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41

That's what it really would look like if you were there with DSCOVR. The albedo of the Moon is only about 0.136, about half of the Earth's average albedo. Of course the part with clouds is higher. I was shocked too, but it was explained in written copy that accompanied the release of the original image. Shouldn't the Moon appear as bright as a full ...

23

Pan, Daphnis, and various other moonlets, I would argue, are inside the rings. If you explicitly discount the Encke gap (which Pan orbits in) and the Keeler gap (which Daphnis orbits in) as being part of the ring system, Daphnis would be your answer, as it is a ~8 km object in a 42 km gap. (for comparison, Pan is a ~35 km object in a 325 km gap) Really, ...

22

Given the date and timing, this could be most like the Starlink satellites in their "stacked" configuration. They are currently in a line, but they will later move to separate orbits. Dr Marco Langbroek has a video of a line of 60 satellites. They won't stay in a line; they have ion engines that they will use to spread out into separate orbits.

16

A lot of satellites are visible under the right conditions. Usually up to 2 hours after sunset and 2 hours before sunrise. This allows the sun to strike the satellite when you are on the dark side. Depending on the orbit, it will take between 1 and 5 minutes to traverse most of the sky. Usually, they will enter the shadow and you lose sight of them.

15

Hubble was in low earth orbit, and was always intended to be serviceable. In fact, the original plan for Hubble was to have the space shuttle carry it down from orbit and take it back up, but they decided that was too risky compared to servicing in orbit. JWST, on the other hand, will be at the Earth/Sun L2 Lagrange point, like WMAP and Planck before it. ...

15

tldr; some modelling has been done with a lot more to do, but generally the impact of these constellations is fairly negative, but potentially manageable. Ok, there's a lot to unpack here. First things first, while people have been aware of Starlink and have thought of it, modelling the impacts to observatories hasn't happened as of yet in quite a few ...

11

I think the confusion comes down to what "brightness" means. In general we don't really measure (either with our eyes or our cameras) the absoloute brightness of objects. Instead we measure the brightness relative to other objects in the scene. In the photo of the earth and the moon taken from between the earth and the sun the moon looks dark grey because ...

9

The other answers have pretty well covered the fact that the moon is made of dark gray stone. But I wanted to mention a way that you can verify this on your own with nothing more sophisticated than a camera with a manual exposure mode. Photographers have a rule of thumb called "sunny 16". This was much more important back in the day when cameras didn't ...

9

The two terms are used in answering different questions. Hill Sphere: given a large mass (eg Sun) and a small mass (eg Earth), can a tiny mass (eg Moon) find a stable orbit around the small mass? (If the tiny mass goes outside the Hill Sphere of the small mass, no.) SOI: given two large mass objects and a small object between them, (eg sending a probe ...

8

People have been thinking about and a articles have been coming out on this topic. This article states that as many as hundreds of satellites may be visible at once during twilight at certain latitudes. The IAU has released this statement, which is based in part on this article. The exact way Starlink will affect observational astronomy is not yet known, ...

7

Yes, the dark side of the moon is mostly hidden from earthbound observers. This is due to tidal locking: A tidally locked body takes just as long to rotate around its own axis as it does to revolve around its partner. This causes one hemisphere constantly to face the partner body. In other words, one moon-year is just as long as one moon-day. The left-...

6

Earth's center of mass must be at one of the two focus points of a satellite's elliptical orbit, or at the center of a circular orbit such as a geostationary orbit. One cannot orbit a certain latitude, except for the equator. But there are clever alternatives for different purposes. Geosynchronous (as opposed to geostationary) orbits mean that the satellite ...

6

This is not really known; while the rotation period of most of the larger satellites is known, and they're mostly synchronized, for the smaller ones it's harder to determine and they're less likely to be synchronized: One conclusion is that, other things being equal ... a large moon will lock faster than a smaller moon at the same orbital distance from ...

5

The astronomy "magnitude" scale works backwards: smaller numbers indicate brighter objects. Back in the days before precision measurements of brightness, stars were categorized by eye, with the brightest being "stars of the first magnitude". When more precise measurement became possible, this scale was retained, and extended into the negative numbers for ...

5

An excellent book by Jean Meeus, "Astronomical Algorithms", provides the calculations for planetary positions and Jupiter's Galilean Moons (Io, Europa, Ganymede, and Callisto). Meeus does not really explain the math to a great degree, so it is relatively simple to follow it as a formulaic approach. He uses VSOP for planetary positions and Lieske's E5 theory ...

