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22

I feel it's a cheap answer but heavy Jupiters can get much denser than Earth because planets with Jupiter's mass stop adding size as they add more mass. A planet with Jupiter's size and 10-12 times Jupiter's mass would be over twice Earth's density. As far as Earth-like planets, there's a cool property of terrestrial planets, more mass means more tightly ...

17

Our own magnetic field is generated by convection currents in Earth's liquid outer core. A useful summary from Physics.org: Differences in temperature, pressure and composition within the outer core cause convection currents in the molten metal as cool, dense matter sinks whilst warm, less dense matter rises. This flow of liquid iron generates electric ...

14

Yes, but they're not very good and they're amazing. The Wikipedia article for Pluto shows a low-resolution map of the surface, generated from Hubble images: And the Wikipedia article for Pluto's largest moon Charon shows a low-resolution map of the Pluto-facing side of Charon (not to scale): Larger image here. Only the Pluto-facing side is shown because ...

9

Super-Earths and Mini-Neptunes are the "in-between" types of exoplanets you're looking for. A sweeping generalization would put most in the range of $\sim1$-$10M_{\oplus}$ (Earth masses), with some outliers a bit above that. They may have significant quantities of hydrogen and helium in their atmospheres, as well as water, in liquid or vapor form. The latter ...

8

From the wikipedia page on Chthonian planet https://en.wikipedia.org/wiki/Chthonian_planet "Transit-timing variation measurements indicate for example that Kepler-52b, Kepler-52c and Kepler-57b have maximum-masses between 30 and 100 times the mass of Earth (although the actual masses could be much lower); with radii about 2 Earth radii, they might have ...

7

I've learned that the Earth's core is hot due to decay of radioactive elements. This is unproven, non-standard geophysics. There are several arguments against this. One is that all of the long-lived radioactive isotopes are isotopes of uranium (two isotopes, 235U and 238U), thorium (232Th), and potassium (40K). The problem: Uranium, thorium, and potassium ...

6

Okay, so that's a 130 mm newtonian with an f/5 focal ratio. Per the manufacturer's website, it comes with two eyepieces, 10 mm and 20 mm, giving 66x and 33x magnifications respectively. First off, make sure your scope is collimated. The procedure is described in their generic Telescope Maintenance document, and also more specifically in the Astromaster ...

5

The observational determination of the chemical abundances in exoplanets is in its infancy. In terms of terrestrial type planets, i.e. those of size less than a few Earth radii, the constraints are confined to comparing the measured densities (obtained from the masses and radii of transiting planets found by Kepler and CoRoT) with models of what planets with ...

4

At the moment there is basically only one way. That is to associate the planetary-sized object with a cluster of stars or moving group of stars of known age. That's basically it. If the planetary-sized object really can't be associated with another object, then only limits can be placed on its age by comparing it's luminosity to theoretical planet cooling ...

4

This is not a characteristic of the solar system. It is a characteristic of the definitions of the names you used. Neptune and Uranus are the bodies you believe to be missing. In fact, with the mass of Earth at 6*10^24 kg, Uranus at 9*10^25 kg and Jupiter 2*10^27 kg, you'll notice that Uranus is only ~15 times the mass of Earth while Jupiter is ~20 times the ...

4

> Do other stars have similar gaseous-to-rocky ratios among their planets? For any given stellar system, are there typically as many gaseous planets as there are rocky planets? With the current instruments and methods we have, we can only access certain population of exoplanets. Terrestrial-mass exoplanets are quite hard to find hence we have more ...

4

The currently-detected planets do not show a clear distinction between rocky and gaseous planets. While there seem to be somewhat two ensembles, the rocky planets of Earth-size and "super-earths" and the gaseous planets (Jupiter-size), there is a broad transition between them. The transition is roughly where we find Neptune and Uranus. Play around with ...

4

The Kepler-20 system has planets with masses in the following order, going outwards from the star: Kepler 20b: $\approx 10 M_\oplus$ Kepler 20e: $\approx M_\oplus$ Kepler 20c: $\approx 16 M_\oplus$ Kepler 20f: $\approx 1.5 M_\oplus$ Kepler 20g: $\approx 20 M_\oplus$ Kepler 20d: $< 20 M_\oplus$ If Wikipedia is to be believed, Kepler 20b may be a ...

