55
Yes.
1st scenario: A spacecraft orbiting the Sun at Earth distance vs. Pluto distance, shedding its orbital velocity
The orbital velocity decreases with distance, according to the following formula, where $r$ is the orbital radius, and $\mu$ is the mass parameter (it's just a shorthand we use)
$$v_{circular} = \sqrt{\frac{\mu}{r}}$$
The orbital velocity ...
19
Why zero excess velocity? Well, with almost zero excess velocity you can stay near Earth, but not too near. For example, the Spitzer space telescope did this to communicate with Earth while avoiding radiant heat from Earth. It's been drifting away, but slowly enough that other factors first reduced its effectiveness.
12
Why would one want to choose zero excess velocity upon escape?
If you aren't in a great hurry, and you have a small delta-V budget, and you want to visit the Trojan points L4 or L5, you can do so by getting just outside of Earth's sphere of influence, then lowering or raising your solar orbit very slightly to get ahead of or fall behind Earth. You wouldn't ...
10
If your jump is perfectly nadir pointing, and all perturbing forces are neglected so we can consider both your orbit and the ISS orbit as perfect two-body orbits, you will meet again after one full orbit, as a low-velocity impulse that is perfectly radial does not affect the orbital period.
From your perspective, relative to the ISS, you will move downward ...
6
You are describing the Local Vertical Local Horizontal (LVLH) frame of reference.
It is used to describe the orientation of the spacecraft in relation to the Earth's surface.
For example, if you wished to point an instrument at the point on the Earth directly below the spacecraft, the craft would fly in a constant LVLH attitude. But its inertial attitude ...
6
The FAQ page from the planetary society said that this is because the craft is not very precise with its pointing.
LightSail 2's attitude control system does not have the precision to maintain a circular orbit. Therefore, as one side of the spacecraft's orbit rises, the other side will dip lower, until atmospheric drag overcomes the forces of solar ...
5
If "tangential velocity" is not tangental to the orbit, what else can it be tangential to? I studied orbital mechanics in Howard Curtis' Orbital Mechanics for Engineering Students, and while I don't claim that it holds absolute truth, it uses the same definition as the diagram you posted (but less clutter):
(Figure 2.8 from H. Curtis, Orbital Mechanics for ...
5
The Oberth effect doesn't give you more delta-V. It gives you a larger change in specific orbital energy for the delta-v you decide to expend because of where you decide to expend it. Spending delta-v in a burn parallel to the direction of your velocity when you're moving fastest gives you the largest change in specific orbital energy.
Specific Orbital ...
5
Parabolic escape trajectory is only theoretical, it only "works" in a two body system, and in a two body system "escape" is a meaningless practice anyways.
In a multi body system the forces from other bodies, especially around the edge of the gravity well, make parabolic escape impossible: before you'd have zero velocity the other body would've already ...
4
I can't think of one. For mission planning there is nothing special about parabolic velocity. There is something special from the perspective of teaching orbital mechanics as it is the boundary between closed and open orbits, but from a practical mission point of view it looks just like a very long ellipse or a barely open hyperbola for many, many years. ...
4
The cited article is talking about a 1984 idea by Robert Forward. (See this English summary). In it, he proposed that solar pressure could be used to move an geostationary satellite a few 10’s of kilometers north or south of the usual geostationary orbit. These are more generally called “statites”.
There’s really nothing here that points to much larger ...
4
Considering only two bodies, if the closest approach of the two bodies is not too close, an elliptical orbit will remain perfectly* stable. An intuitive explanation for this that might help you is that as the bodies approach closer, they move faster, so they spend less time under the higher gravitational influence, and when they're further apart, they're ...
4
But how do probes know exactly by how much to reduce their velocity and by how many degrees to deviate to ensure they hit the right position in the Martian atmosphere for precise entry?
Most of JPL's interplanetary probes don't "know" where they are. What they do "know" what time it is and where they are pointing. It's combination of people and equipment ...
4
This relates to the concept of escape velocity, which is the speed needed to escape an object's gravity well. Even though gravity has an infinite reach, the fact that its force reduces quadratically with distance means that it only takes a finite amount of energy to climb out of a planet's gravity well. For some fixed energy cost, you can get arbitrarily far ...
4
Physically, you can think of a sun-synchronous orbit as having its orbital plane precessing once a year. Like a toy gyroscope’s plane of rotation precesses under the torque due to gravity, so does the satellite’s plane due to the torque of being attracted to the tidal bulge.
Now think of a prolate Earth as having a “negative tidal bulge” (to first order, ...
