New answers tagged uranus
4
I started Stellarium on my computer and pressed F6 to bring up the "Location" window.
Then I changed the planet to "Uranus", and marvelled at the view of the many rings and many moons from the planet's "surface"
For convenience I clicked the buttons to remove the ground and the atmosphere. then I found and clicked on Saturn. It ...
1
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 ...
8
The brightness of a Solar System object, seen in reflected light, depends on how far it is from the Sun, $d_s$, and how far away it is from the observer, $d_o$, (and the angles between them). Both dependencies are "inverse square laws":
$${\rm brightness} \propto \left(\frac{1}{d_s^2}\right)\left(\frac{1}{d_o^2}\right)\ . $$
Both Uranus and Neptune ...
-2
If I have calculated it right then the apparent magnitude of Neptune as seen from Uranus is approximately 4.
The absolute magnitudes of Neptune (-7.11) and Uranus (-7.00) are nearly identical.
The apparent magnitude of Uranus is 5.38.
The distance between Neptune and Uranus is roughly half of the distance between Earth and Uranus.
At half the distance a ...
3
The Sun loses mass all the time. That mass loss will accelerate as it ascends the red giant branch and ultimately the Sun will end up as a white dwarf of about half a solar mass.
Since the solar wind exerts almost no torque on the planets, their angular momentum is a fixed quantity. From Kepler's third law, we know that the angular velocity $\omega$, orbital ...
14
Supplementary answer supporting @PierrePaquette thorough and well-source answer:
I tried the nice new JPL Horizons interface and fired up Excel which I haven't used in a long time.
For years 1800 to 2100 in Observer mode it calculates apparent magnitude using all the bells and whistles (albedo model, phase angle, illumination, etc.) and gives the following ...
18
According to https://arxiv.org/pdf/1808.01973.pdf, the magnitude of Neptune follows the relationship (formula 17, page 25):
$ V = 5 \log_{10} (rd) - 7.00 + 7.944 \times 10^{-3} α + 9.617 \times 10^{-5} α^2 $
Where r is the distance of Neptune to the Sun, d is the distance of Neptune to the observer, and α is “the arc between the Sun and the sensor with its ...
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