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.