5 ms·
> You can put it on a collision course with the sun Counterintuitively, this is far more expensive than launching it into said black hole, or just out of the S
by skykooler 7y ago
> You can put it on a collision course with the sun
Counterintuitively, this is far more expensive than launching it into said black hole, or just out of the Solar System all together.
- ruffrey 7y agoInteresting- why is that?
- 317070 7y agoThe escape velocity from the sun is 42.1 km/s, while the earth is orbiting at 29.78 km/s. The 29.78 km/s is 3 times closer to the escape velocity than to 0 km/s. This is the first order approximation reason.
- ekimekim 7y agoThe Earth is moving around the sun at ~30km/s. The escape velocity of the sun is ~42km/s. So if you start from Earth, it only takes 12km/s of change in velocity (called "delta-V" by rocket scientists / KSP players) to leave the solar system, but more than twice that (30km/s) to slow down enough to hit the sun.
- thebluehawk 7y agoOrbital mechanics. The earth is orbiting the sun at 30 kilometers per second. So if we launched something into space, since it started on earth, it would have that speed (similar-ish to throwing a ball from a moving car). So that object would now also be orbiting the sun at 30 km/s. We would need to slow it down that much in order to "fall" into the sun. Once something was in earth orbit, it would only take about 12 km/s of delta v (change in velocity) to escape the solar system. More info and math here: https://space.stackexchange.com/questions/3612/calculating-solar-system-escape-and-and-sun-dive-delta-v-from-lower-earth-orbit https://space.stackexchange.com/questions/3612/calculating-s...
- mihaifm 7y agoWhy do you have to slow it to 0 to hit the sun? Can't you just cruise at whatever speed you're cruising and redirect it with thrusters towards the sun?
- rileyteige 7y agoOnce you're in orbit (say around the sun), you have to cancel the orbital velocity to fall into the object you're orbiting around. If you point at the sun and accelerate 1 km/s directly at it, you're still moving 30 km/s "sideways". All you'd end up doing is making the orbit more elliptical-shaped. At least that's how I see it, but I am far from being an authority on this topic.
- tux1968 7y agoYou can but then you're using a lot of propellant rather than gravity to reach your target.
- kibwen 7y agoI'm no rocket scientist but perhaps that would cause the object to shift from the Earth's kinda-circular orbit into a highly-elliptical orbit, its existing sideways velocity (relative to the sun) causing it to be flung past the sun at more of a straight line and hence way out into the solar system instead.
- dTal 7y agoIn space (and any frictionless medium), you can't "redirect" an existing velocity vector with thrusters. You can only add a velocity vector. This means that if your desired direction is perpendicular to your current direction, your current speed is no good to you at all. If you're heading due "north" at 10mph, and you want to be heading due "east" at 10mph instead, you have to 1) fully negate your "north" velocity, and 2) come up with 10mph of "east" velocity from scratch. Now, the trajectory of an object in solar orbit is exactly at right angles to the direction it needs to go in to hit the sun. No part of this velocity is helpful for getting to the sun - in fact it actively prevents it! The only vector that takes you directly into the sun is one with no sideways component - if you imagine yourself falling right in, any sideways nudge will cause you to miss it by a hair and go flinging off into a highly elliptical orbit. If you just ignore this and just thrust directly at the sun, hoping to overpower everything by brute force, then like a ballerina pulling her arms in, the more you try to get close to the sun with your thrusters, the faster your orbit will go; the closer you manage to get, the further out you'll be flung when you inevitably miss. All this ignores that the sun is not a point, but quite a large ball - you can get away with some small horizontal velocity. A highly elliptical orbit will still do what you want if its lowest point is below the surface.
- deleted 7y ago[deleted]
- dmd 7y agoHere's a delta-V map of the solar system: http://i.imgur.com/SqdzxzF.png http://i.imgur.com/SqdzxzF.png
- jhrmnn 7y agoInteresting! I would have never guessed that starting from Earth, getting to Mars is harder (by about 5%) than escaping the Solar system.
- Agathos 7y agoWell, there's "I want to match Mars' orbit around the Sun, so I can orbit the planet," and then there's "I want to intersect Mars' orbit just as the planet is passing through so I smack into its surface and leave a big crater." The latter takes a lot less delta v, but it has its drawbacks. Leaving the solar system, you don't have to budget for that rendezvous.
- labawi 7y agoThat seems like a orbit/landing chart. Doesn't seem accurate for head-on collisions. Or am I reading it wrong? As a counter example, someone mentioned nearly leaving and then cheaply coming back directly into the sun.
- deleted 7y ago[deleted]
- thegagne 7y agoEver been on a merri-go-round? While spinning it’s hard to get to the center. Once it stops, it’s easy. The earth is spinning around the sun. To get to the sun, you need to slow down.
- fragsworth 7y agoI assume you have to "slow down" ~107,000 kmph to fall into the sun, how much do you have to "speed up" to escape?
- deleted 7y ago[deleted]
- tfha 7y ago~11 km/s to escape the sun, ~30 km/s to fall into it. Escaping is easier.
- sandworm101 7y agoJust getting into the sun isnt enough. The sun's outer layers are not very dense. A blob of uranium might survive long enough to come out the other side, at least on the initial few orbits.
- labawi 7y agoCan't you use some kind of gravity assist to do the slowdown or adjust the trajectory into the sun?
- Symmetry 7y agoIf you're using a Hohmann transfer, sure. But IIRC a Bi-elliptic transfer will get you on a collision with the Sun for less delta-v than escaping the solar system. That's where you speed up to get high and slow then burn to cut your velocity when there. For ratios of oribital radii over 15 it's more always more efficient than the standard Hohmann transfer and since the ratio is very large in this case it works much better. But in this case you're just trying to get within the radius of the Sun and don't even need the third burn to slow down. https://en.wikipedia.org/wiki/Bi-elliptic_transfer https://en.wikipedia.org/wiki/Bi-elliptic_transfer