# What is Saturn’s apparent magnitude at its brightest when viewed from Uranus?

Considering that the apparent magnitude scale is an inverted scale; what is the lowest apparent magnitude of Saturn that can be seen from Uranus. We are of course talking about when Saturn is viewed from one of the moons of Uranus.

Saturn is in a similar relationship with Uranus as Venus is to Earth. To know this could be helpful when attempting to answer this question.

• The answer to this is far more complex because of the phase angle complications. Nov 11, 2021 at 12:46
• This is very similar to your previous question, if not a duplicate. Nov 11, 2021 at 13:34
• I provided all the elements to calculate that as well in the first reference of my first answer to your question about Uranus and Neptune. Please read it before asking more related questions; a lot is in there. Nov 12, 2021 at 4:49
• additionally, the resource uloh linked you to in his answer, that also is able to answers you this question. Nov 12, 2021 at 12:16
• It is the same question as your previous one. They both come down to “how bright is planet A from planet B?” The identity of the planet changing from one question to another is not significant in this context—not enough to justify two questions. Also, as I said, the first reference in my first reply to your first question gives you the tools to find the answer yourself. READ IT. And USE THE TOOLS. Nov 14, 2021 at 19:18

Leveraging Pierre Paquette's excellent answer and reference to Hilton and Mallama, the magnitude of Saturn can be estimated by:

$$V = 5 \log_{10} (rd) - 8.95 - 3.7\times10^{-4} \alpha + 6.16\times10^{-4} \alpha^2$$

Here, $$r\approx9.5$$ AU is the distance from Saturn to the Sun, $$d$$ is the distance from Saturn to the observer, and $$\alpha$$ is the angle of the Sun/Saturn/Observer triangle.

If the observer is located on a planet with orbit inside of Saturn's orbit, it should be clear that the maximum apparent brightness of Saturn should occur when the observer is closest to Saturn, since both $$d$$ and $$\alpha$$ are at a minimum.

However, if an observer is on a planet with orbit outside of Saturn's orbit, it certainly isn't the case that Saturn is brightest when closest, since it will be backlit by the Sun from the perspective of the observer.

Using the Law of Cosines, we can compute the distance between the observer and Saturn as a function of $$\alpha$$:

$$d(\alpha) = r\cos{\alpha}\pm \sqrt{c^2-r^2\sin^2{\alpha}}$$

Here, $$c \approx 19.2$$ AU is the distance from Uranus to the Sun. This allows us to formulate the expression of magnitude only as a function of $$\alpha$$.

$$V(\alpha) = 5 \log_{10} (r^2\cos{\alpha}\pm r\sqrt{c^2-r^2\sin^2{\alpha}}) - 8.95 - 3.7\times10^{-4} \alpha + 6.16\times10^{-4} \alpha^2$$

Plugging in values from 0 to 180 degrees for alpha, we get the following magnitude phase curve: Here the x-axis is $$\alpha$$ in degrees, and the y-axis is apparent magnitude. It shows that the maximum apparent magnitude of Saturn as seen from Uranus is at opposition (perhaps unintuitively since it is also the maximum distance between the two planets' orbital paths). The value at opposition is about 3.228. As a comparison, Saturn is always much more bright from Earth, varying between -0.55 and 1.17.

Notes:

1. For convenience, we assume in this answer that the orbits are circular and coplanar. Since both planets have low inclination and eccentricity, we wouldn't expect a very different phase curve from a more sophisticated model.

2. The above answer is for the apparent magnitude of the sphere of Saturn only. The rings of Saturn can contribute significantly to the brightness of the overall system. Hilton and Mallama provide a more sophisticated equation involving $$\beta = \sqrt{\beta_1\beta_2}$$, where $$\beta_1$$ and $$\beta_2$$ are the inclinations of Saturn's rings with respect to the Sun and the observer respectively. One could get a higher fidelity answer using this equation, perhaps assuming that the maximum ring illumination can occur simultaneously in opposition with $$\beta_1=\beta_2 \approx 26.7^\circ$$

$$V = 5 \log_{10} (rd) - 8.914 - 1.825 \sin{\beta}+ 0.026 α - 0.378 \sin{\beta} e^{-2.25\alpha}$$

• Thx. So are you saying that the answer to my question is +3.88 in apparent magnitude? Nov 15, 2021 at 23:16
• @Constantthin no, +3.228 Nov 16, 2021 at 0:01
• I don’t understand your chart. Does it say that the smallest apparent magnitude is measured at zero angle? Nov 16, 2021 at 3:00
• yes, the smallest mag is at zero angle Nov 16, 2021 at 3:10
• I don’t understand how that can be, Venus’ lowest apparent magnitude is not at zero angle, but at 39 degrees from the sun. Shouldn’t Saturn be in a similar situation in relation to Uranus? Nov 16, 2021 at 9:15