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4

Partial answer: I don't think it's quite as mysterious as it first seems. Bruno explains that the launch site on Earth only passes through the orbital plane of its first intercept target for a moment, and suggests that that launch window is only 1 second long, otherwise "you can't get there from here" anymore. In reality there are going to be ...

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At 200km altitude, the orbital radius is 6,378km (earth radius) + 200m, = 6,578km. Geostationary orbit is at an orbital radius of 42,164km That's a ratio of only 6.4, not above the limit of 11.94 for some bi-elliptic transfers becoming a viable option. So a Hohmann transfer will be both faster and cheaper. Altitude must not be confused by distance to the ...

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You can use a root solving method to calculate eccentric anomaly As you stated, Kepler's equation for eccentric / mean anomaly and eccentricity is: $M=E-e sin(E)$ and there is no closed form solution for E as a function of M but you can still iteratively calculate E. You can set this up as a root solving problem by subtracting M from both sides of the ...

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@Ng Ph: Yes, this is an algorithm for avoiding one object. It has to be applied iteratively to check every object that might intersect. @Robotex: You are being very optimistic. There are literally hundreds of thousands of objects orbiting Earth today, and at the relative speeds they're moving, they'll orbit the world several times before you're done with ...

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As the wikipedia article states, there is no closed form way to express $E$ in terms of $M$. You would have to approximate it, for example using a series expansion: $$E \approx M + \left(e - \frac{1}{8} e^3\right) \sin{(M)} + \left(\frac{1}{2}e^2 \right)\sin{(2M)} + \left(\frac{3}{8} e^3\right) \sin{(3M)}$$ See Morrison 1882 for details about making your own ...

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"Space is big. Really big. You just won't believe how vastly, hugely, mindbogglingly big it is. I mean, you may think it's a long way down the road to the chemist's, but that's just peanuts to space" - Douglas Adams, The Hitchhiker's Guide to the Galaxy Spacecraft are tiny compared to the vast amount of space you have around a planet. So ...

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Express the position of your craft in your orbit as an equation with respect to the parameter T (time). Express the position of something else in its orbit also as an equation with respect to T. Now you can derive an equation for the distance D between objects on these respective orbits as an equation in T. At the point T where the differential D' of this ...

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I need two solutions: fastest and fuel-efficient. There is actually one in the fuel-efficient case, which involves doing a gravity assist by Venus. It has about the same delta-v cost as a regular Hohmann transfer to Mars, but can potentially be used when the angle isn't right since it has other requirements for planetary alignment. Requiring 3 planets in ...

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You're looking for porkchop plots (and Lambert's problem). These plots are created by brute force solving Lambert's problem for a range of departure times and arrival times. note: The numbers on the slanted cyan lines give the duration of the trajectory in days. For example, Mars 2020 launched on 2020 July 30, which is 30 on the x-axis, and arrived on 2021 ...

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Exactly the same way you avoid collisions when not altering orbit. Altering your orbit does not significantly alter your risk of collisions, other than possibly moving you to a higher or lesser densely populated part of orbital space. Orbits are not neat stacks of perfect circles around the planet. All orbits are ellipses, with the perigee closer to the ...

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Yes, it is possible to determine the orbital inclination from the Longitude of the Ascending Node , and a position vector that is not the Ascending node or the Descending Node, because both the ascending node and your position are points in the orbital plane, and the vectors pointing to each from the body being orbited uniquely define that plane. There are ...

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