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You can use Kepler's Third Law which says $T^2/a^3$ is a constant. Making further calculations using Newtonian gravity for the two-body problem, we can actually compute the theoretical result of this constant: $$\frac{T^2}{a^3} = \frac{4\pi}{G}\frac{1}{M + m} \approx \frac{4\pi}{GM}$$ Where $M$ is the mass of planet and $m$ is the mass of moon. Assuming ...

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orbiting, graviting, means only mass maters. It could be gaz, diamon, a bag of cockroach, or even a black hole, it would not change gravitational behaviors that orbits are. ( NB: yes, black hole too. Nothing different occurs in gravity as well when a star turn into blackhole. And it was attracting distant matter before exactly as well.)

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If we assume all of the rest of the planets are neutral in charge, then only the mass of the sun matters for the Solar system. The gravity between the star and planets and other objects forms the main force that maintains the rotation of smaller objects around the sun and be stable for a long time. No matter what you fill in the sun, the mass of the sun ...

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The only obvious scenario involves placing a moon in the L2 point of a large enough planet. All the Lagrangian points except L4 and L5 are unstable. That implies that no such arrangement can persist without active station-keeping.

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This is assuming the earths magnetosphere is the minimum required to shield an object from cosmic rays which is, in fact, incorrect. Here is the what you need to figure out to calculate a satellite system to do what your proposing .. 1-what is the required minimum strength of a magnetic field so that it deflects virtually all cosmic rays and ...

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I think there is a better way to think about it. To a small factor, you are essentially saying the planet is on the photosphere of the star. The star will fill almost half the solid angle. Therefore the flux (power per unit area) from the star at the substellar point on the planet is $$f \simeq \pi \int B_{\nu}\ d\nu =\sigma T^{4} = 1.5\times 10^{11}\ ... 0 The precise happenings cannot be predicted. However, let's try using the basics of radiation heat transfer to suppose what will happen. As we know, the radiation emitted by a body is directly proportional to T^4, where T represents the temperature of the body. Thus, the increase in emission of radiation increases very rapidly with an increase in ... 0 Larry Niven dealt with exactly this question in his short story Flatlander and as far as I know his details are quite reasonable. While there is nothing intrinsically different about an antimatter planet from an optical point of view, its interactions with local matter such as solar wind or (for deep space) deep-space gas and dust would leave it rather ... 27 Has Musk done his homework? With regard to the basic idea of using nuclear weapons to release CO2 and thereby warm Mars, no, he hasn't. I suspect this was either Bored Elon Musk speaking, or perhaps the Elon Musk who didn't quite deny being a super villain ( 1-900-MHA-HAHA Elon Musk?) in that interview with Colbert. CO2's enthalpy of sublimation is ... 2 This is a Gauss's Law problem using mass. The only mass important in the g=GM/r^2 equation is the mass inside the radius r. The mass outside this radius is irrelevant in the first order. At r<R_E,$$M=\rho (4/3)\pi r^3$$where \rho can be approximated (poorly) by$$\rho=\frac{3M_E}{4\pi R_E^3}.M=M_E\frac{r^3}{R_E^3} so $g$ below the ...

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As a thought experiment, imagine you are sinking through the Earth, from the surface of the towards the center. At any point as you sink, you can divide the mass of the Earth into two parts: the sphere below you and the spherical shell above you. The net gravitational attraction at any position within a spherical shell is zero, as Newton proved. The ...

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At the center of the earth you would experince no gravtational force. (Since the force composants from the surrounding matter will cancel each other.) Reference https://www.physicsforums.com/threads/how-strong-is-gravity-in-the-center-of-the-earth.203955/

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That equation holds for a point mass (or a radially-symmetric mass when you are outside the shells). The earth is not a point mass, so you can't get to r=0 from the whole thing. The equation fails as soon as you drill below the surface.

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It might. From Wikipedia: "The Kessler syndrome (also called the Kessler effect, collisional cascading or ablation cascade), proposed by the NASA scientist Donald J. Kessler in 1978, is a scenario in which the density of objects in low Earth orbit (LEO) is high enough that collisions between objects could cause a cascade—each collision generating space ...

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