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.