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I don't work in astrophysics, but I make a simple guess based on general mechanics. Assume that in the beginning (billions of years ago), the galaxy was spherical, and that was rotating around some axis, maybe the same as today. Then, where is the biggest centrifugal force? The centrifugal force on a piece of material of mass $m$ in the galaxy is $F_c = m ...


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A related question would be: why most of the Solar System planet lie in the same plane? That is, why don't the planets close enough to the Sun rotate at random angle with each other (such as Pluto which rotates in a plane at some angle to the planets' plane)? Similarly, why do the stars in a galaxy tend to rotate around the center of the galaxy roughly in ...


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You don't say what simulation technique you are using, but clearly errors are adding energy to the system. A simple one would be to take a starting position $(x,y)$ and velocity $(v_x,v_y)$. Note that if your velocity is not exactly right for a circular orbit, you should just get an ellipse that is close to the circle you are after, so that is not your ...


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As explained in this article by Neill DeGrasse Tyson, the tidal forces between the Earth and the moon do indeed slow down the rotation of the Earth each year, the same process that caused the moon's rotation to become tidally locked with its orbit of the Earth. This effect would eventually cause the Earth's rotation to be tidally locked with the moon as ...


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It seems to me you may be misunderstanding the problem as stated. You are assuming you are being asked about two different objects (planets?) in different orbits; but I think from reading the question that you are being asked about the same object at different points in its elliptical orbit. For an object in an elliptical orbit, conservation of angular ...


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Your equation relates the period of the orbit to the length of the semi-major axis, not to the absolute distance at any point. You can use the Vis-viva equation if you have more information. But you don't have the semi-major axis length or other details about the orbit. As you suggest, conservation of energy is the simpler way forward.


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The potential $V(r)=\int \frac{F(r)}{m} dr$ as integral of gravitational force $F(r)$ divided by mass $m$ is only determined up to a arbitrary constant. A satellite coming from infinity will get kinetic energy from falling to the gravity center. So it is a convenient convention to say the potential $V(r) = 0$ and therefore also the potential energy ...


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What they are basically telling is that the gravitational potential at an infinite distance from the fixed mass is zero. The total energy of the satellite would be K.E. + P.E. Since, $$V(r)=-\int{E(r)}$$ $$V(r)=\frac{-GM}{r}$$ With M being the mass of the fixed object. As r tends to infinity, it is clear that the potential of the system rapidly decreases, ...


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As shown in a previous answer, Assuming they have the same density (the Sun's average density is not much smaller than that of the moon) , if they had the same apparent size in the sky, then the mass M of the object will grow as r3 (because M=4/3ρπR3 and R=θr), so the force actually grows linearly with r. this implies that for same apparent size and ...


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Tidal forces drop rapidly with distance - the derivative of $1/t^2$ is $-2/r^3$. Further, the difference in radius of the orbits of Earth and Mercury is a little more than a factor 3x and radius of mercury is about 2.5x smaller than that of earth. From the orbits we gather the tidal effect is 27x smaller - from the radius we gather that moment of inertia is ...


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The answer is "No" - unless the city is on the equator. You specified "...it will always remain direct over a city". Satellites in geosync orbit might be visible to cities but not be directly overhead unless they are located on the equator.


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The answer is the same as the answer to the question "why do satellites stay in orbit": the gravitational pull of the earth is just strong enough to keep it in orbit at the altitude it is, given the angular momentum (velocity) that it has. In equations: $$\frac{GM_{earth}}{r^2}=\frac{v^2}{r}$$ where $r$ is the distance from the center of the earth to the ...


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It's moving the orbital velocity for its particular altitude, which means it has the exactly the speed at which Earth's gravity supplies the centripetal force needed to accelerate it in a circle. Now, I'm guessing what you mean by your question is that at the surface of the Earth, we need to boost a rocket to about $v_e(R_\oplus)=11{\rm km s^{-1}}$ (let ...


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Earth is about 100x more massive than the moon, and since $F \propto M / r^2 $, the distance from Earth to the astronaut would have to be about $\sqrt{100}$ = 10x further than from the moon to the astronaut. Therefore, the astronaut falls "up" about 90% of the way to the moon. [The earlier answers go a lot more into detail (and are more technically ...



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