# Tag Info

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Yes, comets spin although measuring it can be tricky due to the coma and outgassing from the nucleus. It's easiest to measure the rotation period when the comet is inactive near aphelion although this is more difficult as the comet is fainter, necessitating a large amount of large (>4m) telescope time which is difficult to obtain. Searching the JPL Small ...

10

There are approved proposals for Cycle 1 to point the JWST at The Jovian system Jupiter's great red spot Mars Saturn and its moons and rings In those PDFs, they describe exactly what instruments they want to point where, but to summarize as far as I can tell they're planning to try to use pretty much all of the available instruments to image Jupiter and ...

8

It is called Tidal love number. The definition is as follows: In Newtonian gravitational theory, a tidal Love number relates the mass multipole moment created by tidal forces on a spherical body to the applied tidal field. The Love number is dimensionless, and it encodes information about the body's internal structure. (Poisson et.al., 2009) For Io, the ...

7

All modes can be used. But for bright targets, observations are limited to specific filters, subarrays, regions of the target planet, or spectral intervals. James Norwood and colleagues wrote a paper on this topic, although the full answer is complicated because of the "subset of the instrument modes" part. The simplest answer to understand is for ...

6

$k_2$ is one of three tidal Love-Shida numbers related to how gravitation of another body (Jupiter in this case) changes a planet-like body's second degree spherical harmonics (Io in this case). Three Love and Shida numbers exist for each degree of spherical harmonic coefficients. The three Love-Shida numbers for a given degree $n$ are $k_n$, which ...

6

The lunar opposition surge has been well studied, likely because we can study it in detail, we have surface samples, and so it serves as a baseline for other bodies in the solar system (as it does for many other kinds of surface studies). It is quite substantial in visible light. Probably Clementine data are the oldest-modern reference, and among others, ...

5

There is a science-based approach to explaining the dark spot on Uranus. In 2009 (the image and sighting of the spot in question are from 2006) a paper was published titled: The Dark Spot in the atmosphere of Uranus in 2006: Discovery, description, and dynamical simulations by H.B. Hammel a, L.A. Sromovsky et. al.. In this work they say that the dark spot is ...

5

What they have done is take each of the 580 Jupiter-Family Comets (JFCs) from our solar system (from the JPL Small Body DataBase; set object kind=comets, Comet Orbit Classes to Jupiter-family Comet) and kept all of the orbital elements the same, except for the semi-major axis which they have scaled by the ratio of the semi-major axis 55 Cnc d (5.74 au) to ...

5

This might not be so hard after all. Below I show the math for the analytical solution for a Kelperian orbit; the catch is that it's only analytical for $t(\theta)$ and not $\theta(t)$ but that shouldn't cause a problem in this case. I will not solve the problem for you but I'll make a recommendation how to proceed: Step 1: draw a diagram on 2D paper (the ...

4

Yes, the spectra from SuperCam along with that from many other instruments is available from the NASA Planetary Data System (PDS) Geosciences Node. The overall landing page for all of the Mars2020 instruments is here and the one specifically for SuperCam is here. Reading the documentation is highly recommended as PDS archive bundles such as this can be quite ...

4

At zeros order such assumption might be made, but a powerlaw relation is more common and accepted. Also a protoplanetary disk is more complex as is the planet formation process which may include radial migration of the protoplanets. So the mass $m(r)$ available at a distance $r$ might not be exactly representative for the planetary mass found at that ...

4

Assuming most of the escaping Martian atmosphere is entrained in the solar wind, it will flow outward until it reaches the termination shock, and then slow down in the heliosheath until it reaches the heliopause, currently at a distance of about 120 AU. (The actual location will change, of course, depending on things like the strength of the solar wind and ...

4

If you know for sure and can prove it: write a paper, submit it to Nature or Science. They will accept it and will give you the cover image. We can only look at the planetary interiors superficially. We know the bulk mass, we can do a bit tomography with seismics to determine boundaries in the speed-of-sound, with magnetotelurics in the magnetic and electric ...

4

The short answer is, at the moment, only for the Moon. I include below a brief description of the JPL DE format to arrive to this conclusion. This link explains quite well the format of JPL DE ephemerides, which are available here. The link is for the DE440, but a directory up we can find previous ones also. In summary, we need 2 files to properly understand ...

4

First things first, but in this case, second things first. When an asteroid or moon, passes the Roche Limit, it breaks apart and forms rings around the primary body. This is a widely believed misconception of the Roche limit. The Roche limit pertains to objects that are held together by gravitation only. Once chemical bonds come into play the Roche limit ...

