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There is no orbit around the Earth that remains in permanent shadow. Such an orbit would need to have a period of one year, and that orbit would be too large to fit within the Earth's Hill Sphere, which represents a broad upper bound on the range of stable orbits around Earth.
In theory, an object near Earth-Sun L2 (at roughly 1.5 million km from Earth) ...
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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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One highly reusable observation is the following: For any plane going through the centre of mass, a satellite must spend some part of its orbit on one side of it if it spend some time on the other side.
And unfortunately, we can draw a plane where all the shadow is on one side:
The only loophole would be that the shadow revolves around over the course of a ...
answered Oct 8 at 19:10
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If you only know two points on the orbit, and the mass and position of the central body, all you can determine for certain is its orbital plane.
An example of two very different orbits that pass through a chosen $P_1$ and $P_2$
More geometrically: Given a focus $F_0$ and any two points $P_1$ and $P_2$, you can define a hyperbola with foci at $P_1$ and $...
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I believe the answer is yes, given one focus, and three coplanar points on the orbit, you can define the orbital shape, and obtain semi-major axis, orbital eccentricity, and the direction of the periapsis.
Since the three points must be coplanar within the orbital plane. I leave the exercise of converting the three coplanar points you have to orbital-plane ...
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Some notes on your approach, which I believe is a good one as a first order approximation.
Doing this I belive I'm overestimating the number of satellites
I would rather say it has to be an underestimate.
Circles do not perfectly tile. To achieve a 1-fold coverage, for instance, there must be some overlap, requiring $\frac{2\pi}{3\sqrt{3}} \approx 1.21$ ...
answered Oct 15 at 14:57
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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 ...
5
Optimal inclination for a long term LEO station serving as an intermediate stop to and from the Moon
Long term, and it must be in LEO?
There are many things to consider:
any LEO orbit's plane will drift, due to the shape of the Earth, so no orbit will stay aligned with the Moon's inclination.
You likely want a somewhat high-inclination orbit, so that your normal trajectory to the Moon arcs far north or south, thus avoiding the worst of the Van Allen belts.
...
4
As already pointed out in @Slarty's answer and other linked answers, 0 velocity (w.r.t Sun) is not required to hit the Sun. I did a search for trajectories that launch from Earth within the next year and fly by Jupiter (without hitting Jupiter) within the next 10 years before hitting the Sun within the next 12 years from now.
The most efficient trajectory ...
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The inner planets can be used for gravity assist to increase or decrease velocity. As an example in order to get directly to Mercury the delta V is around 12.5km/s or more. However if you are prepared to have a spacecraft shuttle around the inner solar system for 7 years or more then the delta V can be reduce to 4km/s
https://issfd.org/ISSFD_2014/...
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Optimal inclination for a long term LEO station serving as an intermediate stop to and from the Moon
The latitude of your launch site would be the optimal inclination, as the relative inclination between the LEO orbit and Moon's orbit matters very little for delta-v cost.
A lunar transfer orbit is very long and skinny, so at lunar encounter, the spacecraft's velocity (< 200m/s) is much less than the Moon's velocity (> 1km/s), which means that its ...
answered Oct 1 at 8:38
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This answer to How to solve the two-body problem in the ECI frame through numerical integration? says:
Instead of a single second order differential equation
$$\ddot{\mathbf{r}} = -\frac{\mu}{r^3}\mathbf{r}$$
We can solve the following pair of first order differential equations in parallel
$$\dot{\mathbf{v}} = -\frac{\mu}{r^3}\mathbf{r}$$
$$\dot{\mathbf{r}} ...
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$state=[rx,ry,rz,vx,vy,vz]$
$\dot{state}=[vx,vy,vz,ax,ay,az]$
$a=\dfrac{\mu}{r^2}$ (Newton's universal law of gravitation / two body acceleration in scalar form)
Division be zero errors are avoided by using the vector equation for acceleration, since any one of the position components can be equal to zero, but the norm will be a positive number (assuming the ...
