New answers tagged orbital-mechanics
1
Reference 1, citing Reference 2, reports that hypothetical trojan-type asteroids are invariably unstable at the proposed Mercury L4 and L5 points, whereas stable Lagrange-point librations at least over millions of years are available at the L4/L5 points of both Venus and Earth(+Moon). The predictions were made in the 1990s. The discovery of Earth and Venus ...
5
Dangerous asteroids are those that can hit the Earth, and are large enough to cause substantial damage. There are currently no such known asteroids. (2020-02-21)
There are two ways an asteroid could end up as considered dangerous:
We discover it. There may be an asteroid bound for Earth at this moment, we just haven't seen it yet. This is fairly straight ...
answered Feb 21 at 11:57
SE - stop firing the good guys
18.3k1 gold badge63 silver badges125 bronze badges
1
A "classical" fairing is a ballistic object, meaning it has no active controls. You'll find that trying to model its tumbling path through a highly variable atmosphere exceeds our current computational capacity.
If you're thinking about the SpaceX fairing recovery techniques, be aware that there are some active controls, as explained in this space.SE ...
2
A colleague of mine from years ago, Jan King, told me that he was a consultant for Iridium in their early planning stages and that their original plan was to place the spacecraft in a precisely 90 deg inclined orbit. This has some special properties in terms of orbit perturbations because it means that most of the biggest gravity irregularities like earth's ...
3
Drag has a lot of more complex factors that would require some simulation to determine exact values, but the simplification that is often used shows drag increasing linearly with air density and with the square of velocity.
Density at the Karman line is 1/2,200,000 that of ground level, so an increase from 0.5 mach to 7 km/s will feel roughly $\frac{(7000/...
1
On the down side, at 2.77 times the distance from the Sun as Earth, the orbital speed of Ceres is $\sqrt{1 / 2.77} = 0.6$ of Earth's orbital speed.
But on the up side, it's also starting higher up in the Sun's gravitational potential so it needs less delta-v to get to Jupiter, and as you point out, Ceres has a far lower escape velocity, which also helps!
...
2
I received a definitive answer from an expert on TLEs.
When the rev number goes above 99,999 you should simply drop the leftmost digit and write the remaining digits into the TLE. For example:
Rev # Digits in line 2, columns 64-68
------- -------------------------------
100,000 00000
156,287 56287
395,468 95468
1
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 ...
3
In general, there are six directions you can burn. You can burn along the path of the orbit (prograde), you can burn in the opposite direction (retrograde), you can burn at toward the center (radian-in) or away from the center (radial-out), you can burn normal to the orbits plane or in the opposite direction (anti-normal) and all combinations thereof.
Thus, ...
1
First rule of orbital dynamics: Whatever you do in orbit will affect the point at the opposite side of the orbit and leave the current point unaffected.
If you accelerate at the perigee, you change the position of the apogee. Respectively, if you accelerate at the apogee, you change the position of the perigee.
One well-known example of this is the Hohmann ...
3
A numerical integration is the best option for me.
Using the JPL’s DE/LE438 ephemerides and the NAIF’s SPICE library, I found the dates when the angle between the Earth-Sun and the Earth-Moon vectors is about 180, 0 and 90 degrees and the Moon/Earth distance is about the same:
Date Angle Distance
2000-11-12 175.54 371281
2008-03-08 4.25 ...
3
This is a supplementary answer with calculations (for 1) two bodies and 2) three bodies) problem which confirms that @MatthewWells answer is correct.
Would the Moon hit the Earth or not?
Yes, it would hit the Earth.
Do we have to do detailed numerical integration to find the answer, or can we use some simple equations that involve energy and/or ...
2
If a satellite in GEO is let drifting and its position is not continuously corrected, its inclination will start to change. The main cause of the perturbation is the influence of Moon and Sun.
Orbits with a positive RAAN tend to have an increasing inclination, while those with a negative RAAN have a decreasing inclination. For reference, refer to figure 2 ...
2
This is easier than it sounds, provided we choose a suitable reference frame. In this case, we'll to consider the problem from the (doomed) perspective of someone standing on Earth's surface directly below the Moon. Since the question specifies that relative velocity between the bodies is zero, we can construct a free-body diagram, treating the gravity of ...
3
If the orbiting body's mass is a significant fraction of the central body's mass, the weak stability boundaries can be more dramatic.
Call the mass of the central body + orbiting body 1. Call the orbiting body's mass µ. Then the central body would have mass 1-µ.
Here are pairs arranged in order of µ
Pluto/Charon 1.043E-01
Earth/Moon 1.216E-...
3
This is all about gravitational maneuvers.
They allow to obtain huge accelerations/deceleration/velocity changes almost without using any energy. More heavy moons - more opportunities for maneuvers.
General idea is that if satellite trajectory at some point goes near heavy body (one of moons), by very small early adjustments from long distance before ...
5
The period of Earth's orbit around the galactic center is about 225-250MYr or about $750 \times 10^{15}$ seconds. The peak velocity is $230 \times 10^3$ m/s. So the central acceleration is:
$$a = 2 \pi v / T = 2 \times 10^{-12} \rm{m/s/s}$$
That's a very, very small acceleration. To put it in perhaps more understandable units, it's about 6 m/s per ...
Top 50 recent answers are included
