Except, as I recall that scene, it’s not even possible with sci-fi level technology. The missile reaches the sun in about 10 seconds and the effect is instantaneous.
It doesn’t make intuitive sense to me either. It’s still yet another sci-fi property I’d kill to see on the big screen.
That’s what I was going to say… someone really should make a video titled “Everything I ever needed to know about orbital mechanics, I learned playing KSP”.
But… to answer the OP in a super-simplified fashion, there are three parts to it.
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When something’s in orbit around something else, they’re being pulled toward it by gravity, but they’re ALSO going fast enough to, in essence, fall PAST the item continually. That’s why getting into space is more a matter of going fast enough, and not as much about going high enough. Another corollary is that since you’re being pulled at a constant rate, adding speed means you orbit wider, and removing speed means you orbit closer. So what happens when a space capsule or whatever decides to re-enter, is that they turn around and burn their engines exactly in the opposite direction that they’re traveling, slowing themselves down, and shrinking their orbit until it intersects with the Earth.
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Everything in space is (for the most part) in orbit around something. We’re in orbit around the Sun. The moon’s in orbit around the Earth.
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What Munroe is getting at is that the best way to get an object out of Earth orbit, and into a solar orbit that won’t have enough speed to continue falling “past” the sun might well be to fire a rocket away from the Sun, because you could get just far enough away so that you’re no longer in Earth orbit, but are instead in Solar orbit, but without much velocity, and you’ll likely impact the Sun at some point in that orbit.
The circumference of the Earth is much, much less than the circumference of a geosynchronous orbit (GSO). One must travel a hell of a lot faster to circumnavigate that orbit in one day than to circumnavigate the Earth in the same period.
Even if launched at the equator, it inherits roughly 1600kph of tangential velocity from the Earth. But the velocity required for GSO is close to 11,000kph. Launching straight up will result in the satellite falling straight back down. Well, not straight down, but generally straight, and regardless, it crashes back into the Earth.
I think it is the way you are recalling it that makes it sound confusing. Like, what does “to slow down, you must speed up” mean? It could mean something like that you will move more slowly if you accelerate into a higher orbit. But what they always teach people at least sounds straightforward and not contradictory, e.g., an object sweeps out equal areas in equal times, energy is conserved, angular momentum is conserved, square of the orbit time is proportional to the cube of the semi-major axis.
Of course not. It’s not pixies; it’s leprechauns. Japanese leprechauns. And when the repairman comes, all he’s really doing is slipping the little guy some sushi and soda bread. Or, for the really expensive problems, Guinness and saki.
At least, that’s the way my mom explained it to me, and she wouldn’t lie.
Cool thread folks. Lots of basic questions here about “if I do this in orbit, what happens?” I suggest to the curious: Get a copy of the game “Kerbal Space Program” (look it up online). It’s a great place to learn orbital mechanics and play around with rockets and spacecraft.
Circling back to OP’s mention of Randall Munroe, here is his comic where he claims that Kerbal Space Program really helped his understanding of orbital mechanics:
This was mind-blowing to NASA and the early astronaut corps too. They had to wrestle with these strange ideas, so different from flying an aircraft, in their first efforts at rendezvous and docking in orbit. Here’s a pretty nice video documentary of the whole idea:
I was a history major. I didn’t take those pre-med science courses. I had to learn my astrophysics on the street.
On the mean streets of Hell’s Observatory?
I need to make a correction to my posts.
If orbiting clockwise at the 3 o’clock position, aiming “straight up” and firing your rockets will raise your orbit at the 6 o’clock and lower it at the 12 o’clock. My previous posts stated that it was the 3 and the 9 respectively. I didn’t think that through carefully before posting, and I apologize.
*13.1 km/s that is
Thanks for the interesting discussion!
Like Little Nemo, I’m a history major and then worked in software development. Reading this thread made me want to understand this better, so I went out and bought Kerbel Space Program to try.
Those poor Kerbels. (Think Galaxy Quest. “Those poor people.”)
I returned it to save as many Kerbels as I could.
I remember a scene in a book I once read - my gut says Space by James Mitchener, but my memory does fail me sometimes. I would have read it in my teens, which was some many years ago.
The astronauts in training (hotshot pilots all) are taken to a racetrack where there are two rigged jeeps. The jeeps are set to go round in a circular path at a constant speed, but if the accelerator is pressed, it turns momentarily inward, and if the brake is pressed, it turns momentarily outward.
The jeeps are set 180 degrees apart, and the astronauts told to use the brake and accelerator pedals to bring the two jeeps together, one behind the other. As pilots, they expect this to be an easy task - of course, it is almost impossible.
This is deliberate, and is presented as a way to teach the astronauts that manually flying a spacecraft is completely counterintuitive for these pilots.
Once in space, they have to rely on ground control to make the burn calculations for orbital control.
Wait…that doesn’t seem correct either. Maybe show us a diagram of what you mean?
If you’re orbiting clockwise at the 3 o’clock position and pointing straight up and thrusting, then you are thrusting retrograde, that is, opposite the direction of motion.
When you slow yourself down while in circular orbit, then the altitude of your orbit on the opposite side from where you are now, will drop. Why is this? Conservation of angular momentum, in other words, the product of your orbit radius (from Earth center) and tangential velocity (along the orbit, perpendicular to your orbit radius) will be the same. So slowing down at your current altitude results in a lower altitude (and higher speed) at the opposite end of your orbit from where you made the maneuver.
So… you’ll be lowering your orbit at the 9 o’clock position, directly opposite 3 o’clock. At 6 and 12 o’clock, you’ll be at a lower altitude than 3 o’clock.
That was in “The Right Stuff” by Tom Wolfe.
I think by “up”, Bear_Nenno meant “away from the body being orbited”. That is, in a direction opposite to the local gravitational force. So looking at the diagram on a piece of paper, “up” at 3 o’clock would be to the right.
Got it. Yes, clarifying your definitions is the first step in solving a problem. In this case, a diagram would have helped!
Hohmann orbit simply means "orbit around sun at lowest level possible?