Why does the Earth follow an elliptical trajectory rather than a parabolic one?

Now that doesn't apply on the orbit of the Earth. The gravitational force can be thought of as constant since the distance fron Earth to Sun can be thought of as constant too

You are correct that the strength or magnitude of the sun's gravitational field is very similar over the length of the earth's orbit, but the direction is not. In a uniform gravitational field, the direction would be the same everywhere.

Over the path of the earth's orbit, the sun's gravitational field points in different directions. This significant difference from a uniform field means that the earth's orbit is quite far from a parabola.

The gravitational force can be thought of as constant since the distance fron Earth to Sun can be thought of as constant too, which by Newton’s second Law means the acceleration of Earth is also constant. Wouldn't that mean that the Earth should just follow a parabolic path ?

No, the gravitational force of the Sun on the Earth is not a constant, for two reasons:

• it is changing direction all the time, that is, it is always toward the Sun as the Earth (in the Sun's reference frame) revolves around it, and
• it is changing in magnitude as the Earth gets closer and farther away. This is because the kinetic energy of the Earth due to its orbital motion at the aphelion is not large enough to let it move in a circular orbit of that radius. And its too large at the perihelion to move in a circular orbit of the perihelion radius. (And everywhere in between the velocity vector is not perpendicular to the radial vector between the Earth and Sun.)

If the Earth moved parabolically around the Sun, it would not be in a closed orbit. It would pass by the Sun once and never return. That's because in order to have a parabolic orbit with Newtonian gravity, $$|\vec{F}|=\frac{Gm_Em_S}{r^2},$$ the kinetic energy of Earth would be to large to stay in orbit.

Is there a mathematical proof (similar to the one I mentioned about projectiles) giving the elliptical orbit as a result ?

Yes, and it can be found in multiple places, usually in university sophomore level classical mechanics (and even engineering mechanic) books. See books by Symon, Marion, Beer & JOhnston, Barger & Olsson, Taylor, just to suggest a few. It is a standard derivation involving calculus, and it too long to detail here.

And actually, a projectile on Earth is following an elliptical path, too, around the center (roughly) of the Earth. We approximate the Newtonian gravity as a constant magnitude, constant direction force for small areas (like football fields), and get the parabolic shape, which is actually a good approximation of a short elliptical path.

Projectiles' paths are not actually parabolic near earth. A parabola does not imply constant total velocity. A parabola is constant velocity in one direction and acceleration in a perpendicular direction.

On the small scale of most ballistics problems, the earth is much larger than the trajectory of the projectile. On such a scale, the earth can be approximated as a plane. Whereas gravity pulls projectiles towards the center of the earth, and therefore is a direction that is changing for any moving object, this direction doesn't change from straight down, again on such small scales. The acceleration is straight down, there is no acceleration across, so the result, to close approximation, is parabolic. If you were to observe the path closely and it were able to "fall"through the earth, it would follow an elliptical path.

The actual equation is:

$$\frac{1}{r}=c_0+c_1\sin{\theta}+c_2\cos{\theta}$$ where $r$ is the distance from the center of the earth.

The c's are constants depending on the mass of the earth, the angular momentum(a constant of the motion), and the mass of the projectile.

$r$ is the distance of the projectile from the center of the earth. Theta is the same theta from polar coordinates.

Let: $L=mr^2\dot{\theta}$.

$L$ is the angular momentum of the projectile about the earth. $\dot{\theta}$ is the angular velocity along the trajectory.

The angular momentum is constant in time. Taking the derivatives of both sides gives you some information of the time evolution of the system.

If you set up the zero coordinate of your theta correctly, you can assume $c_1=0$.

rearranging:

$$r=\frac{1/c_0}{1+(c_2/c_0)\cos{\theta}}$$

One might recognize this as an equation in polar coordinates of a conic section with the origin of the coordinate system at a focus.

Doing this, $c_2/c_0$ determines the eccentricity of your trajectory. This parameter tells you the shape of the projectiles path: Orbital Eccentricity