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Let's say we have a cannonball that can shoot at various angles. How would we find the minimum velocity for this projectile to leave the Earth when the cannonball is shot at an angle $\theta$. I think the answer relies on $\theta$ but I am not sure what $\theta$ is.

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  • $\begingroup$ I am not sure what $\theta$ is. Are you saying that you have a conceptual misinderstanding of it? Or just that you want to consider it an unknown launch parameter? $\endgroup$ – G. Smith Feb 14 '20 at 5:17
  • $\begingroup$ In addition to the comment from G. Smith, please define what you mean by "leave the earth". Do you want to enter lunar orbit, orbit around another planet, enter an orbit that is beyond the orbit of Pluto, leave this solar system and orbit the nearest star, leave this solar system never to return, etc.? $\endgroup$ – David White Feb 14 '20 at 5:50
  • $\begingroup$ I would want to define an angle $\theta$ as a parameter for launch. For example, like a projectile. I would like to define escaping the Earth as moving away in a parabola, hyperbola or a line. I believe that if you shoot a projectile at an angle $\theta$, you can escape the Earth with a velocity less than escape velocity. The Earths gravitational pull would make the projectile curve around the Earth until it launches the projectile away. Similar to how a rocket is launched towards the moon. $\endgroup$ – Soiwo Feb 14 '20 at 12:35
  • $\begingroup$ The angle makes a difference in general relativity, but not under Newton, so you should choose either the Newton or Relativity tag to clarify which framework you are talking about $\endgroup$ – Gendergaga Feb 14 '20 at 16:47
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The answer is surprisingly independent of $\theta$.

All orbits are either circles, ellipses, parabolas or hyperbolas. If you shoot a cannonball into space, and it doesn't enough energy to enter a parabolic or hyperbolic trajectory, it will be a circle or an ellipse (or, as G. Smith points out, the degenerate ellipse which is a line). This means that the orbit always intersects where you shot the cannon from. All circular or elliptic orbits of a cannonball cannot "leave earth." They will always come right back to where they were shot from.

You need enough energy to enter a hyperbolic trajectory. The velocity required is known as "escape velocity", and for earth it is roughly 11.2km/s. The key to this number is that your kinetic energy is greater than your gravitational potential energy, so you can climb forever out of Earth's gravity well without ever slowing down to 0m/s. The energy turns out to be what matters, not the direction of the velocity (unless you point the cannon down. That's a very short experiment).

For a rocket to enter orbit, it can't rely on a single impulse like your cannon ball. This can be done in one long burn, but its common to do 2 burns. The first gets you most of your energy on a very elliptical orbit, and the second circularizes the orbit so that you aren't on a terminal trajectory right back through your own launchpad.

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    $\begingroup$ parabolas show up when an object approaches the Earth, rather than when launched from it That’s not true. A parabolic escape gets to infinite distance with zero speed. A hyperbolic escape gets to infinite distance with positive speed. $\endgroup$ – G. Smith Feb 14 '20 at 5:58
  • $\begingroup$ All orbits are either circles, ellipses, parabolas or hyperbolas. Since this list includes circles, which are one form of ellipse, it should also include lines, which are a degenerate case of ellipses, parabolas, and hyperbolas. $\endgroup$ – G. Smith Feb 14 '20 at 6:00
  • $\begingroup$ en.wikipedia.org/wiki/Parabolic_trajectory “Parabolic trajectories are minimum-energy escape trajectories, separating positive-energy hyperbolic trajectories from negative-energy elliptic orbits.” $\endgroup$ – G. Smith Feb 14 '20 at 7:13
  • $\begingroup$ I believe that if you shoot a projectile at an angle 𝜃, you can escape the Earth with a velocity less than escape velocity. The Earths gravitational pull would make the projectile curve around the Earth until it launches the projectile away. Similar to how a rocket is launched towards the moon. $\endgroup$ – Soiwo Feb 14 '20 at 12:36
  • $\begingroup$ @Soiwo That would violate the conservation of energy. Rockets get away with it, of course, because they can do an additional burn mid-flight. This lets them get into orbits like a circular orbit around earth as slow as 6.9km/s. You might be interested in Kepler orbits. If you only have a two body problem (Earth and your cannon ball), those are the orbits which can occur. $\endgroup$ – Cort Ammon Feb 14 '20 at 16:06

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