4

It's simple geometry: A planet absorbs energy from its star with its geometric cross-section $\pi r^2$. The same planet re-radiates the energy via its complete surface $4\pi r^2$. In equilibrium, in the incoming energy equals the re-radiated energy: $$E_{in} = E_{out}\\ \pi r^2 \cdot S \cdot (1-A) = 4\pi r^2 \sigma T^4$$ Divide by the surface area of the ...

3

Understanding what is out there in the universe, where it is, how it is moving and why is what space exploration is about. Chaotic orbits and trajectories are very important for trajectory planning of spacecraft since one can use them for low-energy transfers. If one wanted to go to an object in a chaotic orbit it is also valuable to know that it is in fact ...

3

tl;dr Maybe it made rings, but they’re certainly not around today. The whole reason the Roche limit exists is because of tidal forces, and the whole reason tidal forces exist is because we choose to work in non-inertial reference frames. If you’re not orbiting an object, there’s no tidal forces from that object. Just as a meteor or asteroid can pass the ...

3

While I couldn't find any quantitative information about views from Earth or orbit down to zero phase angle (especially since views direct from Earth are limited in phase angle due to Eclipse), this website gives an unreferenced data point suggesting Apollo astronauts observed that a zero-phase full moon is approximately 30% (0.2 magnitudes) brighter than we ...

3

According to NASA Meteor Watch Facebook page, they: obtained infrasound measurements from 3 nearby stations - the amplitudes and durations of the signals put the energy of the fireball fragmentation at 440 pounds (200 kilograms) of TNT. Here is an image of the infrasound: From this news story, A man in Canton, Maine said there was a delay of several ...

3

The photo shows a mix of dark basalt, light anortosite from under the basalt, and a thin layer of impact melt (shiny and rippled) that the Boulder is sitting in. On the slope of the peak towards the lower right, there’s a huge block that has slid down the slope from the rim. You can see blocks of light anorthosite in it, and you can see grooves running down ...

3

There are three possible fates for small particles: They fall into the Sun. They escape the solar system. They fall into a planet. Small particles larger than 1 micron across tend to fall into the Sun due to Poynting-Robertson drag. These slowly infalling particles are the source of the zodiacal light. Particles a few nanometers or so across (i.e., ...

3

The $\mathrm{\frac{S}{4}}$ represents the area- and time-averaged incident solar flux and the whole term on the LHS represents solar flux emitted by the planet. The factor of 1/4 comes from the fact that only a single hemisphere is lit at any moment in time (creates a factor of 1/2), and from integrating over angles of incident sunlight on the lit ...

2

I agree with your thinking as far as it goes; it is probably not possible to learn about chaos theory nor celestial mechanics by observing these moons, because: We (believe we) know the underlying laws and mathematics that govern motion (in general at least) We won't be able to study such objects long enough and careful enough to make any tests that haven't ...

2

Scientists (and science fiction writers) have speculated about the possbilities of life under the surfaces or on the surfaces of large moons in our solar system or large exomoons of large exoplanets in other star systems. So a good place to find any limits on the possible properties of moons that can have magnetic fields is a scientific discussion of the ...

2

When the neutron collides with water, it collides with the water molecule as a whole, not just the hydrogen atom. This is no different from colliding with a heavy atom of the same mass. In that case, the neutrons shouldn't slow down much, at least, that's what I think. Help? The de Broglie wavelength of a massive particle is given by: \lambda = \frac{h}{p}...

2

Before I tackle this problem, I will assume that the size of the planet is negligible compared to the size of the star, so I won't take into account partial transits. I will also assume circular orbits. I will also assume uniform random probability distribution of orbital revolution axes on the unit sphere. Suppose the orbital revolution axis of the planet ...

1

This is an amusing (and tough) math quiz. We are confronted with 3 numerical results and various mathematical reasonings (of which one is not given): the OP's calculation gives $Pt=0.8/4\pi$ (about 6.4%). the OP’s announced correct result $Pt=1/8\pi$ (about 4%), w/o demonstration or source. @Connor Garcia calculation gives $Pt=1/8$ (12.5%). The tutorial ...

1

The neutron collides with a nucleus, and it doesn't matter what atom or molecule that nucleus is part of. Collisions between particles of equal mass lead to more energy loss than collisions with a heavier particle, and therefore hydrogen nuclei slow down the neutrons the fastest. The method detects elemental concentrations, not chemical species. However, if ...

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