New answers tagged orbital-mechanics
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How to determine the delta V required in order for the satellite at pt. A to hit the earth at pt. B?
I interpret this as "minimum delta-v as a function of an arbitrary elliptic orbit and an arbitrary point on the surface".
Furthermore, as a pure orbital mechanics problem, I assume the Earth to be perfectly spherical and ignore the atmosphere. However, I will still allow the Earth to rotate.
The general case can always be solved by increasing ...
answered Oct 27 at 8:50
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2
How to determine the delta V required in order for the satellite at pt. A to hit the earth at pt. B?
The answer is much depending on how realistic you try to model "your world" and how small your error margin (are you trying to hit a "POINT" or an "AREA"?) is. Do you need the solution for minimum delta V or "a possible" delta V?
Most simple: reduce it to a 2D scenario with no atmospheric drag - simple hohmann transfer ...
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$$v_{in} = v_{out}$$
That's the rule when entering a single body system. You are going to leave with exactly as much velocity as you entered with. There are two notable exceptions:
More than one body in the system. The flybys of both bodies can very well change how much velocity you leave with. Earth-Moon is such a system. Venus is not.
Very small ...
answered Oct 21 at 15:23
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29
The diagram you show is the digital version of a drawing by someone with an Etch-a-Sketch: completely inaccurate. The diagram below is accurate, showing Pioneer 10 & 11 and Voyager 1 & 2 trajectories in a heliocentric, inertial reference frame, of course with the ecliptic N-S dimension collapsed. No retrograde, no dog-legs between planets.
Every now ...
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Converting between different frames of reference is pretty straight forward.
You subtract the velocities. That's it.
If you have the spacecraft's velocity relative to the Earth, you can just subtract the Moon's velocity relative to Earth. Then you have the spacecraft's velocity relative to the Moon.
In the case of a lunar transfer orbit, the spacecraft is ...
answered Oct 20 at 18:17
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0
The Lagrangian points are gravitationally so close to each other that "free" transfers between them, and nearby orbits, exist.
Both "transfers between libration point orbits" and "libration point gateways" refer to this fact.
The term "libration point gateways" is rare enough that you would have to presisely define it ...
answered Oct 20 at 8:48
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1
While the overall scope is too broad, let me address the "low energy transfer" part.
Firstly, the idea of "gradually increasing apogee" doesn't save you any fuel. Chandrayaan-2 did that because of limited thrust. The only saving to be had here is picking an engine with a slightly lower mass.
Secondly, a "ballistic capture", as ...
answered Oct 20 at 8:36
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3
Transfer
Regardless of how escape from the Earth system is achieved, orbital mechanics pose some restrictions on travel time.
The minimum velocity transfer possible is an elliptical transfer orbit touching Earth's orbit when closest to the Sun, and Jupiter's orbit when farthest from the Sun. The transfer time is then 940 to 1060 days, depending on where in ...
answered Oct 20 at 7:32
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5
As $\Delta v$ is just change in velocity, we can just integrate the norm of the acceleration function over time:
$$\Delta v = \int|\mathbf{a}(t)| dt$$
You're out of luck getting a closed form of that integral though.
As far as analytical solutions goes, we can note that at $t = \frac{\pi}{2}$, all of $a_x$, $a_y$ and $a_z$ are maxed out, and hence $\Delta v &...
answered Oct 19 at 8:22
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2
The calculation uses the following model for "total propulsive delta-v":
$$\Delta v_{total} = \Delta v_{spacecraft} + \Delta v_{launcher}$$
Here, $\Delta v_{spacecraft}$ is what propulsive capabilities the probe has by itself after leaving the Earth system entirely, and is presumed to be a known value that can be looked up.
$\Delta v_{launcher}$ is ...
answered Oct 18 at 9:14
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1
This is a late answer; a closely related question was recently closed as a duplicate of this.
Does mass of orbiting body affect the orbital speed?
tl;dr: Yes it always does, about half as much. If it's small, like one millionth the mass of the primary, the change in speed is one half of one millionth for example. In the extreme case when the two masses are ...
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This is a late answer; a closely related question was recently closed as a duplicate of this.
Does trajectory of an object orbiting a planet depend on the object's mass?
Yes, it does.
Several of the answers correctly invoke the principle of the universality of free fall, which dictates that the acceleration from the perspective of an inertial frame of ...
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Does mass of orbiting body affect the orbital speed?
tl;dr: Yes it always does, about half as much. If it's small, like one millionth the mass of the primary, the change in speed is one half of one millionth for example. In the extreme case when the two masses are equal though the trend breaks down and the speed is now 70.7% ($\sqrt{1/2}$ ) rather than half....
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Your user need to supply you with information about mean anomaly at fixed time (epoch) $M(t_0)$, and then mean anomaly at the moment $t_1$. Or about known mean motion $n$ instead of current mean anomaly $M(t_1)$. You can also get the mean motion as square root of gravitational parameter $\mu$ divided by cube of semi-major axis, and also as $2\pi$ divided by ...
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You don't have enough information. You need a minimum of one more value to determine the true anomaly such as the time of perihelion passage.
In general six values and a timestamp are needed to fully specify a Keplerian orbit. The reason six values and a timestamp are needed in the general case is that the two body problem is a second order differential ...
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While it's a very loose lower bound, it does perhaps have some value to present one most trivial such bound.
With basis in the fact that under no circumstances there exists any more efficient way to increase apoapsis than a prograde burn at periapsis, the following bound exists:
$$\Delta v \geq \sqrt{\frac{2}{r_{P1}} - \frac{2}{r_{P1} + r_{A2}}} - \sqrt{\...
answered Oct 5 at 14:54
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2
I had read many years ago in one book the rough empirical rule for orbital maneuvering and rendezvous: suppose that two spaceships follow the same traectory (circular) with distance between them, say, 100 km, and the second ship try do intercept the first one in the one orbital period. The second ship need to change orbit from circular to elliptical with the ...
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Different methods of low thrust optimization are discussed in Chapter 2 of "Low Thrust Trajectory Optimization in Cislunar and Translunar space", a dissertation published in 2018 by Dr. Parrish.
In short, there is no analytical solution. There are several methods for low thrust optimization depending on the problem you'd like to solve. For orbits ...
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For more complicated scenarios like these, you probably need to use GMAT's Python interface.
I would recommend the following:
Setup both spacecraft to be propagated together: I'm not sure this is available in the GUI, but in the script, change the Propagate statement to Propagate Synchronized (cf. the docs).
Instead of propagating for the whole orbit, ...
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It's true that you could calculate how much a spacecraft would have to slow down by to enter a spiral decent for a reentry from frist principles.
A "spiral" descent isn't an effect of gravity alone (things don't orbit in spirals), but an rather an effect of atmospheric drag.
As such, the part to calculate is really "how do we get low enough to ...
answered Oct 3 at 13:44
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Was Queqiao in a halo or Lissajous orbit?
It is hard to say which, or even yes or no, because there are so many kinds of Lagrange point orbits possible that can be maintained with a little help from propulsive station keeping, and a specific, definitive answer would require either plenty of orbital tracking data or more information from the spacecraft's ...
answered Oct 2 at 9:00
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