# Tag Info

42

Retrograde orbits have multiple use-cases. First of all you should note that "retrograde" doesn't mean 180° inclination - everything > 90° is considered retrograde. This places all sun-synchronous satellites which operate at about 98° inclination in retrograde orbits. The usefulness of sun-synchronous orbits should be obvious. Retrograde orbits of course ...

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The center of the Earth is, for any reasonable approximation, in one of the focus points of an elliptical orbit. For a circular orbit, there is only one focus point, so the center of the Earth is in the center of the orbit. The plane of the orbit thus would intersect both the center of the Earth as well as the launching site. If the launch site was on the ...

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Because the lunar landings happened at some latitude, the landing sites were subject to longitudinal drift due to the Moon's rotation around its own axis. Due to the small latitude of the first landing[1], less than 1 degree for Apollo 11, and the short stay on the Moon, The LM lunar landing site did not drift far from the CSM orbital plane. Hence, Apollo ...

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The short answer is: The $51.6^\circ$ inclination of the ISS's orbit is measured from the equator, not from the latitude of the launch site (for KSC, $28.6^\circ$). Intuitively, as the orbit moves toward a higher latitude, the angle it makes with those lines of latitude decreases, until it is parallel with them, then tilts negative, heading to lower ...

13

The short answer is that a spacecraft is attracted to the center point of the earth, not to the earth's rotational axis. [I]t would make sense to me that launching east would result in a 0° inclination with the orbital plane raised so it's parallel to the equator but above or below it. Here's one explanation of why that wouldn't happen that you might ...

10

The latitude of a launch site determines the minimum inclination that can be directly reached; launching from 28.5º latitude in the due-East direction achieves a 28.5º orbital inclination. Launching to any higher inclination is straightforward, simply by steering continuously North-of-East during the ascent. In the extreme case, if you fly due North, you ...

7

Earth's gravity pulls you towards the centre of the Earth, so if you're above Kennedy, that pull has a Southwards component, as well as the component towards the Earth's axis. So your path curves South, so that in the end the orbit spends equal amounts of time North and South of the equator, and the pulls in that direction balance out over time. All orbits ...

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A retrograde orbit means that you're moving over the surface quite fast, which would probably induce some motion blur in photos and reduce the window of time for which the satellite is over a given target. Note that (for low Earth orbit), orbital speeds are typically somewhat above 7000 m/s relative to the center of the Earth. At the equator (which ...

6

In the case of the space shuttle: These two images show that the flame trenches at KSC point due south, and that the Orbiter's tail when installed on the pad also points due south. (source: Google Maps) (source: NASA) Therefore if the shuttle launched and went into a gravity turn in the desired heads-down attitude without rolling at all, it would have ...

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My lightly-edited question The strange orbit of Ofeq 11 - how does it (actually) do this? has become at least a supplemental answer to @Polygnome's. I assume the retrograde orbit ($\theta_{inc}=142°$) is a side-effect of geography - ocean overfly access only lies West of the secured launch site - (The Spaceflight 101 article Israel Launches Advanced ...

5

UCS publishes a database (text and excel) of "active satellites" which have publicly available data. It directly lists inclination for each, so analyzing them from that source would be pretty easy. Pixalytics published a summary using that database that showed inclination for just the 63% of satellites that are in LEO orbits. From that summary: 57.5% sun-...

3

I know this is quite an old post, but might be worth having a read on this paper "Efficient Orbit Propagation of Orbital Elements Using Modified Chebyshev Picard Iteration Method" by J.L. Read1, A. Bani Younes2 and J.L. Junkins3 (http://www.techscience.com/doi/10.3970/cmes.2016.111.065.pdf) Equation (7) is your inclination change. Additionally, citing ...

3

Some candidates: The Deep Impact spacecraft's impactor, matching Tempel 1's inclination of 10.47 degrees upon collision. The Dawn spacecraft, matching Ceres' inclination of 10.59 degrees. The NEAR Shoemaker spacecraft, matching 433 Eros' inclination of 10.83 degrees upon landing. Solar Orbiter is currently beating all of them, at ~14 degrees.

3

Reductio ad absurdum If you could choose freely on which circle to orbit, the most convenient place to take off from would be the North pole. That would set the circle diameter to zero. You would then climb to whichever altitude you pleased and remain there, immobile in space, for as long as you wanted. How cool would that be? An attempt at analogy In ...

2

If i read your quote correctly, it doesn't state that a burn at the inertial y-axis is cheaper than a burn at the LON, but it explains that your satellite will encounter an acceleration towards north at +y and south at -y, so your LON will in fact line up with the y-axis after a while because of the inclination change due to this perturbation, so it's still ...

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Here is a rough description, it doesn't give you the exact answer but it does provide the bare minimum conceptual understanding that you need to make a start. TLDR: scroll down to "Key points" to get to the point quickly. All non-inclination controlled geosynchronous objects, starting off at inclination = 0, exhibit an increase in inclination from 0deg to ...

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The proposed location is actually further west than the map above - on the A'Mhoine peninsular in the Sutherland area of Scotland. Therefore the range of easterly launch angles is slightly increased.

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There is no maximum inclination for that. As you point out, and as "Fundamentals of Astrodynamics" by Bale et al. confirms in section 3.2.1, an inclined geosynchronous orbit traces out an analemma over a sidereal day. An equatorial geosynchronous orbit then simply traces out an analemma of height and width zero, i.e. a point-shaped one. At e.g. inclination ...

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