# Tag Info

37

You've hit on a really interesting question. To answer this, I'm going to look at JPL Horizons, using the center of the Earth and the center of the Moon as the distances provided. I'm going to look at each of the Apollo missions, with the time that they were orbiting the Moon, showing the max distance, with 10 minute increments included. All distances in kms,...

24

It doesn't really work that way. We can use the Sun to change direction, but we need rocket thrust to increase speed with the msneuver. To begin with, the closest stars (apart from the Sun) are not close. If we were somehow to reach escape velocity from the Solar System (which this method won't do, see below), we would still be moving at only a small ...

23

The "gravitational" (slingshot) maneuvers space probes are performing are actually not so much about gravity. The gravity is method to "tie" temporarily these two bodies, but you could (purely hypothetically of course) use something else, some superstrong tether or so ... "Slingshot maneuver" is in fact much better name in this regard. What actually happens ...

23

I found it. It was NOSS 3-1, a satellite pair. Found it through heavens above. The pair also explains the apparent tumbling. However, it seemed much brighter than 4.1, but it‘s definitely it.

22

To answer the question literally: you'd be looking for NASA Apollo Trajectory (NAT) data files. The report Apollo Mission 11, Trajectory Reconstruction and Postflight Analysis Volume 1 (PDF) provides a summary for Apollo 11 and mentions that the raw NAT data is available in Volume 2 of the report. I have yet to find Volume 2 though, perhaps because The ...

22

I believe the answer is yes, but just barely. The distance from the Earth to the moon varies significantly over time, from 356,400 to 406,700 km. I plugged the dates of orbital entry and departure for each of the lunar Apollo missions (8, 10-17) into pyephem to find the ranges of lunar distance. At Apollo 13's flyby, the moon was one day past apogee and ...

20

Movies are misleading. Space is enormous, and almost entirely empty. Even our ”asteroid belt” is mostly empty space; we have flown several missions straight through the belt to the planets beyond without hitting anything. Low Earth orbit has collected quite a bit of space junk over the years, but even so, interplanetary probes spend only a very short time ...

12

Approximately, yes. The gross gravitational effects on the trajectories of the spacecraft and the other object will be the same. The force of gravity between two objects is proportional to the product of their masses; by $F = m a$, the acceleration of each object cancels out its own mass ( $a = \frac {F} {m}$ ) and so depends on the mass of the other object....

12

I think the question is based on a misconception about how gravity assists work. If you just let yourself get pulled to a distant object then continue out the other side, the same gravity that attracted you to it will then begin pulling you back again. You'll just oscillate around it like a bouncing ball. Gravity assists work because the target itself (e.g....

9

This paper by Dan Adamo states that their fate is unknown. Other Apollo Program hardware certainly accompanied some of the components cited here into interplanetary space. Unfortunately, there are no empirical data relating to these objects' trajectories. Likely the largest such undocumented disposed components are four spacecraft/LM adapter (...

8

This question turns out to be surprisingly involved to answer. 1. "Brachistochrone" is a 17th-century term for a particular physics problem. The term appears to have originated in 1694[src] (or several years later, by source), with Johann Bernoulli, in the course of originating (or at least popularizing) the Brachistochrone Problem (whose solution is the ...

6

There are two major issues that I can see. Whatever you're using to calculate the arcsin is indeed giving you a value in degrees. The parameters you've provided are basically that of a satellite racing by an object ad a moderate distance, moving far above escape velocity all along its trajectory. It neither gets close enough, nor hangs around long enough ...

6

The rotating reference frame is usually preferable over an inertial reference frame when analyzing three-body motion. Two main reasons for this: The motion is determined by the two primary bodies, so it makes sense to use these to define the reference system The typical three-body dynamics for which you would use such a reference system, such as Libration ...

6

What the heck are "space-fixed coordinates"? To what in space can a coordinate system be fixed? That's two questions. The answer to the first is that those are Earth-centered inertial or Moon-centered inertial coordinates as indicated by the "Ref. body" column. Look at the velocities. 25600 ft/sec is orbital velocity for a vehicle in low Earth orbit while ...

6

In constrast to the other answers here, I would like to point out that the Oberth Effect does allow you to gain kinetic energy from a gravity well without needing to rob it of momentum ... if you fire your rocket motor at periapsis. Atomic Rockets has an excellent discussion of what the Oberth effect can do. So by burning 6 km/s of Δv, you get an actual ...