3
You seem well aware that there is a reference implementation for SGP4 in Matlab. Yet you choose to make your own. I've sadly been there, and done that.
Your first problem is to compute XKE, but from the reference you've posted on page 88:
And on page 89:
Since G = 6.67408 × 10-11 m3 kg-1 s-2, GM has units of length³/time². Convert these units to Earth ...
3
If you look at this delta-v map of the solar system, you can see that to get from the surface of the Moon to Neptune transfer requires about 7.67 km/s of delta-v.
To get to Neptune from the surface of Mars requires 10.56 km/s of delta-v. So even if you were magically transported to Mars first, it would still take more energy to get to Neptune than it would ...
3
I think each person will visualise this problem differently. Personally, whilst formal papers describing separation strategies can be interesting I often find that they miss out on the basics (though I couldn't comment on the article you mention).
Here is a simple explanation that applies between any pair of orbits but is particularly used at GEO, ...
3
Neither.
In the old times, geostationary satellites had their longitude windows, each with around 1°, and tried to keep themselves at the center os this window. This strategy was good enough for its time, when few satellites with poor attitude controls were located in this orbit.
Nowadays, geostationary spacecraft are usually in a "collocation window", ...
3
As everyone else has mentioned, there doesn't seem to be a mission for a true parabolic escape, especially since an exact parabolic trajectory is a target of zero size and therefore there is a zero chance of hitting it exactly.
Also, a true parabolic orbit only makes sense in a two-body model. Once you consider the gravity of anything else, the orbit ...
2
I think your problem is just a matter of terminology. Almost all of the orbits you hear about are elliptical. Some are almost exactly circular, but that counts as a special case of elliptical. Orbits like that keep going round and round, and stay within a certain distance of the primary. You might have the impression that's part of the meaning of "orbit".
...
2
note: The question was modified while I was writing this answer, so I will have update(d) this (with a new) answer with more math shortly finally, here!
So let's think about what an orbit is. One way to think about it is a combination of moving forward while falling down. At each moment the object keeps moving forward, but it also accelerates towards the ...
2
Since you have all three positions in the same frame (in this case ECI but it doesn't matter which one for the purposes of your question), you calculate the angle the same way I did in this answer.
The cosine of the angle between any two unit vectors $\mathbf{\hat{a}}$ and $\mathbf{\hat{b}}$ is their dot product:
$$ \cos(\theta)=\mathbf{\hat{a}} \cdot \...
1
I like to play with the notion of cyclers. One type of cyclers is a circular orbit crossing an elliptical orbit, both with the same period. On this comic book page the Cheng Ho is such a cycler.
The geometry of orbits about the sun and around the earth are the same so I'll leave the labels at A.U. and years.
Here for simplicity's sake I set both the semi ...
1
Okay, I had said that I would post another answer based on your clarification, but then I got temporarily distracted. Here it goes now.
Here is just what you've asked for; How to calculate the radii of high orbits above the parallels of 45 and 80 degrees and then some even higher ones!
Just fyi I have confirmed by a simple 3D orbit simulation that this ...
1
If the aim is to hover a solar powered ion satellite the basic numbers come out as:
Weight for 1KW of solar panels = 5.33kg
Force of gravity at at 400km up = 8.6 Newtons per Kg
Mass 5.33 kg so that = 46 Newtons of gravitational force
Thrust of a 0 mass Xenon engine per KW = 0.034 Newtons
So pretty clearly it will not be possible to make something that ...
1
Partial answer. Here's what I have so far. I use Python instead of Matlab, and I "roll my own" dot products, but these plots do look a lot like your plots! I think there might be a missing "2" in the expression for angle, but now that you mentioned you're using 1550 nm light we seem to agree on the size of the Doppler shift though there is still a sign ...
1
If $\vec{R}_1$ and $\vec{R}_2$ are the positions of the two satellites in any consistent coordinate system, including an Earth-centered one, then $\vec{R}_1 - \vec{R}_2$ is what you’re looking for: the vector distance between them. You don’t need to worry about the motion of one vs the other, just how that vector lengthens/contracts and rotates.
For same-...
1
I'm not an expert on the subject, but here's an analysis based on basic physics.
Since you've used 2D diagrams where it looks like the two orbits are in the same plane, I'll stick with that as well, but remember that orbits are 3D and you'll need to calculate the radial and perpendicular velocities using the 3D velocity vectors of each satellite.
The ...
1
I can't identify the object, only describe it: A comet that is orbiting retrograde, as short a period as you can find.
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