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Seasons are mostly caused by Earth axis inclination relative to Sun. The closest point of the Earth's elliptical trajectory does not coincide with the winter nor summer solstice. Accidentally, we live in an epoch when Earth is closest to Sun in January, thus the northern temperature swing is a bit milder than the southern. However the Sun's tidal force ...


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You have identified the problem as the factor $2$ between your calculation and the book. You have read the problem correctly, so it seems the book is wrong.


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It seems Wikepedias article "Two-body problem in general relativity" adresses the subject in detail. Obviously things are very much more complicated than I imagened as it involves GR and Einsteins field equations to which there are no closed form general solutions. To sign off on the matter I have found the following supposedly correct answers to my 7 ...


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If you compare the Earth-Sun distance in the summer and in the winter, you will notice that it differs by only about 4%. Given that the radiation strength of the Sun is proportional to 1/distance^2, the radiation intensity is roughly 8% STRONGER in the WINTER (on the northern hemisphere) than in the summer. But the distance to the Sun is not the only ...


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It is a very general doubt and really important. Slanting sun rays coming to earth from sun are refracted and rarely reach earth. They deviate by refraction. At equator and tropics refraction is less. As you know for $\theta = 0$ no refraction take place and as angle increase refraction increase. As poles do not get much sun rays throughout year and ...


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The picture of a black hole being like a huge vacuum is pretty misleading, not to say wrong. The gravitational field around the black hole is exactly the same as around any other object of the same mass. Like any other object, it's perfectly possible to orbit a black hole, just like we orbit the Sun and don't fall in. In fact, if you were to magically ...


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You've the answer in your statement: ...black holes are giant vacuums that will absorb/consume the area around it. The key is that the area around it is a few times the event horizon of the black hole and we are very far from the event horizon of the supermassive black hole at the center of the Milky Way. The event horizon for Sag A* (the name of the ...


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You have got your forces and potentials mixed up. If the force is given by: $$ F(r) = \frac{GMm}{r^n} $$ then the potential is the integral of this: $$ V(r) = -\frac{GMm}{(n-1)r^{n-1}} $$ In the question you link the force has an inverse cubic dependance, $n=3$, while the gravitational force has an inverse square dependance, $n=2$. We get a circular ...


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The trajectory will not be sinusoidal. Depending on the initial speed it will be one of elliptical , parabolic and will "reach" infinity with zero speed or hyperbolic and will "reach" infinity with a finite speed. So the trajectory will very much depend on the initial speed of the satellite. So how have you set up the computation?


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Here is the proof: Please refer to the wikipedia page on eccentric anomaly for a diagram and a couple of intermediate formulae. For an ellipse with the usual formula $x^2/a^2 + y^2/b^2=1$, it is the case that $\sin E = y/b$, and also by studying the figure on the wiki page you can see that $\sin (\pi-\nu) = \sin \nu = y/r$. Thus the two results you wish to ...


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How long time does it take before three planets achieve the same relative position? The answer is never, except for the case when their orbital periods can be expressed with low integers, like the 4:2:1 resonance of Io, Europa and Ganymede However, what you are asking about is when they are going to be in almost the same position again, a quazi-period. To ...


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To understand this, you have to understand how the spin affects the gravity. The spin creates a sling shot effect of pushing an object in the direction of the spin. With that in mind, in a retrograde orbit, there are two forces pushing in opposite directions keeping an object in a purgatory state like. Essentially it's like a game of tug of war between the ...


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Let's simplify the scenario slightly - imagine an object at the distance of the Moon, that has had its angular momentum slowed sufficiently that it will approach the Earth to a distance less than the radius of the Earth plus the radius of the Moon. Distance Moon-Earth ~ 400,000 km Radius Moon ~ 1,700 km Radius Earth ~ 6,300 km Mass Moon ...


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Meteors are essentially bits of rock that are independently in orbit around the Sun and which cross the Earth's orbit. If the Earth happens to be there at the same time then it will enter the Earth's atmosphere and we will see a meteor. The velocity of meteors is related to how fast they we going in their orbit around the Sun, combined with how fast the ...


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All speed is relative. But an object that starts from rest at infinity will reach a velocity of about 11 km/s when it hits Earth, if Earth is the only thing pulling on it. At the same time, Earth is moving with an orbital speed of about 30 km/s. Their relative importance will depend on the direction from which the meteor is approaching - but on the whole ...


