# How close an orbit could you get to Earth with a planet-sized object/ship without severely disrupting Earth's orbit?

My idea involves an alien spaceship inside a roughly Earth to Mars sized/massed planet, but the whole planet is displaced when the drive is activated. If you wanted to travel back to our solar system but didn't want to destroy Earth or send it out of orbit, is there a temporary orbital path that wouldn't throw our solar system's orbital mechanics out of whack? A general idea would be sufficient. Ideally, it would:

1. Be close enough that 23rd century ships (specs to be decided but not super-tech) could fly back and forth in a time-frame of a few months.
2. Allow communication back and forth that was in hours or less (at least initially).
3. The closer and better matched the orbit, the shorter the duration of the orbit would need to be. (if we have to get weird and exotic, we could probably do a rendezvous between planets at the front end, then again after a year to 6 months.
4. We can have almost perfect control of the starting position of the planet as well as direction and velocity. However,it would be very hard to alter the trajectory of the planet once it arrived.
5. Once the story goal is achieved, the planet would be displaced to elsewhere and solar mechanics would need to be able to resume something close to normal orbits.
• I suppose that depends a lot on the time frame. Within astronomical time frames even little perturbations can have significant impact. – Peter - Reinstate Monica Apr 8 at 15:33
• Here I was hoping someone would bring up horseshoe orbits. Sigh. – candied_orange Apr 9 at 6:37

Not that far, honestly.

## Newton's Law of Gravitation is Your Friend

$$F = \frac{G * m_1 * m_2}{d^2}$$

So the moon has a pretty substantial effect on the Earth, jostling us about as it whips around us. Its mass is $$7.4 * 10^{22} kg$$, while Earth's mass is $$6.0 * 10^{24} kg$$. We're about (on average) $$3.8 * 10^8 m$$ from the moon, so plugging in the gravitational constant $$6.67408 × 10^{-11}m^3kg^{-1}s^{-2}$$, we get a force of approximately $$2 * 10^{20}N$$.

Let's say we want a thousandth of that, or $$2 * 10^{17}N$$. It seems like a lot of force, but given the mass of Earth, it really isn't. We're increasing the numerator by the ratio of the mass of earth to the mass of the moon (given your earth-mass spaceship), so a factor of ~81.2. We need only increase the distance by the square root of that, or a factor of ~9. So we drop the spaceship into an overtaking orbit inside Earth's which has a close approach of $$3.4 * 10^9m$$.

That's 3.4 million km, compared to Earth's orbital radius of 149.6 Mkm. This planetship would then have an orbital period equal to earth's year times the cube of the ratio between their minor orbital axes, or approximately 93% that of ours.

If, then, the planetship is dropped in "behind" Earth in its year, the time it takes for it to overtake us substantially (and make travel less convenient) is going to be more than long enough for ships to go back and forth for a while.

The effect on Earth's orbit is going to be minimal, and on the rest of the solar system, likely impossible even to detect. If it sticks around for geologic time, you might want to find a safer place for it, but if it's only there for a little while, you may as well park it close.

• The last period doesn't take into account the chaotic behavior of N-body systems. – L.Dutch - Reinstate Monica Apr 8 at 3:35
• I don't know more than the basics of celestial mechanics, despite two semesters of calc-based physics. If this is accurate it's about what I was hoping for.I don't understand chaotic N-body systems (sorry). – DWKraus Apr 8 at 3:39
• The solarsystem is chaotic. That means that we don't know where in its orbit the Earth will be in a billion years. But it is pretty stable insofar as orbit radii are concerned (There is a minor concern about the a potential Jupiter/Mercury resonance). The planet-ship will change the long term anomaly of the earth, but not the semi-major axis by much. – James K Apr 8 at 9:11
• Alien typing...: Thanks! That clarified my doubt. – a25bedc5-3d09-41b8-82fb-ea6c353d75ae Apr 8 at 11:44
• @costrom. Nope, no typo. 3x10^9m, 3 million kilometres. – jdunlop Apr 8 at 16:04

You said it was the size of a planet... but is it as dense as a planet? If the ship is a hollowed-out sphere (engines gotta fit somewhere, living space is useful, and no ship is going to be hauling around useless mass if it doesn’t need to), its mass may be far less than a planet and therefore able to get very close indeed. Take the gravitation equations in the other answer, but run them again with a body 1/10th the mass of Earth.

• Good idea, but I should have been more precise and said size and mass of a small planet. – DWKraus Apr 8 at 3:40
• Not so close. Hollowed-out and very big would mean that earth gravity would pull with a different force on the closest and farthest parts of the ship. That's how big objects in space are ripped apart when they get close to bigger ones. – FluidCode Apr 8 at 11:06
• @FluidCode True, but it is still closer than would otherwise be, and some engine thrust can be used to compensate for Earth's pull to keep the ship coherent. But, alas, DWKraus has put the kibosh on this plan. – SRM Apr 8 at 18:16

The best option would be to put the planet into an orbit around the suns poles at 1 astronomical unit from the sun so that it only passed through the plane of the ecliptic twice a year, the planet would have an orbital period of 1 year like the Earth.

With careful timing and positioning each time the plane of the ecliptic was crossed the Earth could be on the other side of the sun and so far away from any significant influence.

Of course any object the size of the Earth would have an effect upon the orbit of the planets, but in the above configuration I believe the effect would be minimal and non disruptive on a temporary basis (years) although eventually there would be problems.

• I would put it maybe a little further or a little closer than 1 AU, so that its orbit never intersects earth's. You know, just for safety's sake. – jdunlop Apr 8 at 1:21
• While close approach distance may be the same as the overtaking orbit in @jdunlop answer, this will make the relative velocity between Earth and Planet Spaceship much greater, requiring more delta-v for shuttle runs. Divergence will also be faster. – KerrAvon2055 Apr 8 at 4:29
• That's going to need a lot of delta-v for the trip. – Loren Pechtel Apr 8 at 4:56
• @jdunlop: Actually, you want to make its orbital period exactly the same as the earth's so that the closest-approach distance remains constant from orbit to orbit -- at least, over a period of a few years. (You will get deviations over geological time scales.) – Charles Staats Apr 8 at 13:23