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In 2091 the Tesla Roadster is said to maybe get closer to Earth than the Moon is. At Starman's current speed, would its location in 2091 be sufficient for the Earth to re-attract the Starman to let it reenter the Earth's atmosphere? How close must the Tesla be in order to get attracted by Earth strong enough to fall on Earth? And as for falling onto other planets: is the critical distance determined by the distance according to the Earth's and their masses times the Earth's? Such as: Mars has about 0.11 the Earth's mass so the critical distance for Mars would be about 0.11 that of Earth for the Tesla to get attracted by Mars?

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    $\begingroup$ Roadster needs to be in contact with earth's atmosphere for a sustained period of time to slow it down sufficiently and make it reenter. Roadster's orbit has made it impossible to contact earth's atmosphere at all so it will never reenter. $\endgroup$ – user3528438 Apr 20 at 7:17
  • $\begingroup$ @user3528438 I don't think so. Just because it reached the Earth's escape velocity doesn't mean that it can't get re-attracted when close enough. Comets have even higher speeds in the vicinity of Earth, nonetheless there's the danger that a comet might get attracted by the Earth, thus posing a threat to it. $\endgroup$ – user35272 Apr 20 at 7:43
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    $\begingroup$ Comets impact the earth not because earth's gravity pulls it in, but it's orbit and trajectory intersects with the earth even without earth's gravity affecting it. $\endgroup$ – user3528438 Apr 20 at 8:13
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    $\begingroup$ @user30007 hang on it's a little more complicated than this; I'll post another answer in a bit. $\endgroup$ – uhoh Apr 20 at 8:43
  • $\begingroup$ Op you should check out how incredibly hard it is to get a satellite close to the sun. Much much harder than it is to fling it out of the solar system completely. People's general conception of how orbits work is very wrong - a common recommendation is that you should take a look at the game "kernel space program" as it's one of the best ways to gain an intuitive feel into how orbits work. $\endgroup$ – eps Apr 20 at 15:27
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According to Keplerian orbital mechanics, it will never come so close to earth "to get attracted by Earth strong enough to fall on Earth" (unless it will hit the earth directly).

Everything entering Earth's Hill Sphere from "outside" is on a hyperbolic orbit, so it is going to pass earth no matter how close it passes Earth (here also: except a "direct hit").

EXCEPT: You take into account Earth's atmosphere: Earths atmosphere can slow down an object.

This is not directly connected to Earth's mass but to atmosphere density. So for earth, the minimum altitude is about 80 km (depending on many factors). For mars it will be much lower due to less atmospheric density. For the moon, you can theoretically pass a millimeter above surface without touching it (assuming it's a perfect sphere).

EDIT: As mentioned in one comment:

Just one additional clarification: of course earths gravity affects the objects path, and of course more mass means the object is affected more. An object that collides with earth would not do so without first being affected by the gravity of earth or other celestial bodies. BUT the point is: objects from "outside" move in hyperbolic orbits. As long as the periapsis (closest point) does not hit earth nor it passes the dense part of the atmosphere, it will always pass.

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  • $\begingroup$ So why do asteroids like Apophis get a "x% chance of hitting the Earth"? I thought it's due to the question whether the Earth's gravity affects them enough to pull them towards it? $\endgroup$ – user35272 Apr 20 at 8:23
  • $\begingroup$ Yes: they are affected and Earth's gravity pulls them, that is why they pass earth on a hyberbolic orbit and not on a staight line. But still: as long as the closest part of this hyperboling orbit is not a direkt hit or within the earths atmosphere, it will pass. $\endgroup$ – CallMeTom Apr 20 at 8:28
  • $\begingroup$ And why do they get a percentage of a chance then? Is it simply because one can't foresee their path exactly? Like, maybe it will intersect with the Earth and maybe not? $\endgroup$ – user35272 Apr 20 at 8:56
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    $\begingroup$ Yes, it is as simple as you have said: It is impossible to measure anything exactly. Even when you take a measuring stick and measure something you will have a uncertainty of +-0,1mm left. In Orbital mechanics you will have uncertainties in position and velocity all three directions and uncertainties in the mass of your object and in solar activity and so on. Result: you can only say: object XY will hit earth with an x% chance. $\endgroup$ – CallMeTom Apr 20 at 9:06
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    $\begingroup$ @user30007 There's no probability involved in gravity. We can exactly calculate the path an object will follow given its location, velocity (speed and direction), and the time of measurement. The reason we give percent chances is because while our calculations are 100% reliable, our instruments aren't. You'll notice the x% chance gets updated from one day to the next; this is because astronomers are taking additional readings on the object and have a more precise idea of the trajectory it is on. $\endgroup$ – Mikkel Apr 20 at 19:32
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How close would the Tesla Roadster with Starman have to get to Earth in order to become attracted and fall on Earth?

tl;dr: If Roadster approached the Earth with $v_{inf}$ of 3000 m/s and an impact parameter of about 24,000 km from Earth's center (about 3.8 Earth radii), Earth's gravity would deflect it just enough for a grazing collision.