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Plenty of others made it clear that babies aren't made out of matter from somewhere else. But if the earth's mass were to increase ten percent by magic, its orbit would not change. Its momentum—its tendency to fly off into space in a straight line—would increase by ten percent, and the counteracting force of the sun's gravity would increase by ten percent. ...


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As others have said, the mass of the Earth doesn't go up because we are eating food produced on Earth. Suppose we somehow were importing really large amounts of extra-terrestrial food, now what happens? We roast. Food contains a fair amount of carbon. Carbon from extra-terrestrial food is just as much a greenhouse problem as burning fossil fuels. ...


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Mass is conserved (up to whatever small contribution nuclear decay has to the overall loss of mass). All biological matter is just created from materials from the environment (we do eat, right?). With exception of an occasional space probe, shooting stars and solar wind effects, the earth can be considered a closed system in terms of matter exchange. ...


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Life forms are made up from materials already present in Earth. Thus, increasing population would not alter the overall mass of the planet, and can't impact its orbit.


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If the mass of the moon is $$ M=7\cdot 10^{22}\text{ kg} $$ and its orbital velocity is $$ \Delta v = 1000 \text{ m/s}$$ while the time in which you need your job done is $$ t = 1\text{ yr}=365.25\cdot 24 \cdot 60 \cdot 60 \text{ s}$$ then the force you need is in Newton $$F = \frac{M\cdot v}{t} = 2.2\cdot 10^{18} \text{ N}$$ or a power in Watt of ...


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A spinning space station does not provide artificial gravity away from Earth, it provides artificial gravity toward the center of the station. The first effect the spinning surface has on an occupant is to provide a tangential acceleration through friction. Then, because the tangential direction runs into the wall, the occupant is "thrown against the wall" ...


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I have been using Orbital Mechanics for Engineering Students by Howard Curtis available on Google Books has a great derivation of the Universal Variable formulation, see chapter 3.7, as well as step by step algorithms and example Matlab code for calculation. I have been using it as a basis for a python program I am developing (as a hobby) and highly ...


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In order to arrive to this answer you have to assume that the size of the planet and star is negligible compared to the size of the orbit of the planet. When solving this question you have to understand what each symbol stands for: $T$ stands for the orbital period of the planets orbit; $r$ stands for the semi-major axis of its orbit; $GM$ can be combined ...


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I mean, if you believe what you've just written, $$\theta(t_2)-\theta(t_1)=\theta(t'_2)-\theta(t'_1)$$ Then I think you have already proven it, with a little more formal calculus. First rewrite that expression above as $\Delta \theta(t)=\Delta \theta(t')$, with the prime indicating the other interval. If $$t_2-t_1=t'_2-t'_1,$$ Then just set $\Delta ...


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Imagine that the wheel is stationary. A force is applied which accelerates the wheel horizontally. This add translational momentum to the wheel. Now, since the wheel does not slide, a frictional force is produced at point P which produces a torque on the wheel. The torque causes the wheel to start rolling, adding rotational momentum. Once the wheel is ...


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If at any time the speed of the planet in the reference frame of the star exceeds the escape velocity $\sqrt{2GM_\star/r}$, where $M_\star$ is the mass of the star and $r$ is the distance from the star to the planet, it will escape in a hyperbolic trajectory (or straight line if $M_\star\rightarrow0$). As noted in the other answers, the result of the ...


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The scenario you suggest is of course hypothetical, but in all cases you must conserve angular momentum and mass/energy. So for example: If you have a way of removing mass from a star in such a way that the mass disappears outside the orbit of the planets (in astrophysics this is accomplished simply by mass loss - either the star has a wind that expels mass ...


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Your final question very much correlates with a famous thought experiment.If the Sun was suddenly removed the planet s will still continue to stay in orbit. For 8 minutes and 20 seconds. This is because the speed of the space time fabric or simply putting gravity travels at the speed of light. That is, the earth will be devoid of sunlight and will move ...


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A coarse correction would have to be made because of the difference in the curvature of space in that section of space when a planet is deleted. The sun holds the planets in orbit and they would fly away as if the string to a tether ball was cut to all the planets.


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The motion of a comet can be understood in terms of a couple of principles. First - Inertia. Newton's first law states that an object will remain at rest or in uniform straight line motion unless acted upon by an external force In other words - assuming that "something" had given a comet a velocity, it will keep going unless something changes that. ...


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Assuming that the twin star system is isolated, there is no external force acting on it so the center of mass should not move if the system was at rest initially. Centre of mass doesnot move Centre of mass moves although no external force is applied and the system was at rest initially



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