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IMHO a dust-particle-sized black hole, roughly of the mass of Pluto, moving at few hundreds km/s, could produce effects similar to those are pictured, namely: Ejecta above the entry point (destruction of the crust by the hole’s gravity); Except vicinity of aforementioned point, an almost undamaged leading hemisphere (tidal forces become too weak to break ...


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The Earth's rotation rate and the location of the rotation axis change over time. These collectively are called the Earth orientation parameters. On very short time scales, a day or less, the changes in the Earth orientation parameters result predominantly because of the ocean tides. On the scale of decades to a century or so, the dominant driver is exchange ...


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As David Hammen points out, individual interactions and collisions are extremely important in determining the individual spins of planets and the explanation involving a very large moon-forming collision is almost certainly a major factor in determining the Earth's original spin. However, there are more fundamental reasons why all the planets should spin and ...


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Yes, but only very slightly as the Earth is much heavier than the atmosphere. If you were accelerating at the equator you would exchange angular momentum with Earth, slowing or accelerating it's rotation, depending which way you accelerate. Same goes for air masses, as they accelerate when they warm up or cool down and move more one way as the other way is ...


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The dominant hypothesis regarding the formation of the Moon is that a Mars-sized object collided with the proto-Earth 4.5 billion years ago. The Earth is rotating now because of that collision 4.5 billion years ago. As the linked question shows, angular momentum is a conserved quantity. Just as something has to happen to make a moving object change its ...


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No, you don't need to account for the mass of humans. We're a part of the total mass. Children who grow and gain weight are merely transferring mass from other parts of the Earth to them. The mass of humanity versus the mass of everything else on the Earth is a zero sum game. To see whether the Earth is changing mass, you need to look outside the box (or in ...


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Lots of places state that the Earth's gravity is stronger at the poles than the equator for two reasons: The centrifugal force cancels out the gravity minimally, more so at the equator than at the poles. The poles are closer to the center due to the equatorial bulge, and thus have a stronger gravitational field. TL;DR version: There are ...


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There is approximately 7 billion people on earth. If we take some average of about 75 kg each, that means $$ m_{\rm population}=7\cdot10^9\times75\,{\rm kg}=5.25\cdot10^{11}\,{\rm kg} $$ The mass of the earth is roughly $$ m_{\rm earth}=5.97\cdot10^{24}\,{\rm kg} $$ which is about 13 orders of magnitude bigger than the total population of the world. So no, ...


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Yes you need to include the mass of the 7 billion or so humans in the total mass of the Earth, though since their total mass is something like $5 \times 10^{11}$kg and the mass of the Earth is around $6 \times 10^{24}$kg humans make up only about 0.00000000001% of the total mass. We don't make any great contribution to the escape velocity. The point of your ...


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Here's a simple argument that doesn't require any knowledge of fancy stuff like equipotentials or rotating frames of reference. Imagine that we could gradually spin the earth faster and faster. Eventually it would fly apart. At the moment when it started to fly apart, what would be happening would be that the portions of the earth at the equator would at ...


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The point is that if we approximate Earth with an oblate ellipsoid, then the surface of Earth is an equipotential surface,$^1$ see e.g. this Phys.SE post. Now, because the polar radius is smaller than the equatorial radius, the density of equipotential surfaces at the poles must be bigger than at the equator. Or equivalently, the field strength$^2$ $g$ ...


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The key to answering this question is the Goldschmidt classification of elements. Thirteen of the long-lived elements are siderophilic; they preferentially bind to iron. Those thirteen elements are significantly depleted in the Earth's crust compared to their prevalence on meteors, asteroids, and the Sun. This list of thirteen does includes rhenium to gold, ...


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Actually, they are still currently sinking to the core. Earth's internal heat comes from a number of sources, and one of these is the release of gravitational energy from the heavy elements migrating further toward the center. A similar statement holds for other planets. This isn't the majority of the source of heat. Other sources are the original thermal ...


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Floris's answer gives you an excellent description of the forces that would be present in both a magnetic and gravitational field, whilst MaxGraves's Answer gives you a clear and careful discussion of how you should use the word weight. In more the spirit of MaxGraves's Answer, something that seems a little pedantic but may be interesting to you is the ...


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Yes, the gradient in spherical coordinates contains angular terms. It has to, because the force need not point to the center. The relevant equation is $$\Delta U=\frac {\partial U}{\partial r}e_r+\frac {\partial U}{r\partial \theta}e_\theta+\frac {\partial U}{ r \sin \theta \partial\phi}e_\phi$$ You have no $\phi$ variation, so can ignore the last term.


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The apparent radial accelleration due to moving in a circle is ω2R. For earth, ω = 2π rad/24 h = 73 x 10-6 rad/s. Earth's radius is about 6.37 Mm. Therefore the upwards accelleration at the equator is 34 mm/s2. That's pretty small compared to the 9.8 m/s2 downwards accelleration due to gravity, so we generally ignore it. Then there is ...


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There are two parts to this question (even when you cut out the bonus bits). How much energy is stored in the earth's magnetic field (ramp up the magnet) How much power to keep that field going (drive current through big loop) The former is given by the $\frac12 L I^2$ - so we need to estimate the inductance of the coil needed and its current. A single ...


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The picture drawn by docscience is basically correct, but the math that goes along with it was wrong - as a result, the "height of 370000 km" part of the answer is completely off. The effect you describe is well known by navigators: when you measure the height of the sun or star with a sextant, you actually have to take into account the height of the eye ...


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The reason it may not happen in a plane is probably because you are tilted with respect to the horizon (even if it may not feel like it due to fictitious forces). "Touching the Earth", you are "forced" by gravity to be straight up, and thus the triangle described in the comment above works. Also, these kinds of strange properties usually come from an ...


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The construction and calculations below show that if the altitude is very small compared to Earth's radius, the line of sight as measured from local vertical is very near 90 degrees. The Earth's radius is about 6371 km. For the line of sight to fall 1 degree you would have to elevate your point of view by about 370000 km!


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These videos practically demonstrates what happens when launching an object from a moving object. Projectile Launch from moving object Horizontal velocity remains constant Projectile Motion ( Skit video to 3:25 ) Anyway, thanks to all for your valuable and logical answers on this question.


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In spherical symmetry, the gravitational force only depends on the mass inside of a sphere of radius r, in an image where you pretend it was concentrated in the center. The gravitational force falls off as $\frac{1}{r^{2}}$, always. When you are inside the sphere, though (if you assume constant density) then the mass included increases by a factor of ...


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Could someone give me a relatively straight forward answer why the gravitational field strength follows a straight line (linear) relationship with distance from the center of the Earth but when you get to the surface it follows a 1/r^2 relationship? Basically, it doesn't. A better model is that gravitation is more or less constant at 10 m/s2 to halfway ...


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Besides Mars having a weaker gravity field than Earth, it also has a much weaker magnetic field. Therefore, Mars' atmosphere was not protected from the solar wind. NASA scientists put forth the theory that the small sparse magnetic field that Mars does have actually helped the solar wind to drive off the atmosphere. ...


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The much more important PHYSICS question is, where can life NOT exist? The answer to that is trivial: it can not exist in thermodynamic equilibrium. Life is defined by perpetual change, thermodynamic equilibrium is the case of no change, at all. Beyond that it is scientifically prudent not to rule out ANY non-equilibrium environment, at all, until sufficient ...


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As stated by Kyle Kanos, we only have statistics from one planet. You make the point that there may be undiscovered organisms that can survive without water. Based on our current knowledge, every life form we know of so far has had its components suspended in an aqueous environment. While it is possible there is another form of life that does not require ...



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