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22

Why shouldn't the orbits of stars be Keplerian? The answer is simple. Keplerian orbits are predicated on a single central point mass. That assumption fails to some extent even in a solar system. It fails massively in a galaxy. A galaxy is not a point mass.


11

Elliptical orbits are direct consequence of orbiting entirely outside a spherically symmetric mass. Even if you assume that a galaxy has a spherically symmetric mass distribution, the amount of mass at a radial distance less than that of the star would be changing (assuming some eccentricity). Once that happens, the orbit is no longer an ellipse.


5

Not Keplerian, because it is not a conic-section. It is not even explained by Newtonian gravity. In contrast, Kepler's laws are explained by newtonian gravity. The lowest orbital-energy from Keplerian orbit is circular. And the orbits of stars are observed to be approximately circular. Hence: $$ \frac{mv^2}{r} = \frac{GMm}{r^2} \quad\Longrightarrow\quad v = ...


3

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.


3

If you are in a circular orbit what you need is a Hohmann transfer, from Wikipedia: In orbital mechanics, the Hohmann transfer orbit /ˈhoʊ.mʌn/ is an elliptical orbit used to transfer between two circular orbits of different radii in the same plane. It works like this assuming the planet is in a circular orbit. Then the amount of delta v needed to ...


3

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.


2

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 ...


2

The equivalence principle tells us that we can evaluate $\nabla_u u$ in a co-moving reference frame and that for geodesics we should find no acceleration (to the occoupants of an elevator in free-fall, the contents seem to be experiencing no acceleration). Therefore, if we evaluate this when we are not along a geodesic (elevator sitting on earth), we find ...


2

There seem to be three kinds of slingshot manoeuver. You can bleed some kinetic energy off a moving body, sort of like an ancient slingshot; that allows a spacecraft to either increase or decrease its own kinetic energy. Or you can get more "bang for the buck" with the assistance of a gravity well, be it moving or fixed, by expelling mass after having ...


2

This basically is a specific case of Lambert's problem. I will cover the maths involved solving the problem in your case. When looking at the velocity of mass C, 10000 m/s radial outwards relative to the sun, it can be noted that its movement therefore will basically be one dimensional along the radial direction. This means that its angular moment is zero ...


2

You've pretty much covered everything, except that below the geostationary orbit the orbit is elliptical - there is no transition to a hyperbolic orbit at small $r$. Let's draw a picture of the pole and get some preliminary formulae out of the way: If you are on the pole at a distance $r$ from the centre of the Earth (NB from the centre of the Earth not ...


2

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 ...


2

In general the planes of solar systems are not aligned with the plane of the Galaxy, but are oriented in all different directions. The size of a solar system is so much smaller than the size of the Galaxy, that the Galaxy's structure has no impact on the orientation of a solar system. What determines their orientations is the direction of the angular ...


2

A good estimation would be an Hohmann transfer from LEO to MEO. It might be slightly more efficient if the burn would be performed in the upper atmosphere, due to the Oberth effect. When you calculate the total required $\Delta$v for the two burns of the Hohmann transfer, you get about 2.3 km/s, so the total $\Delta$v to get from Earth to MEO would be about ...


1

The "pole" you are referring to is also known as a space elevator. Creating a cable of sufficient tensile strength is currently unfeasible, but carbon nano- and macrotubes are promising. Your question can be generalized, ignoring Earth's non-spherical shape, to apply to any position in space around earth by replacing "height above the equator" with ...


1

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 ...


1

The exobase is defined as the effective end of the atmosphere, and is a gray area between 500 km and 1,000 km. Presumably, once a craft's orbit is outside of the exobase, drag is negligible and stationkeeping basically not a necessity (as in won't return to earth for a century or more, until we can repark it or utilize it). From my time in Kerbal, I would ...



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