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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 ...
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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, ...
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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/...
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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 ...
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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 ...
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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-...
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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 ...
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A detective story, not an answer
I pulled the historical TLEs for the ISS from Space-Track.org. On 16 May 2016, it experienced a rev rollover. The TLEs around the rollover have rev's 99,989, 99,992, 1, 6
I then pulled up the historical TLEs for Explorer 7 in the time frame you suggested around 21 Aug 2019 - 27 Aug 2019. It includes the two TLEs you ...
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1) and 2) are easy to show, the bonus is very hard and I will not attempt it.
A $L$iberation point can be seen as a balance between three accelerations in a rotating frame of reference.
Gravity from $M_1$
Gravity from $M_2$
Centrifugal acceleration.
For $L_2$, the first two are $-\frac{(1 - \delta)M_1}{(R + r_2)^2}$ and $-\frac{M_2}{r_"^2}$ respectively. ...
answered Jan 22 at 12:13
SE - stop firing the good guys
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Given the radial distance $r$, velocity $v$, flight path angle $\gamma$ in radians, gravitational parameter $\mu$, and specific orbital energy $\epsilon$,
the specific relative angular momentum is
$$h= \|{\overrightarrow{r} \times \overrightarrow{v}}\| = rv\sin\left(\frac{\pi}{2}+ \gamma\right) =rv\cos\left( \gamma\right)$$
Then the orbital eccentricity ...
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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
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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 ...
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There are a couple of potential solutions to this problem. One would be to place the mirrors at a distance just a little closer to Mars than the L2 point and let the radiation pressure counter the weak Mars net gravity field. One doesn't have to be right in line either, but a "halo" orbit circling the Mars/Sun line would allow you to balance forces while ...
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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 ...
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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!
...
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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 ...
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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 ...
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