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A spacecraft in a circular LEO orbits at ~7.7km/s. If it performs a retrograde burn and cancels this velocity (within the Earth reference frame) it will start falling towards the Earth center along a straight line and hit the Earth's surface.

A spacecraft in a circular orbit around the Sun at 1Au radius would orbit at ~29.78 km/s. If it performs a retrograde burn and cancels this velocity (within the Sun reference frame) it will start falling towards the Sun center along a straight line and, unless Venus or Mercury would happen to perturb its trajectory, will burn after getting close enough to the Sun's "surface".

The Solar System is traveling at an average speed of 230 km/s within its trajectory around the galactic center.

Let's imagine a spacecraft that is launched from Earth in direction opposite to Earth motion around the Sun, and performs a burn to cancel both velocities: a) due to Earth's spin and b) the ~29.78 km/s around the Sun, hence becoming "stationary" in the Solar system.

Straight after this the spacecraft performs another burn in direction opposite to direction of Solar system motion in the Milky Way galaxy (and I'm not sure what this direction is, as this related question remains unanswered) in order to cancel the galactic 230km/s speed, hence becoming "stationary" in the Milky Way galaxy reference frame (I know it's practically not possible yet).

Question(s):

  • Will the spacecraft in this case start falling towards the Galactic Center?

  • Will it reach the Galactic Center if happenned to be not perturbed heavily by nearby stars on its way? Or maybe motion in galactic scale is affected by the nearby stars so much that the perturbation effects can not be neglected?

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    $\begingroup$ This is an interesting exercise. Questions about the interstellar and galactic gravitational potential and its lumpiness are probably more likely to receive authoritative answers in Astronomy SE than they might here, and galactic trajectories might not even be on-topic here since they don't reflect aspects of Space Exploration normally discussed here. But let's see what happens! $\endgroup$ – uhoh Feb 2 at 6:17
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    $\begingroup$ @uhoh we already have 5 spacecrafts that are bound to the galactic gravitational laws, so I hope that if not the gravitational potential, than at least galactic trajectories might be loosely on topic here. And if not, might need to transfer to Astronomy. $\endgroup$ – Sergiy Lenzion Feb 2 at 6:42
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    $\begingroup$ Those spacecraft are not bound to the Sun, but the Sun's gravity is still by far the strongest force affecting their trajectory. All of them are still constantly slowing down with respect to the Sun, it's just that it's not enough to stop them completely. $\endgroup$ – uhoh Feb 2 at 6:50
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    $\begingroup$ The first burn is wasteful, you just need to cancel your galactic velocity, which would be more efficiently done in a single burn. The spacecraft would immediately receive a gravitation assist from Earth (briefly) and the Sun, so it would no longer be at rest WRT the the galactic centre.. $\endgroup$ – JCRM Feb 3 at 9:41
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    $\begingroup$ @uhoh Astronomy has ruled that questions about spacecraft are only allowed when they are about doing astronomy with a spacecraft, or questions about observing satellites' orbits. In this case, I don't think the question should be moved there. $\endgroup$ – called2voyage Feb 3 at 14:28
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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 thousand years. Almost any other effect on a spacecraft will be larger: The much-studied Pioneer anomaly is a factor of 100 larger.

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  • $\begingroup$ Good start; how does this compare with the acceleration towards a random star, say of 1 solar mass at 10 ly distance? $\endgroup$ – Russell Borogove Feb 2 at 16:38
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    $\begingroup$ Well at 1AU it's $6\times 10^{-3} ms^{-2}$ and 10ly is roughly 640000 AU. So we are looking at $1.5\times 10^{-14} ms^{-2}$ or about 1% of the overall galactic acceleration. Conveniently that means 1 solar mass at 1 ly matches the galaxy. $\endgroup$ – Steve Linton Feb 2 at 17:52

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