Otdfbt

I was taught that when the acceleration experienced by a body is constant, that body follows a parabolic curve. This seems logical because constant acceleration means velocity that is linear and position that is quadratic. This is what I learned from projectiles: Bodies are thrown with an initial velocity near the surface of the Earth, they experience constant acceleration and the result is a parabolic curve.

Now that doesn't apply to the orbit of the Earth. The gravitational force can be thought of as constant since the distance from the Earth to the 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?

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

My question is, in a word, why can't the Earth be treated as a projectile? And if it can then why doesn't it behave like 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,
$$|vecF|=fracGm_Em_Sr^2,$$ 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:

$$frac1r=c_0+c_1sintheta+c_2costheta$$ 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:
$L=mr^2dottheta$.

$L$ is the angular momentum of the projectile about the earth. $dottheta$ 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:

$$r=frac1/c_01+(c_2/c_0)costheta$$

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

I was taught that when the acceleration experienced by a body is
constant, that body follows a parabolic curve

That last part is wrong.

Under central force, the body will follow a conic section of some sort. A parabola is one type of conic section. An ellipse is another. A circle is yet another.

Whether you follow a circle, an ellipse or a parabola depends on the initial conditions - the amount of the force and the angular velocity.

But very generally, body's don't follow a parabola, but a conic.

Gravitational acceleration acts toward the center of mass and follows the inverse square law. This will make all true trajectories either elliptical or hyperbolic (depending on the object's velocity relative to the escape velocity of the body acting on it).

When we examine the motion of objects close to the Earth's surface and traveling over short distances, Earth's center of mass is very far away, so gravity appears to act uniformly - falling objects will follow parallel paths and acceleration does not change with height. Under these conditions, trajectories will be parabolic. But these are just approximations because the distances over which motion is observed are so small in relation to the distance to Earth's center of mass. In other words, the quadratic formula and parabolic motion holds when you make assumptions about the direction and uniformity of the gravitational field, but these assumptions are only useful on the small scale (relative to the gravitational source/distance).

Earth's motion around the Sun could also be approximated as parabolic, but the approximation would only be valid over short distances. If you consider a comet in a highly elliptical orbit around the Sun, the segment of its path at or near apoapsis could be approximated as parabolic, but it is only an approximation.

Similarly, projectile motion "over the horizon" cannot simply be approximated as a parabolic path. For example: battleship cannons evolved to the point where they could fire shells tens of miles. Setting aside aerodynamic/wind effects, a parabolic arc is not an adequate representation of the shell's path because the curvature of Earth's surface and the difference in direction of "down" becomes significant over the shell's distance of travel.

A math/geometry perspective

Consider that the ellipse, parabola, and hyperbola are all conic sections distinguished one from the other by eccentricity (the circle is a special case of an ellipse with eccentricity of zero). The circumstances of a given trajectory problem will determine the eccentricity and therefore the appropriate conic. In the case of "everyday" physics, the distance to Earth's barycenter is approximated to infinity; this sets the eccentricity to 1 and the trajectory takes the form of a parabola. For the "over the horizon" problem, distance to Earth's barycenter cannot be assumed infinite, so the trajectory takes the form of an ellipse.