026 and using Stellarium to determine its position 24 hours before and 24 hours after, and this tool, θ, the angular distance Jupiter is seen to have moved in those two days, is approximately 0.27° of arc. This is smaller than was the case with Mars, now about half the angular size of the full moon, but perhaps still measurable with reasonable accuracy using the naked eye, a good star chart and a quadrant, say?

Using these values, rⱼ ~  5.1 AUs, which is very close to the accepted figure of about 5.2 AUs, which we would likely have arrived at by averaging the results from many opposition events.

Again, checking Kepler's third law:

• the cube of its orbital radius is 5.1³ ~ 132.7
• the square of its orbital period is 11.85² ~ 140.4
• the former divided by the latter is ~ 0.94

If we instead use the accepted value for the orbital radius, then this ratio becomes ~1, as we should expect.

For Saturn, the angle θ over just two days will be smaller again, so in that case it really will make sense to increase the time interval Δt and hence the value of dₑ.

Determining the orbital radius for Mars

As mentioned in an earlier post, the (average) orbital radius for an inner planet can be readily determined from its (average) maximum elongations observed from the Earth. However, another way is needed for the outer planets and the sketched diagram below illustrates one such method.

As an approximation, we assume orbits are circular.

In a short interval of time, Δt, centred on the opposition of Mars (corresponding to position 2 in the diagram) seen from the Earth, the Earth will move a short distance, dₑ, in green, approximated as a short straight line segment of length rₑ.ѡₑ.Δt, where rₑ is the radius of the Earth's orbit and ѡₑ is its angular velocity.

Similarly, in the same interval of time, Mars will move a short distance, dₘ, in red, approximately equal to rₘ.ѡₘ.Δt, where rₘ is the radius of Mars' orbit and ѡₘ is its angular velocity, which we know (see earlier post) is around ѡₑ/1.9. Hence dₘ is about dₑ.rₘ/1.9 AUs.

Let's use the Mars opposition of January 16, 2025. It happened at 00:48 UT (position 2 in the diagram). Using Stellarium to determine its position 24 hours before (position 1) and 24 hours after this time (position 3), and this tool, from the Earth, Mars is seen to have moved approximately 0.8° of arc (0.014 radians) in those 2 days.

In this case, distance dₑ is 4π/365.25 AUs.

We see that the angular distance, let's call it θ, is the angle between the line joining the Earth and Mars 24 hours before opposition and the line joining the planets 24 hours after opposition.

These lines meet at a distant point, which is at distance l₁ from Mars and distance l₂ from the Earth. The distance between the Earth and Mars at opposition is therefore l₂ - l₁.

We can see that the distance travelled by the Earth in that time, dₑ, is approximately equal to l₂.sin θ.

Also that the distance travelled by Mars in that time, dₘ, is approximately equal to l₁.sin θ.

Putting this all together:

rₘ ~ (dₑ + sin θ)/ ((dₑ/1.9) + sin θ) ~ 1.5 AUs.

Being based around a single astronomical event (unlike the results determined for the inner planets), this result is fortuitously close to the the average of Mars' perihelion (1.38 AUs) and aphelion (1.66 AUs) and also to the semi-major axis of its orbit, according to wikipedia.

Let's check Kepler's third law for Mars as we did for the inner planets and the Earth:

• the cube of its orbital radius is 1.5³ = 3.375
• the square of its orbital period is 1.9² = 3.61
• the former divided by the latter is ~ 0.93

Actually, the accepted figure for the orbital period is 1.88 (years). Squaring this gives 3.53 and the ratio is now 0.95.

We could use the same method for the other outer planets, but as the angle, θ, would be smaller, and less easy to measure, it would be better to increase Δt, perhaps even extending it to the full period of retrograde motion. If we did so, however, the distance dₑ, at least, would need to be calculated as being the chord of a circle centred on the Sun.

Checking Kepler's third law for the inner planets

Updated 31 May 26

(You'll need to view this in landscape mode if using a small screen).

It's easy to see from the diagram in the previous post that the distance from the Sun of an inner planet can be found by measuring the angle of greatest elongation (East or West) from the Sun and then taking the sine of that angle.

There will be some variation, so we'll need to take an average over many measurements. We will consider orbits to be more-or-less circular and hence will loosely refer to this distance as the orbital radius.

For Earth and the inner planets:

Planet	Orbit radius (r) in AUs	r^3	Period (t) in years	t^2	r^3/t^2
Mercury	sin 22.5° * = 0.38	~0.06	7/29	0.06	0.96
Venus	sin 46° *  = 0.72	~0.37	8/13	0.38	0.98
Earth	1	1	1	1	1

(* from approximate average greatest elongation data here)

So far, so good, but to extend this to the outer planets, I'll need a method of determining the orbital radii for those. 🤔