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Partial answer because this is a hard question and because clarifying some of the terms may require more than the exchange of a few comments in this particular case. Kudos for taking on a challenge!
What "same orbit" might mean is hard to pin down
The term "same orbit" is really hard to pin down, different people use it for different ...
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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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Thanks everyone for your help and advice!
After some troubleshooting, I found out that the large position error of the "orbit propagator with J2 perturbation" is due to the bad initial position and velocity.
Apparently the initial position and velocity at TLE epoch time generated from the MATLAB Aerospace toolbox SGP4 is off by a few kilometres, ...
3
Perhaps you are looking for a relationship in the form of (or wondering whether such a relationship exists):
Nmin(A_cap,n) = μ(n) * (A_Earth/A_cap)
A_cap: area of an instantaneous coverage by a single satellite, constant in time (circular orbits), modelled as a spherical cap.
Nmin: smallest number of satellites in a practical constellation that can provide ...
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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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If two position vectors and the time between them are known, a conic section orbit may be determined using Lambert's method.
If the position and velocity relative to the observer are available
(as is the case with radar observations), these observational data can
be adjusted by the known position and velocity of the observer
relative to the attracting body ...
3
Jupiter's L2 point is 53.94 million km behind it.
The L2 point "orbits" the planet on a 1-Jupiter-year cycle, exactly
(just about by definition)
From there Jupiter is 0°8'55" wide, and the sun is 0°5'45" wide.
Jupiter casts a permanent shadow on its L2 point.
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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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You can use SPICE and JPL data. https://naif.jpl.nasa.gov/pub/naif/generic_kernels/spk/planets/
JPL publishes SPICE kernels of planetery ephemerides for these types of calculations. This plot is just for years 2040-2050, but their DE441 kernel has data all the way to year 17191. You can calculate the position and velocity vectors of Earth and Mars at any ...
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A burn is required to make up a difference in velocity.
In the case you have already figured out: The difference in velocity is between the velocity of the lower circular orbit, and the periapsis velocity of the transfer orbit (which you get from vis-viva).
In this case: The difference in velocity is between the higher circular orbit and the apoapsis ...
answered Oct 5 at 13:15
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A Hohmann transfer requires two burns.
You already appear to know how to calculate the burns going out. For coming in do exactly the same calculations, just reverse the order and signs on the burns.
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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)}$$
See Morrison 1882 for details about making your own ...
answered Oct 13 at 12:47
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I need two solutions: fastest and fuel-efficient.
There is actually one in the fuel-efficient case, which involves doing a gravity assist by Venus.
It has about the same delta-v cost as a regular Hohmann transfer to Mars, but can potentially be used when the angle isn't right since it has other requirements for planetary alignment.
Requiring 3 planets in ...
answered Oct 11 at 22:31
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Express the position of your craft in your orbit as an equation with respect to the parameter T (time).
Express the position of something else in its orbit also as an equation with respect to T.
Now you can derive an equation for the distance D between objects on these respective orbits as an equation in T. At the point T where the differential D' of this ...
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I'll adapt ai-solutions.com/_freeflyeruniversityguide/hohmann_transfer.htm:
I'll go from 20,000 km to 7,000 km
First find the velocity of the starting orbit:
$$\begin{align}
v_{starting} &= \sqrt{\mu \left({2\over r}-{1 \over a}\right)}\\
&= \sqrt{\mu \left({2\over 20000km}-{1 \over 20000km}\right)}\\
&=4.464 km/s\\
\end{align}
$$
(Unsurprisingly,...
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The rotation method uses a 3-1-3 Euler rotation: rotate by Z, then X, then Z again. For each rotation, one axis (in the new frame) is fixed. A quaternion based rotation requires finding a vector which is immobile between the initial frame and the final frame (see here for an excellent visualization and explanation of quaternions). One method to compute the ...
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