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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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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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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 ...


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Oke this link is a nice reference sheet for elliptical orbits. we are interested in two things the speed at perihelion The time it will take to reach aphelion The initial velocity is what OP wants to now. The time it will take to reach aphelion will be compared with the position at time t of object C So lets call $r_c(t)$ the distance of Object C to ...


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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 ...


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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 ...


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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 ...


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Yes, all central forces can produce circular orbits given an appropriate initial velocity. However, if some radial velocity is introduced, these orbits need not be stable (i.e., remain in orbit indefinitely) nor closed (repeatedly returning to exactly the same path). There are only two kind of forces that produce exactly closed orbits: the inverse-square ...


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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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You can approach this problem from different angles, using different parameter which would stay constant. In your case you tried to use the semi-major axis and eccentricity, however you could also use specific orbital energy and specific angular momentum. Depending on the problem one approach might be faster/easier than the other, however both should lead to ...


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Orbital life of satellite relates to the loss of angular momentum due to drag. So you could change aspect ratio: reduce the forward-looking area, reducing drag increase the mass (more angular momentum for same drag) regularly boost it (propulsion that counters the drag) put it in a higher orbit (less drag) I considered whether making the orbit more ...


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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 ...


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The answer to this is really very close to Ted Bunn's answer to this question, but I think there may be one additional point .... a disk the most stable configuration Maybe it could be thought of as the 'most stable configuration', but I think it should rather be thought of as the natural state that accreting matter will form for the following ...


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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 ...


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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 "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 ...


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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 ...


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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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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.


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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.


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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 = ...


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I think the answer lies in the fact that the variational principle says is if there are two fixed points in space-time then there exists an open subset of space-time containing these two points such that amongst all curves contained in this open subset, a geodesic will be the curve of longest length between these two points. So, the variational principle ...


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Rosetta initially went round the comet in a roughly triangular orbit using its thrusters to change direction. Eventually in September 2014 it entered a true orbit at a distance of about 30km going round once every two weeks. The orbit can hold in such weak gravity because it is very close and slow compared to satellites orbiting Earth. It has now moved ...


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A stable orbit requires that the orbital insertion velocity be just right. Too high and the spacecraft flies away on a hyperbolic trajectory. Too low and it eventually falls and impacts the planet (or comet in this case). Upon approaching a planet or comet the spacecraft will perform a delta-v maneuver, a burning of thrusters that insert the spacecraft into ...



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