5

The Sun's gravity does perturb the Moon's orbit but more subtly than you imagined. The Moon's perigee and apogee migrate eastward in an 8.9 year cycle called apsidal precession. Also the plane of the lunar orbit, inclined about 5.1 degrees to Earth's orbit, shifts westward in an 18.6 year cycle called nodal precession. Lunar theory also addresses various ...

4

Do geostationary satellites need to have the equator as the plane of rotation, and the earth's centre to be the centre of rotation? To be stationary above a point, yes. Can it rotate over, say, the Tropic of Cancer, focusing on a single city? If the satellite's orbit touched the Tropic of Cancer, it would not be geostationary since the orbit about the ...

4

I do not know for sure, but you are viewing the satellites by the reflection of sunlight. As the satellite moves across the sky, the Sun-satellite-you angle changes, therefore there is no reason to expect that the brightness should stay the same. As for why it should fade and brighten again - if you imagine the satellite to consist of a number of surfaces, ...

4

The best way to figure it out would be to use the site "In the Sky" (https://in-the-sky.org/). You can enter time, date and location to identify satellites. This would at least be able to tell you if what you saw was a satellite.

4

There are over 1000 operational satellites, and probably as many again defunct or non-operational satellites in orbit, plus a great deal more small pieces of space junk. About half of these are in a low earth orbit (LEO). Satellites in LEO have a period of between 90 minutes and 130 minutes. But you would not expect a satellite to follow the same path in ...

4

The Moon has a "Roche lobe", where the Moon's gravity dominates, and all you need for the situation you describe is that the Roche lobe lies physically outside the Moon. That will be true, because the Lagrange points are on the outside edge of that Roche lobe, and it's outside the Moon. The Moon's Roche lobe is shown here: http://hyperphysics.phy-astr.gsu....

4

I'm treating this as a homework question. You have made a clear attempt, so it is not off topic, but you should clarify the source of the problem. You will first need to establish if the rotation of the Earth is to be considered (it makes a big difference, but in a homework, sometimes the Earth stops spinning!) You have then correctly found the two possible ...

4

Kepler's third law in this situation tells us that $$\frac{R^3}{P^2} = \frac{GM}{4\pi^2},$$ where $R$ is the orbital radius (or semi-major axis for an elliptical orbit), $P$ is the orbital period and $M$ is the mass of the Earth (actually, it is the mas of the Earth plus the mass of the satellite, but we can neglect the latter). Thus the orbital period \$P \...

4

That sounds like a high altitude nonfunctioning satellite to me. For example, a rocket body or upper stage that is tumbling. I have seen these types of satellites flash many times before. The pattern is irregular because of the spinning and changing angle to the Sun.

4

Definitely a UFO - Unidentified Flying Object, not necessarily aliens :-). A few possibilities I can think of (would be nice to know how fast the objects were going, and whether or not they disappeared), arranged by most plausible to least plausible: Birds If it was just after sunset, or just before sun rise, birds could have been flying up far above, ...

4

Starter answer: Wikipedia's Local Group lists the following items as galaxies, dwarf galaxies, or possibly galaxies. You should refer back to the Wikipedia page, but here is a summary: Galaxies: Sagittarius Dwarf Galaxy, Large Magellanic Cloud, Small Magellanic Cloud Dwarf Galaxies: Ursa Minor, Draco, Carina, Sextans, Sculptor, Fornax, Leo I, Leo II, Ursa ...

4

Orbits are all ellipses, so when you throw a baseball you change the orbit of the base ball from your circular ellipse to another ellipse. If you throw it prograde (in the velocity direction), then you baseball will have an elliptical orbit. It will be closest to the Earth at the point at which you threw it, and furthest from the Earth (and higher than the ...

3

Because if the photograph were taken with an exposure setting that made the moon appear shiny and bright white, the Earth is so much more bright that it would completely wash out the scene. There is no way to objectively measure "brightness" of objects in a photograph without knowing how the image was captured.

3

These are known as two-line elements (see also nasa.gov, satobs.org). On both lines, the first field after the line number is a NORAD ID, which you can use to query a database such as N2YO. The second line gives geocentric orbital elements for the epoch given in the first line. The last line looks like an error estimate by observation analysis software. The ...

3

Im not sure of the number, upwards of 2k probably, Godards lists it at 2271, but who knows how accurate that is as some are inactive, or not listed due to government privacy. The period would depend on the satellite, its orbutal distance from the earth and its velocity.

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