4
$\begingroup$

Imagine that in addition to Earth, there was a second planet, say of one Earth mass, orbiting the sun - but on the opposite side of the Sun, and at the same orbital radius.

My question is - given today's technology, would it be feasible to try to launch something to reach that planet? A rover, a capsule, or something of the sort is what I have in mind.

There are a couple reasons I initially don't see a solution. First off, the escape velocity of the Earth is about 11.2 km/s. The Earth's orbital velocity around the sun is nearly 30 km/s. This means that if I were to launch a rocket in the opposite direction of Earth's orbit to reach this "second Earth", not only would I have to counter this 30 km/s orbital velocity, but I would have to provide enough thrust for double this velocity in order to maintain a stable orbit around the Sun at the radius that Earth orbits at - but in the opposite direction. I'm not quite sure if rockets can get up to velocities that high - I know that the Juno probe is moving at around 73 km/s, but I believe that was after a series of maneuvers involving a gravitational slingshot.

Perhaps there is another solution? One that popped into my head was to launch it at an angle to the plane of Earth's orbit, but I don't quite know how that would work out, given that unless some serious trajectory corrections were made, the rocket would not arrive anywhere close to the planet - as the planet would have moved in the time the journey took.

Maybe it involves taking a path that is closer to the Sun, and more elliptical? At this point I'm just throwing out everything that comes into my head.

Basically, I'm looking for some sort of general explanation here, with a rough idea that I could build on. And if it isn't possible with today's technology, what do we lack?

$\endgroup$
  • 1
    $\begingroup$ "Imagine that in addition to Earth, there was a second planet, say of one Earth mass, orbiting the sun - but on the opposite side of the Sun, and at the same orbital radius." We can do better than just imagining it :-) Why would Hollywood's Planet X (at earth's L3) be unstable? $\endgroup$ – uhoh Mar 7 at 10:53
  • $\begingroup$ This question is perhaps better suited to Worldbuilding SE, where, in fact, there are already several question dealing with this : see this question and the linked questions on that page. $\endgroup$ – StephenG Mar 7 at 12:16
  • 1
    $\begingroup$ There's some good info on Space Exploration: How much delta v does it take to get to the Sun-Earth Lagrange 3 point? $\endgroup$ – PM 2Ring Mar 7 at 14:31
  • 5
    $\begingroup$ Ignoring the fact that said planet cannot exist (the L3 point is unstable), there exists an infinite number of two burn solutions to reach the L3 point that have the spacecraft making $m$ orbits in $n+\frac12$ years, where $m$ is a positive integer and $n$ is a non-negative integer. How long are you willing to wait to reach a planet that cannot exist? The longer the wait, the cheaper it gets, at least in terms of delta-V. $\endgroup$ – David Hammen Mar 8 at 5:56
  • $\begingroup$ Similar question here. space.stackexchange.com/questions/32286/… Reaching a theoretical planet at L3 isn't that hard, just drift to a higher orbit or a lower orbit, and wait. After that, aero-break. $\endgroup$ – userLTK Mar 8 at 16:43
5
$\begingroup$

Venus has roughly the same gravitational potential as earth and the Venera probes successfully and softly landed there. Similarly, the Apollo space craft landed safely on earth on return. In both cases aerodynamic friction with the atmosphere can be used to get rid of excess velocity in order to enter orbit and eventually land. Thus landing will work cheaper, if there is atmosphere. But at the expense of much more fuel also without (see moon landing again... you'd need stronger descent rockets than Apollo for an exclusively rocket - driven landing)

Getting there is not a big deal either. Look at the Stereo mission of two satellites which trail earth's orbit more and more to get a better understanding of the sun, gaining more and more separation from earth while (nearly) staying at 1 AU: You get to another position in earth's orbit by just lifting your orbit slightly and waiting for earth to go and the other to catch up (or go slightly towards the sun and be faster yourself).

| improve this answer | |
$\endgroup$
  • $\begingroup$ There's more to read about STEREO's orbits here and here and to see in youtu.be/VzhMvEkK0gA $\endgroup$ – uhoh Mar 7 at 10:56
  • 1
    $\begingroup$ So from this answer, what I'm getting is that the best way to go about reaching this hypothetical planet would be to place the rocket/satellite/probe into an orbit slightly offset from that of the Earth in terms of distance from the Sun, which would gradually allow it to reach the planet - at which point we could use a bit of fuel to get ourselves into the gravitational reach of the planet. Does that seem about right? And thank you @uhoh, the first link you posted had a really helpful animation regarding the STEREO mission. :) $\endgroup$ – Calc-You-Later Mar 7 at 22:06
  • 1
    $\begingroup$ @Calc-You-Later I don't want to speak for planetmaker's answer, but I think that might be correct. Since your question is not really about Astronomy as much as it is Space Exploration, I'd recommend that you ask a follow-up question in Space SE and ask "For a given amount of delta-v what's the fastest way to travel between opposite points at 1AU from the Sun beginning from and ending in low orbit around Earth-like planets?" Or you could ask what fraction of your spacecraft would be fuel for travel times of 1, 5, and 25 years. $\endgroup$ – uhoh Mar 8 at 1:33
  • 1
    $\begingroup$ I'm not a prioriconvinced that gradually drifting as I described is always the fastest way. There might be faster paths through Venus or Mars swing - bys. But if you send an un-manned probe, lower delta v is preferable over quick. However I didn't try any calculation on this issue $\endgroup$ – planetmaker Mar 8 at 11:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.