# Trajectories of projectile based on different speeds of projection [duplicate]

So my teacher was teaching gravitation and an interesting fact that he mentioned was the trajectory of a projectile projected from Earth at a speed equal to escape velocity, is parabolic.

Also, he mentioned that the path would be elliptical with Earth at one of the foci if the body is projected at a speed greater than orbital velocity but less than escape velocity.

I'd like to know how these results were mathematically derived.

• Here a link to hand-written surfable (link in light blue, if not broken) notes linking conics and gravitation basics.altervista.org/test/Math/analitic_geometry/…. Here some more detail about gravitation basics.altervista.org/test/Physics/Me/actions_gravitation.html Commented Jun 22 at 8:54
• Note that an escape speed trajectory is still called parabolic even if the motion is in a straight line. en.wikipedia.org/wiki/Radial_trajectory Commented Jun 22 at 9:22
• Also, if the projectile is launched with less than escape speed, its elliptical path must return to the launch point, so it can't actually go into orbit. (Assuming the planet isn't perfectly smooth & airless, so a perfect tangential trajectory is impossible). Commented Jun 22 at 9:32
• Conic sections in polar coordinates are discussed in more detail here. Commented Jun 22 at 10:41

According to classical dynamics (Marion and Thornton, Central Force motion chapter), the eccentricity $$\epsilon$$ of the trajectory of two bodies in central motion is given by:

$$\epsilon = \sqrt{1 + \frac{2El^2}{\mu k^2}}$$

where:

• $$E$$ is the total mechanical energy,
• $$l$$ is the angular momentum,
• $$\mu$$ is the reduced mass,
• $$k$$ is a constant related to the gravitational force, $$k = GMm$$ for gravity.

If the projectile is launched at escape velocity, then the total mechanical energy $$E$$ is zero:

$$E = 0$$

Substituting $$E = 0$$ into the equation for eccentricity yields $$\epsilon = 1$$. An eccentricity of $$\epsilon = 1$$ corresponds to a parabolic trajectory, which is the boundary case between elliptical ($$\epsilon < 1$$) and hyperbolic ($$\epsilon > 1$$) trajectories.

If the speed $$v$$ is greater than the orbital velocity but less than the escape velocity, the total mechanical energy $$E$$ is negative (bound orbit), and the trajectory is elliptical:

$$- \frac{\mu k^{2}}{2l^{2}} < E < 0$$.

Oh, the polar equation for conic sections is discussed here in detail, along with the role of eccentricity.

For a complete mathematical derivation, more time is needed, and I might return to it later.