# Tag Info

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ESA has Space Rider in development. First launch is planned for 2022 on a Vega-C. Planned payload is 800 kg.

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eRosita is on its way to an elliptical orbit around L2 (with L2 in the centre of the ellipse). (image source: https://www.slideshare.net/esaops/wilms, p. 19) According to Merloni et al. (2012, https://arxiv.org/abs/1209.3114), the semi-major axis is planed with about 1,000,000 km and the orbital period should be about 6 months. Taking a look into the ...

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The expression on the right is meant to give the eccentricity vector but the vector notation has been lost. Here it is in this answer: $$e = {v^2 r \over {\mu}} - {(r \cdot v ) v \over{\mu}} - {r\over{\left|r\right|}}$$ and the vector nature is not clear either. We should write it as $$\mathbf{e} = {v^2 \mathbf{r} \over {\mu}} - {(\mathbf{r} \cdot \... 3 No, there's no plan. Tragedy of the commons. ESA have trialed some solutions, with the RemoveDEBRIS mission. The Journal of the British Interplanetary Society has some papers on the Necropolis system 4 Wikipedia sez that for a body starting at rest at distance r, the time it takes to reach a distance x is given by:$$t(x) = \frac{r^{3/2}}{\sqrt{2 GM}} \left( \arccos(\sqrt{b}) + \sqrt{b(1-b)} \right)$$where b=x/r. The Sun's standard gravitational parameter GM is about 1.327E+20 m^3/s^2. The time to fall all of the way (x=0) is just$$t(x) = \...

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If you have a direct cosine matrix (also called a "rotation matrix") which converts from ECI to LVLH, then the transpose of that matrix will perform the opposite rotation: LVLH to ECI.

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To a pretty good approximation, the angle must be between 82.75° and 97.25° from the Sun's pole, and the distance doesn't matter. This is because if you're starting from rest relative to the sun, you will fall straight in relative to the Sun. To hit the Earth on your trajectory, the straight-line path from your starting location to the Sun must intersect ...

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Trajectories are reversible. So if you had a spacecraft heading into the solar system at the same velocity that NH is currently heading out, it would follow the same trajectory but in reverse. If you time things correctly, it could narrowly miss Jupiter and be deflected into a solar orbit which could then hit the Earth and it would reenter at the same ...

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Could a spacecraft fly in a direct line from the outer solar system to Earth? Ignoring tiny deviations due to long-range attraction to the larger planets, sure! absolutely! no problem! you betcha! And you will need zero propulsion to do it. Sit back and enjoy the ride! At the right time of year you can position yourself at say 100 AU from the Sun in the ...

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You are confusing aerobraking with aerocapture. Aerobraking is used to circularize an elliptical orbit into a circular one after orbit insertion, and it has been used a few times on the following missions: Hiten: this was a demonstration mission in Earth orbit Magellan: Around Venus Mars Global surveyor Mars Odyssey Mars Reconnaissance Orbiter Venus ...

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Theoretically: Yes If the net force on an object is zero, then Newton's 1st Law says that it will continue in a straight line. By continually adjusting the amount and direction of your thrust, so that the thrust cancels out the force of the Sun's gravity, you can do just that. Practically: No Doing the above takes ridiculous amounts of energy -- which in ...

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It depends on if you are doing a direct to leave the solar system or doing a flyby, but probably. If you are directly leaving the solar system, then the close you are to the Earth's inclination around the Sun, the more your velocity will count. If you go completely perpendicular to that, it will take quite a bit more fuel, as you have to do an inclination ...

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Their solar sailing algorithm is described in the Planetary Society blog: Solar sailing: The spacecraft is attempting to raise its orbit using the solar sail. To do this, it must make two 90-degree turns each orbit. When flying towards the Sun, the sail orients itself edge-on, effectively turning off the thrust. When flying away from the Sun, the sail ...

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Grady Hillhouse was able to get it working on a Nucleo F401RE development board: link. I improved on that to get it on an ESP8266 wifi module. And added some extra stuff to calculated satellite overpasses and see if the satellite is visible: project , library This is focused on embedded devices, but it shouldn't be to hard to get it running in other ...

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If you are trying for Geostationary orbit an equatorial launch site is better, but if you are stuck with launch from inside the UK but are not prepared to drop spent stages on voters then you can still do low altitude polar launches . As with so many things this is much more about politics than physics. So Israel launches ...

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In 1964, the Soviet scientist A.D. Kuzmin, together with the American scientist Barry Clark, began observing Venus using a movable radiointerferometer consisting of two 27-meter paraboloids (Owens Valley Radio Observatory, California). The radius of the hard sphere of Venus was measured: 6057 km (before that, astronomers measured only the radius of ...

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This was indeed quite a challenge. The main advantage was having the mariner probe there. By determining the effect of Venus on the orbit of mariner, Having the exact distance to the surface of Venus using radar on-board Mariner and tracking data of Mariner itself allowed for a pretty good estimate. In particular the difference in distance between where ...

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With engines! Orbiting at L1 is completely feasible, as long as your satellite regularly uses little bursts from its engines to keep it there. L1 is "unstable", meaning that a satellite without engines will eventually drift away from L1. But the closer your satellite stays to L1, the less fuel it requires to stay in place. Low thrust, high specific impulse ...

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Note: the question has been radically re-written since this answer was written. Consequently, it is no longer relevant to the question. If you want an object to stay between the Sun and Earth, it has to be at the Earth-Sun L1 point, which is about 1.5 million km away. We already have stuff there: https://en.wikipedia.org/wiki/...

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If I understand the question as it is evolving, you are looking for an orbit that produces a solar eclipse; a complete shadow of the Sun on a small area of the Earth, and further that the object casting the shadow not be in an orbit around the Earth as in the question Is a sun-blocking orbit possible? but instead be in a heliocentric orbit. That would mean ...

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The only stable points that orbit at the same speed as Earth are the L4 and L5 points, as you mention, but there are some unstable ones as well. See this pic from NASA: L4 and L5 remain ahead of and behind the Earth, whereas L1, L2 and L3 are inherently unstable. From your question, I'd suggest L4 and L5 would be best suited, unless you really need ...

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You're on the right track looking up Lagrangian points, orbits where a small object can stay in the same relationship with two celestial bodies, one orbiting another. The one you are describing is the earth-sun L2 point, a point outside of earth's orbit around the sun. This Wikipedia page will tell you more.

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Not even remotely enough delta-v to make the TLI burn. The apollo burn was 3+ km/sec. The space shuttle OMS system was good for about 300 m/sec, so about 10% of the amount needed.

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@asdfex's answer is correct. I'll just add an illustration. In the Circular Restricted Three-Body Problem or CR3BP (where the Lagrange points are defined) there are some conventions to make the math easier. With $m_1 + m_2$, the distance between $m_1$ and $m_2$, and the rotation rate all being unity (1.0) the reduced mass is defined as \mu = \frac{m_2}{...

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The 5 points you show are the only ones in the Sun-Earth system, there is no other set "with respect to Sun". The set of points seems to be Earth-centric, just because the points are closer to the smaller of the bodies and we choose Sun as the center of rotation here. You can take the exact same drawing and revolve it around Earth or around the common ...

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