From Wikipedia's Hyperbolic Trajectory; Impact parameter and the distance of closest approach

$$r_p = -a(e-1) = \frac{GM}{v^2_{inf}} \left(\sqrt{1 + \left(\frac{bv^2_{inf}}{GM} \right)^2} - 1 \right) $$

See also What's the planetary exploration word for “impact parameter” (distance of closest approach if gravity were “turned off”)?

With Earths standard gravitational parameter $GM$ of 3.986E+14 m^3/s^2 and an approach velocity of 3000 m/s we can find out what impact parameter $b$ would result in a distance of closest approach $r_p$ equal to Earth's radius.

On earth when things fall into things they start from rest, or from a low speed.

"Falling into" isn't exactly the right way to think about moving bodies in the solar system because they are all generally going fast with respect to each other.

When Roadster was launched from Earth Elon Musk mis-tweeted that it had achieved 12 km^2/sec^2 excess C3 ("energy") relative to Earth after launch. He got the number wrong but the point is that once Roadster had this positive energy relative to Earth it could never "fall back" to Earth.

See Starman/Roadster in a=1.795 AU orbit, now what's the method to this madness? and this answer.

So Earth orbits the Sun at about 30 km/sec and when Roadster passes 1 AU from the Sun it will be going 33 km/sec (also around the Sun) in roughly the same direction, so if it was moving towards Earth it would slam into it at about 3 km/sec.

It's true that if it were going to miss the Earth by a diameter or two, then Earth's gravity could "pull it in" somewhat so that they still hit. We would probably call that a deflection rather than a "falling in".

Now the deflection could happen just before impact, or hundreds or thousands of years earlier! For more on that see

closest approach to Earth calc

import numpy as np
import matplotlib.pyplot as plt
vinf = 3000.
GM = 3.986E+14
Re = 6378137.

def r_closest(b):
    term = np.sqrt(1 + (b * vinf**2 / GM)**2)
    return (GM/vinf**2) * (term - 1.)

b = np.linspace(0, 10*Re, 101)[1:]

r = r_closest(b)

if True:
    plt.figure()
    plt.plot(b/Re, r/Re)
    plt.plot(b/Re, np.ones_like(b), '--k')
    plt.title('approach velocity (v_inf) 3000 m/s', fontsize=14)
    plt.xlabel('impact parameter (Earth radii)', fontsize=14)
    plt.ylabel('closest approach (Earth radii)', fontsize=14)
    plt.show()
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    $\begingroup$ Thank you. It's a pity that one cannot accept multiple answers on SE. I don't want to de-accept Tom's answer which is about as good as yours. Perhaps I should make clear only that one viewpoint of your answer is from a heliocentric reference frame while the other one (3 km/s or ~2 mi/s) is from a geocentric reference frame. $\endgroup$ – user35272 Apr 20 at 9:57
  • $\begingroup$ @user30007 yes "Roadster passed the Earth at about 3000 m/s and about 24,000 km from Earth's center" seems geocentric enough, but maybe "Earth orbits about 30 km/sec and... Roadster... will be going 33 km/sec in roughly the same direction" can have "around the Sun" added to it. $\endgroup$ – uhoh Apr 20 at 10:18
  • $\begingroup$ Do not worry my ego would cope with beeing de-accept for such an good answer^^. Actually I thnik uhohs answer is quite good showing/explainting how to get what I called a "direct hit". $\endgroup$ – CallMeTom Apr 20 at 10:24
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    $\begingroup$ @DrSheldon definitely we should wait before accepting answers, I do for days or weeks sometimes. Quick acceptance discourages others from posting more answers. $\endgroup$ – uhoh Apr 20 at 22:57
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    $\begingroup$ @DrSheldon It is good that I accepted it because now with three good answers I'd have a hard time which answer to accept. I would have to do a lottery or something. $\endgroup$ – user35272 Apr 21 at 5:38
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First of all, the 2091 date is based off an old set of orbital predictions, the next time it will be close to Earth is actually 2047. For more information, see the article I wrote for Ars Technica.

For it to orbit Earth, you would have to have a particularly close flyby of the Moon, which changed the orbital energy just right to orbit the Earth in a pretty unstable orbit. To hit Earth, it would have to be within about 100 km of Earth, close enough where the atmosphere would start to slow it down, and really more like 80 km.

I don't know exactly how close it would have to pass by the Moon, but I'm guessing really close if you want to do it in one go, or somewhat less close but more times.

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  • $\begingroup$ +1 for an authoritative and well-sourced answer! ;-) $\endgroup$ – uhoh Apr 20 at 22:45
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    $\begingroup$ Perhaps I should tell you that 2091 is the date at which the Tesla shall get closest to the Earth out of all predicted close encounters. I dunno whether it's an outdated value. $\endgroup$ – user35272 Apr 21 at 8:05
  • $\begingroup$ It is most indeed outdated. From memory, that is based off of the 6th set of ephemeris, but using the latest (10th), it does not appear. $\endgroup$ – PearsonArtPhoto Apr 21 at 9:58

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