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Given that $$|\vec{g}|=\frac{GM}{|\vec{r}|^2}$$ and $\vec{g}$ always points to the Earth's center of mass, can the "almost parabolic" elliptic trajectory be derived provided the initial velocity $\vec{v(0)}$ of a rigid body launched into the air?

The answers to this question touch on this matter when explaining the intuition: Throwing a Football. Is it truly parabolic?

However, the answers to that question only seem to consider derivations with respect to a parabolic approximation (I'm referring to the math and not the intuition; one of the answers does indeed provide intuition about the elliptic trajectory)

I'm already quite comfortable with the oversimplified version of translational kinematics which treats the direction of $\vec{g}$ as though the Earth were flat, and I know how to perform the derivations. I imagine the derivations for the true, accurate non-flat Earth are a lot more mathematically challenging, but if I'm pointed in the right direction I may be able to calculus myself through the rest of the derivation.

Simon M
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Ok so it's an ellipse. I'm still curious about the derivation, though. Would it be best to switch to a celestial mechanics point of view for the derivation? – Simon M

That is probably the best approach. The orbit of a body with less than escape velocity is an ellipse according to Kepler's first law. A projectile such as a thrown football follows an elliptical orbital path (if we ignore air resistance), with the centre of the Earth at one focus of the ellipse. Imagine you can hover at your exact location when you throw the ball, but the earth has shrunk to a tiny sphere about the size of a pumpkin, so all the Earth's mass is concentrated there. (Don't worry, it is not small enough to form a black hole). The ball will orbit the pumpkin. We only see part of the ellipse because the ground surface intersects the ellipse. The derivation of the equation of motion for a Keplerian orbit, can be found in the Wikipedia link for Kepler's laws.

KDP
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