4

The acceleration due to gravity will be identical regardless of mass, assuming the mass of your spacecraft is negligible compared to mass of the object you're orbiting. For example the Earths moon is large enough to effect the motion of the earth so it doesn't orbit the centre of the earth, but instead it orbits the shared centre of mass of the Earth and ...

4

An estimate would be from trajectory simulation, but that doesn't seem satisfying... Why doesn't that seem very satisfying? That is exactly what was done, many times over. From the perspective of the New Horizons operations team, they didn't need to know where Pluto or Arrokoth were 100 years ago, or where they will be 100 years from now. They needed to ...

4

Let's go through this one thing at a time. First off, your value for $\delta$ is clearly in degrees. With an eccentricity of $200$, the reciprocal $1/e$ is so small it's nearly equal to it's own inverse sine -- in radians. $\delta$ is basically $2/200$ radians $=0.573°$. Next, realize that geometrically, the eccentricity of a hyperbola equals the center ...

3

It seems like you want to generate some Monte Carlo trajectories. I would recommend a tool which is designed for Monte Carlo analyzes: such a tool would allow you to write a script which quickly changes the inputs and run the scenario without any human intervention. nyx: a mission design, orbit determination and Monte-Carlo tool Important disclaimer: I am ...

3

Some good practices I'm aware of are: As you mentioned, maneuvers are simulated before they are commanded and their effect is evaluated on ground so that thruster parameters and tank filling are updated, so if anything funny is happening during maneuvers this can be identified. If propulsion is electric (which is still not so common), then thrusting is ...

3

The New Horizons mission tutorial you refer to may come from our blog here. You'd have to do something similar. Astrogators Guild Blog For multiple flyby missions, Marty Ozimek and Justin Atchinson at APL did a great job of describing a similar problem with multiple flybys here: APL Example

3

Partial answer: And also, is the three-body problem theory only used to solve for periodic orbits around Lagrange points or are there any other implications? The CR3BP or CRTBP or Circular Restricted Three Body Problem assumes circular orbits of two massive bodies around their common center of mass, and a third massless body that responds to their ...

3

Partial answer so far... For the trajectories in your drawing the objects will miss the Moon. They pass much closer to Earth, so their orbital motion will be substantially faster than the Moon's so they will pass through the interception points days before the Moon does and definitely miss it. However if those are elliptical orbits with the same semi-...

3

The four pieces of the SLA fairing would follow a similar path as the third stage S-IVB of the Saturn V. The SLA pieces were separated by explosive devices only, so their additional acceleration was very, very small. See apollomaniacs. After Wikipedia, the S-IVB stages of the missions 8 to 12 are in heliocentric orbits now and of the missions 13 to 17 were ...

3

Only the mission operations team for each mission can answer that question. During an interplanetary mission, spacecraft perform Trajectory Correction Maneuvers (TCMs). These are used to better target the B-Plane target (as explained in the documentation for FreeFlyer). The selection of the B-Plane depends on many factors, such as the thrust capability of ...

2

Yes! The good folks at AGI (Makers of STK) simulated the entire Apollo 11 trajectory in STK: "Based on original NASA documentation, our engineers have built a high-fidelity, eight-day digital mission in Systems Tool Kit (STK). They modeled precise trajectories using STK Astrogator" The web site has tutorials on how to do some of the simulation using STK/...

2

What the other answers fail to mention is that the mass of your orbiting object actually cancels out. It does not matter. See these two equations: (1) F1 = F2 = G*m1*m2 / r^2 (2) F1 = m1 * a1 Where F is force, G is the universal gravitational constant, m is mass, and r is distance between centers of mass of the orbiting and orbited bodies in question. The ...

2

In-the-sky.org doesn't show anything that quite matches your specification. There was an H-2A rocket body (i.e. upper stage) overhead at 7:46 going from NE to ESE; there was an SL-16 (Zenit) rocket body a couple of minutes later going ESE to N. Tumbling rocket bodies can vary considerably in brightness.

2

Earth Satellites: get TLEs from Celestrak or Space-Track and propagate them with SGP4. One way to do that and to also calculate their ground tracks (latitude and longitude) is the Python package Skyfield but there may be others. However I'm not aware of proper databases that you can download constantly updated positions already calculated each second. Things ...

2

The distance from the focus of attraction of an orbit can be expressed as a function of the true anomaly (angle) given by $r(\theta)=a\frac{1-e^2}{1+ecos(\theta)}$, where $a$ is the semi-major axis and $e$ is the eccentricity.

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