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Imagine a spherical planet with no atmosphere. A ball is launched radially with a velocity $v$, maintaining $\frac{v_e}{\sqrt2}<v<v_e $ and there is no other objects in the universe (Ideal two body problem.)

My question is, will this projectile orbit the planet? Note that, the projectile is launched radially.

As far as I know, the trajectory doesn't depend on launch angle for large velocities. All that matters is the energy of the projectile. But yet, I might be wrong. I also can't explain if the answer of my question is 'Yes', because I don't know how gravitational force will bend the projectile. And if the answer is 'No' then do other angles except $90°$ have any effect on trajectory?

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No. Assuming the planet is not rotating (so the launch velocity is truly radial), it will go up and then come straight back down - unless its velocity is greater than the escape velocity from the planet's surface, in which case it will travel away from the planet forever.

As far as I know, the trajectory doesn't depend on launch angle for large velocities.

This is not true. The escape velocity is angle-independent, which means that if the ball has speed greater than $\sqrt{\frac{2GM}{R}}$ then it will never return to Earth regardless of the angle at which it is launched; however, that doesn't mean the trajectory is angle-independent. Obviously a ball launched radially will follow a different path from a ball launched at a $45^\circ$ angle.


For an object projected from the surface with no other external influences, it is not possible to achieve a stable orbit. Exactly one of two things will happen - either the ball will fall back down and impact the surface of the planet, or it will travel away from the planet forever.

If you want to achieve orbit, you need to provide a secondary boost after the ball has already left the surface. You can understand this by noting that the bounded orbits of objects under the influence of $1/r^2$ forces are ellipses, which come back to their starting point. If you want the orbit to not intersect the surface of the planet, you need that starting point to be higher than ground-level.

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    $\begingroup$ I feel like it's worth noting here that even a tiny boost at the apoapsis of the post-launch orbit can do this. Small changes at high altitudes have massive changes on the effect of the orbit's periapsis. I'd love to know the term for this kind of "leverage", if that's a thing. $\endgroup$ – William Oct 25 at 4:58
  • $\begingroup$ Can you assume that the planet is not rotating? It is an ideal two body problem. $\endgroup$ – user456350 Oct 27 at 20:18
  • $\begingroup$ @user456350 I’m not sure what you mean - I did do that. If you add rotation, all you do is change the launch velocity, but the main point of the answer is unchanged. $\endgroup$ – J. Murray Oct 27 at 20:32
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All trajectories of a projectile in an inverse-square field like the gravitational field of the Earth are either elliptical or open (parabolic or hyperbolic). Open trajectories are not orbits; the projectile escapes. Elliptical orbits always return to the same path relative to the Earth. If the projectile goes straight away from the center of the earth at less than escape velocity, it will eventually turn around and fall back down, hitting the Earth with the same speed it had as it was leaving.

If we assume the Earth is rotating so that the projectile has a tangential velocity added to its radial velocity, and the sum is still less than escape velocity, then the projectile follows an elliptical trajectory and will return to strike the Earth's surface at the same angle and speed at which it was launched. It will not be in an orbit.

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