Timeline for Why are rockets launched vertically?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 18, 2016 at 11:16 | comment | added | David Hammen | Why all the upvotes? This answer is quite incorrect. | |
| Sep 14, 2015 at 16:35 | comment | added | Jess Riedel | Sorry if I used "elliptic trajectory" where "elliptic instantaneous orbit" would have been more precise, but I thought the meaning was clear. I am well aware that the rocket does not follow a fully elliptic trajectory. Rather, the highly elliptical orbit is clearly the instantaneous orbit the rocket has a moment after it leaves the launch pad, i.e., the orbit it would follow if it cut it's engines and the Earth were compressed. And this is clearly the relevant initial orbit for orbital transfer. If you want to discuss this further, we'll need to move to chat. | |
| Sep 14, 2015 at 16:29 | comment | added | Luboš Motl | No, there is no reason why the trajectory of the rocket before it gets to the orbit should be elliptic. It surely has nothing to do with Kepler's orbits which you seem to assume. On the contrary, when the rocket is accelerating, it should have as huge acceleration (contributed by the jets) as possible - which also means that it should be as far from any Kepler's orbit as you can possibly get. So your assumptions are exactly the opposite of what is actually needed. | |
| Sep 14, 2015 at 16:27 | comment | added | Jess Riedel | The highly elliptic trajectory is the starting orbit of a rocket on the launch pad, and the problem of getting from this orbit to a LEO circular orbit is exactly the form of the idealized problem where the Earth and atmosphere have been removed. | |
| Sep 14, 2015 at 16:26 | comment | added | Luboš Motl | Your highly elliptic trajectory has absolutely nothing to do with the orbits that rocket scientists want to achieve as the final state which are, on the contrary, almost exactly circular (i.e. fixed height i.e. fixed distance from the Earth's center). You're solving a totally different problem. Obviously, if there is no well-defined "final height" one wants to achieve, all arguments and derivations assuming such a final height - or ways to commute in between two different heights - are irrelevant. But in the rocket science, they're always relevant. | |
| Sep 14, 2015 at 16:22 | comment | added | Jess Riedel | Imagine the Earth is compressed to a point and that the rocket is given an infinitesimal amount of horizontal motion. Now the rocket is on a highly elliptic trajectory with apogee at the Earth's former radius and perigee near the center. What is the most efficient way to quickly raise the rocket's orbit to a circular orbit at LEO altitude? I believe it's by burning horizontally at first. Even if I'm wrong, it's definitely not obvious that vertical is the way to go based on the argument you gave ("connect these two potential levels ...by the shortest possible path...clearly vertical"). | |
| Sep 14, 2015 at 16:16 | comment | added | Luboš Motl | Another way to think about the bent trajectories. At the end, you want to get the satellite or something else on an orbit where no fuel has to be spent anymore. Basically, the faster one gets to that state, when $(x,p)$ is already on the desired orbit, the less fuel will be spent. Now, the point is that the main difference between points near the surface and points on the orbit is the altitude - the horizontal coordinates may be anything and the final vertical speed is zero on the desired orbit - and to change the height $h$ as quickly as possible, one wants to start vertically. | |
| Sep 14, 2015 at 16:12 | comment | added | Luboš Motl | No. It is the simpler part of the job to change the horizontal part of the velocity, and it may be done at the end. The harder part is to get away from the Earth's gravity, to get the sufficient height, and it has to be done as quickly as possible. So the optimum trajectory even if one demands some horizontal velocity at the end is to start almost exactly vertically and bend the trajectory later. All these statements are true and provable even if the air resistance is completely neglected. Such issues were already understood perfectly by Tsiolkovsky. | |
| Sep 14, 2015 at 16:09 | comment | added | Jess Riedel | But the goal is not just to get to that height, it's to get to that height going a certain horizontal speed. In particular, if the Earth and the atmosphere weren't in the way, it would be better to launch at an angle, which is exactly what is done when a craft wants to get from a low-Earth orbit to a higher one. (Hohmann transfers and that sort of thing.) | |
| Sep 14, 2015 at 16:06 | comment | added | Luboš Motl | My argument where I talked about 10,000 km works for 250 km, too. The goal is to get to a certain height and it's still true that the vertical trajectory is the shortest one to get to that height. The reason why the optimum direction is vertical has nothing to do with the drag in the atmosphere and the basic rocket equations designed to roughly get all the answers - as well as preferred trajectories - completely neglect the air resistance. | |
| Sep 14, 2015 at 16:04 | comment | added | Jess Riedel | This really doesn't answer the question well since rockets just going to low-earth orbit (a mere 250 km) aren't trying to escape the Earth's gravitational field, but they also launch vertically. (Criesto doesn't mention which sort of launch he's talking about.) Any satisfying answer is going to have to discuss the effects of drag from the atmosphere. | |
| Sep 14, 2015 at 16:03 | history | edited | Luboš Motl | CC BY-SA 3.0 |
added 317 characters in body
|
| Sep 14, 2015 at 15:58 | history | answered | Luboš Motl | CC BY-SA 3.0 |