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

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?

• Comments are not for extended discussion; this conversation has been moved to chat. May 11 '19 at 11:34
• Just to supplement the other answers, there is a really great video on this topic: youtube.com/watch?v=xdIjYBtnvZU May 11 '19 at 16:48
• – rob
May 13 '19 at 14:15

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.

• This is because the kinetic energy of the Earth due to its orbital motion is not large enough to let it move in a circular orbit. – that is only true in apohelium; in perihelium it is too high and in between it is just right, but the velocity vector is not aligned. May 11 '19 at 19:41
• @JanHudec Yes! You win the prize! May 12 '19 at 22:23

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

• If... it we’re to fall through the earth...: Oh I know, tempting. But not quite. Once you “fall through” you start to leave part of the mass of the earth behind; that changes the problem and the curve is no longer an ellipse. (Literally “leave behind”; under these idealized conditions, the spherical shell of the earth that is farther from the center of the earth than the bullet exerts no gravity at all and the force gets smaller the closer you get to the center) May 10 '19 at 3:11
• Yes I'd forgotten about that. The trajectory above eatth is part of an ellipse which has the center of Earth at a focus. Below the surface, gravitational force is linear with radius. I'll need to factor that in. May 10 '19 at 3:18
• Parabolic trajectories are essentially a flat earth model...
– jk.
May 10 '19 at 12:45

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.

• Conic sections are what you get from inverse-square central forces. That's actually true (to a good approximation) for the earth-sun system, but is not what the OP is describing when he says "when the acceleration experienced by a body is constant". May 10 '19 at 0:28
• Note - a circle is a special case of an ellipse. May 10 '19 at 2:12
• "body's don't follow a parabola". Bodies often follow a parabola, they just don't do so for very long.
– Aron
May 10 '19 at 9:22
• @DavidWhite Note - a parabola is a special case of an ellipse, too ;-). May 10 '19 at 9:46
• @"That last part is wrong" If "constant force" means "constant vector force", then it isn't wrong (the statement as a whole isn't wrong, as it is true that if the Earth were experiencing constant force then it would have a parabolic trajectory, but the premise that the Earth is experiencing a constant force is wrong). Only if we take "constant force" to mean "constant magnitude force" is it wrong. May 10 '19 at 15:38

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.

I'm trying to arrive at the crux of your question. There probably is a mathematical way to get from a parabolic projectile trajectory to an orbit, whether it's basically circular or elliptical, but I haven't heard of it. Atmosphere aside, as projectiles velocities get higher and higher it will get to a point where the curve that the projectile is taking would be pretty much the same as the curve of the Earth, circular. Even with a ballistic trajectory, I think if the bulk of the earth wasn't in the way of it, it would have an elliptical path around the center of the Earth.

The earth does not follow an parabolic trajectory because we could not observe it if it would not. We would not exist to observe it. We could not observe it because other trajectories do not allow the earth stay long enough in a distance to the sun that allows life to evolve.

This is using the Anthropic principle, which is more commonly used to explain why the physical constants have values that allow life to evolve.

There is some level of absurdity to this idea, but I think it is less than completely absurd.

• I think you're commenting on why the Earth doesn't follow a highly eccentric elliptical trajectory. The Earth does in fact follow an elliptical orbit, with enough eccentricity that our distance from the sun varies by about 3 million miles over the course of its orbit. May 10 '19 at 22:08