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The Sun is pulling Mercury. If the Sun were not pulling Mercury, it would go out of the Solar system in a straight line. It's a balancing act between the Sun's pull inwards and Mercury's tangential speed. It just so happens that the Sun's pull and Mercury's speed balance out just right to keep Mercury in a stable orbit. Actually, it's a bit more ...

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The sun does pull Mercury, that's why its orbits the sun. Internally the sun does work in a push and pull network called a hydrostatic equilibrium. The larger planets do attract smaller things and have helped clear the way for Earth and other inner planets.

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This is an interesting question. Basically it's very similar like any meteorite collision. The gas planet makes here no other difference, but that there will be no crater. Assuming a Jupiter-like planet and an Earth-like planet (Except, say... half the mass of Jupiter), what would happen when the two collide? For clarification: What would the actual ...

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This question has been asked and answered (by me) on Astronomy Stack Exchange: It's brighter on Pluto than you think. NASA developed a tool called Pluto time, which tells you when at your place the ambient light conditions are similar to the ones on Pluto. This occurs when the Sun is only 2° below the horizon! That's quite shortly after sunset, ...

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In this answer we assume a spherically symmetric spacetime, no cosmological constant $\Lambda=0$, and signature convention $(-,+,+,+)$ for the metric. I) Birkhoff's theorem (BT) only works for a vacuum branch of a spherically symmetric spacetime, i.e. in a radial interval $r_1<r<r_2$ without any matter, cf. e.g. this Phys.SE post. Therefore BT would ...

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you could do that by replacing gravity (that is ultra-weak) by another similar force (i.e. attraction in $\frac 1 {d^2}$ ), like electrostatic. It's easy to act on small charged objects. (but if you want a liquid to be attracted, it's more difficult :-) )

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The geometry of spacetime is described by a function called the metric tensor. If you're starting to learn GR then any moment you'll encounter the Schwarzschild metric that describes the geometry outside a sphrically symmetric body. When you go inside the body the geometry is described by the (less well known) Schwarzschild interior metric. The exact form ...

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What is the general relativity explanation for why objects at the center of the earth are weightless? At that location spacetime is locally flat. See the Wikipedia Riemann curvature tensor article and look at the schematic on the right: CCASA image by Johnstone, see Wikipedia Let's imagine we could take away the Earth and look more closely. ...

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The explanation is Birkhoff's theorem, which states that the Schwarzschild solution is the unique spherically symmetric vacuum solution in general relativity. An immediate result of this is that, just as in Newtonian gravity, a spherical shell does not contribute to the gravity experienced by an object within it. If this were otherwise it would suggest the ...

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I agree with John's comment. The termination shock was first crossed by the Voyager 1 spacecraft, and it lies between ~75-90 AU from the sun. Thus, at those distances the solar photon flux would be reduced by factors of ~5,625-8,100, thus solar power received would only be ~0.17-0.25 $W \ m^{-2}$. There would be slightly higher levels of galactic cosmic ...

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The New Horizons spacecraft has an internal heat source. It is a radioactive material - plutonium. The best known celestial body is Earth. Its internal heat source are radioactive elements. Pluto is not as large and not as dense as Earth. But it also has a lot of rocks. Those rocks contain radioactive elements which generate heat.

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Parameters The surface gravity of Mars is ~0.376 g, where g ~ 9.81 $m/s^{2}$ for Earth. The surface pressure of the atmosphere on Mars is ~0.636 kPa, which is roughly 0.63% of Earth's atmospheric pressure (i.e., ~101.325 kPa at sea level). The density of air at STP on Earth is ~1.2 $kg/m^{3}$, compared to Mars at ~0.020 $kg/m^{3}$. Background Typical ...

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About the only relationship worth considering is whether the planet orbits close enough to its parent star so that tidal forces lock the rotation period to the orbital period. Even this is fraught with problems because we currently don't know the exact "tidal friction" coefficients for exoplanets. This will depend greatly on the structure of these planets ...

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Rotational speeds of planets cannot be calculated/predicted because planet formation seems to be highly chaotic. The spin of planets (both rocky and gas) is determined by many factors, including: the angular momentum of the material which was accreted on the planet, gravitational interactions with other planets, the history of collisions as the ...

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Check out this chart as very rough baseline. Solar energy, even with 88 days of Mercury level sunshine, Wouldn't reach nearly as far into Mercury as you suggest. A few KM, perhaps 10 or 20, but not 4,000. I could back that up with a thermal energy calculation of Mercury's mantle and compare it to annual solar energy it gets hit by, but I'm quite sure ...

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The plot below shows a model of how an isolated mass of gas (planet, brown dwarf) cools down with time, taken from Baraffe et al. (2003). The cooling tracks are labelled with mass in Jupiter masses. The time axis is logarithmic in years, the luminosity axis is lograrithmic in units of solar luminosities. Young brown dwarfs and giant planets are governed by ...

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I'm not an expert but I believe the following is correct. The object PSO J318.5-22 is referred to as a "young L dwarf." An L dwarf is a type of brown dwarf, meaning a mass of hydrogen and other elements that is not large enough to fuse hydrogen. PSO J318.5-22 is Jupiter-sized, but I guess there is no particularly important difference between "failed stars" ...

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the answer is no: for now there is a high correlation between the properties of planets (size, distance to their star) and their probability to be detected, which totally bias the observed distribution.

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The Earth has an average density of about 5500 kg/m^3. For a small planet, the density would be pretty similar throughout. Therefore the force of gravity you would experience on a planet with Earth density is: Your Mass * 5500 * 4/3pi * r^3 / r^2 * 6.6723 * 10^-11 This is equal to about 1.456 * 10^-6 * Your Mass * Radius of Small Planet. (I know, if you ...

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The problem is in your assumption that "we fall into open space" unless the planet is large enough. Even if there were no planet at all, we would not fall. In open space you just stay where you are - unless you are affected by some star or planet. In other words you will always drift slowly towards something or other. Now the earth's gravity is so large ...

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Lets just start with the cosmic velocity assuming you dont want to escape fully. $v \le \sqrt{\frac{2GM}{r}}$, where M is the mass of the planet and r is the distance to the center of Mass. We assume a spherical planet. We know the average density $\rho$ of our planet earth and since you will probably want to live on the new planet we will assume it has ...

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