4 ms·

No, you would just be in a different kind of orbit. You would be in Earth orbit until reaching escape velocity, at which point you would be in Solar orbit.

alistproducer2 9y ago
so are you saying that even though I could escape Earth's gravity going at sub-orbital velocities that I would still need to be going fast enough to avoid being pulled into the sun?
- aphextron 9y ago
  Disclaimer: the entirety of my orbital mechanics knowledge derives from Kerbal Space Program. I cant recommend it enough if you're really interested in building an intuitive understanding for this stuff. But no, it's not that you would be pulled into the sun. You would simply be in a solar orbit just like the earth, unless you continued accelerating and reached solar escape velocity (~600km/s). You can think of gravity like successively larger "wells" which you must escape from if you want to go anywhere. You can't just point a spaceship at a planet and go there.
  - kuschku 9y ago
    > You can't just point a spaceship at a planet and go there. Disclaimer: You can, you just need so much power it’s like flying with a helicopter from your bedroom to the kitchen, absolutely wasteful.
    - a_t48 9y ago
      The main question at that point becomes: how can you carry that much fuel?
      - Florin_Andrei 9y ago
        Depends on how efficient the engine is. If you're riding a nuclear rocket, you don't need that much fuel.
      - kuschku 9y ago
        Theoretically? Fusion reactor + ion drive. Throw enough power at those and you can accelerate for a long time.
  - isostatic 9y ago
    One thing you learn from KSP is it requires less energy to fly into to alpha centauri than fly into the sun.
    - mannykannot 9y ago
      Getting the Parker Solar Probe down to the sun will take 7 flybys of Venus to use its gravity to slow down, over 6 1/2 years. http://parkersolarprobe.jhuapl.edu/The-Mission/index.php http://parkersolarprobe.jhuapl.edu/The-Mission/index.php 'Journey to the Sun.'
  - pdonis 9y ago
    > solar escape velocity (~600km/s) That's for escaping from the surface of the Sun. At Earth's distance from the Sun, the velocity needed to escape from the Sun's gravity is only 42 km/s.
- garmaine 9y ago
  You can’t escape Earth’s gravity going at sub-orbital velocities. Earth escape velocity is higher than any orbital velocity.
- marcosdumay 9y ago
  You don't escape Earth's gravity at sub-orbital velocities. Escape velocity is always larger than orbit velocity. What the GP said is that if when you reach Earth's escape velocity, there is still a long way to go until you have Sun's escape velocity (so you are on a solar orbit), and after you reach that, there is a huge distance until Milk-Way's escape velocity, and an even huger way until you escape the Local Group.
- pdonis 9y ago
  > even though I could escape Earth's gravity going at sub-orbital velocities That's not what he said. Escape velocity is higher than orbital velocity (as ythn pointed out upthread). And escape velocity from Earth is still slower than Earth's orbital velocity around the Sun, which in turn is lower than escape velocity from the Sun's gravity at Earth's distance from the Sun.

alistproducer2 9y ago

so are you saying that even though I could escape Earth's gravity going at sub-orbital velocities that I would still need to be going fast enough to avoid being pulled into the sun?

aphextron 9y ago
Disclaimer: the entirety of my orbital mechanics knowledge derives from Kerbal Space Program. I cant recommend it enough if you're really interested in building an intuitive understanding for this stuff. But no, it's not that you would be pulled into the sun. You would simply be in a solar orbit just like the earth, unless you continued accelerating and reached solar escape velocity (~600km/s). You can think of gravity like successively larger "wells" which you must escape from if you want to go anywhere. You can't just point a spaceship at a planet and go there.
- kuschku 9y ago
  > You can't just point a spaceship at a planet and go there. Disclaimer: You can, you just need so much power it’s like flying with a helicopter from your bedroom to the kitchen, absolutely wasteful.
  - a_t48 9y ago
    The main question at that point becomes: how can you carry that much fuel?
    - Florin_Andrei 9y ago
      Depends on how efficient the engine is. If you're riding a nuclear rocket, you don't need that much fuel.
    - kuschku 9y ago
      Theoretically? Fusion reactor + ion drive. Throw enough power at those and you can accelerate for a long time.
- isostatic 9y ago
  One thing you learn from KSP is it requires less energy to fly into to alpha centauri than fly into the sun.
  - mannykannot 9y ago
    Getting the Parker Solar Probe down to the sun will take 7 flybys of Venus to use its gravity to slow down, over 6 1/2 years. http://parkersolarprobe.jhuapl.edu/The-Mission/index.php http://parkersolarprobe.jhuapl.edu/The-Mission/index.php 'Journey to the Sun.'
- pdonis 9y ago
  > solar escape velocity (~600km/s) That's for escaping from the surface of the Sun. At Earth's distance from the Sun, the velocity needed to escape from the Sun's gravity is only 42 km/s.
garmaine 9y ago
You can’t escape Earth’s gravity going at sub-orbital velocities. Earth escape velocity is higher than any orbital velocity.
marcosdumay 9y ago
You don't escape Earth's gravity at sub-orbital velocities. Escape velocity is always larger than orbit velocity. What the GP said is that if when you reach Earth's escape velocity, there is still a long way to go until you have Sun's escape velocity (so you are on a solar orbit), and after you reach that, there is a huge distance until Milk-Way's escape velocity, and an even huger way until you escape the Local Group.
pdonis 9y ago
> even though I could escape Earth's gravity going at sub-orbital velocities That's not what he said. Escape velocity is higher than orbital velocity (as ythn pointed out upthread). And escape velocity from Earth is still slower than Earth's orbital velocity around the Sun, which in turn is lower than escape velocity from the Sun's gravity at Earth's distance from the Sun.