3

No, it's an illusion. Probably the ancient one. Simple experiment you can do is, set grid on telescope, measure the angle subtended when moon is at horizon and when moon is atop. You will see angle subtended by moon is same, hence the size are same. It's merely an illusion. For more details: https://en.m.wikipedia.org/wiki/Moon_illusion

3

It's a big question, but kind of a favorite subject of mine, thinking about exoplanets, so I can give a ballpark answer, and I invite anyone to give correction or give a more technical answer if they like. Ice(s) formation An ammonia-water ocean wouldn't be friendly towards ice formation because water ice would sink in the ammonia-water solution and ...

3

The transfer of heat would be insignificant. Consider the Earth, the core is as hot as the surface of the sun. Some of this heat does get transferred to the surface, but it is only 0.03% of the energy received from the sun. Rock is a good insulator, most of the heat is transferred by convection. The far side of a tidally locked planet would be very cold. It ...

3

I think the answer to your question lies with Jupiter, be it directly or indirectly, the gas giant is now believed to have had a large influence on the way the inner planets formed. Many planetary scientists believe that the reason we only have 4 small inner planets today and not any "super-Earth" sized ones is because of Jupiter, and how it interacted ...

3

The flattening of a planet is a function of both its spin rate and its structure. But for a series of planets of homologous structure the flattening depends on the spin rate. the faster it spins the greater the flattening (for spin rates typical of planets any way). So if a planet is spinning slowly it will display little polar flattening, similarly if it ...

3

You may have a hard time finding definitive answers to these questions. I'm after the same answers and hope this will contribute as a starting point to something more specific and comprehensive . Generally the abundance of elements appears to be closely related to their mass. In effect the heavier the element the more scarce it will be. This is due to the ...

2

From the article: "In fact, its age may be 4.36 billion years old." This is still older than the estimated age of the Late Heavy Bombardment, so it wouldn't affect that hypothesis much. In fact, it may actually help the hypothesis, as it means the Moon would have more of its interior molten (though probably not by much). It's suggested that the magma from ...

2

No known planet except Earth can be colonized by a human civilization. There are at least three serious issues: temperatures at around 300K, an atmosphere of appropirate pressure, and damaging cosmic radiation (low gravity is also a worry for long-term human presence). Minerals are less of a problem (and water can be synthesized). Mars and the Moon are close ...

2

By lens in this case I presume you mean eyepiece. The magnification you get from an eyepiece is given by dividing the focal length of the mirror (650) by the focal length of the eyepiece. You did not say what eyepieces you have but if, for instance, you had a 10mm eyepiece you would get a magnification of 650/10=65x. A 20mm eyepiece would give a ...

2

The current ideas are that both terrestrial planets and giant planets start their formation in a similar manner. Dust settles towards the mid-plane of a predominantly gaseous disk, starts to stick together and eventually small (km-sized) planetesimals are formed. This process may be quicker in the outer parts of the solar systems where the gas is colder and ...

2

I'm inclined to say no (and footnote, I realize Wikipedia isn't a good source for scientific proof as it's not always right, but I'm using it more to demonstrate a point than than use it as an authoritative definition). Wikipedia: A terrestrial planet, telluric planet or rocky planet is a planet that is composed primarily of silicate rocks or ...

2

From another question on the stack exchange I just found out about an exoplanet with a much higher density than Earth: Kepler-131c. https://es.wikipedia.org/wiki/Kepler-131c Sorry for the WP page being Spanish, there is no English one.

2

There are two main factors that control if planets have magnetic field. There must be a fluid conducting medium (liquid iron for Earth, liquid metallic hydrogen for Jupiter), and the faster the core rotates the stronger the field. Mercury rotates slowly,, not sure how much of it's large metallic core is liquid: weak field Venus probably similar structure as ...

2

Rogue planets have two formation mechanisms: Independent formation and ejection. An independently formed rogue planet would have condensed out of the nebular material by itself and not formed from a young star's protoplanetary disk. We understand how individual stellar-mass objects condense and which have been tested by observation, but I'm not aware of any ...

1

(Some of) These plots are incorrect. Whilst Rayleigh scattering has a steep dependence on wavelength $(\propto \lambda^{-4})$, it cannot scatter what is not there. There is almost no blue light at all coming from stars with $T_{\rm eff} < 3500$ K. A detailed description of the problem with your calculations(?) is given here with regard to a red giant ...

1

I think the star,its halo and cloud can't be darker than the sky color even early M type still look blindingly bright orange-red on the sky not dim orange-red. The sky color, if heavier i think it will look more "desaturate"(brighter too but not bright to the point it's brighter than the star) and thecolor shift toward the red side(you are correct) that ...

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