# Parabolic Gravity?

nhmllr
Ok, so if you throw a ball up in the air, it comes down in an (almost, slight deviation due to jerk) parabolic trajectory. Now, if you look at the dance that two (let's keep it simple) planets make when effected by eachother's gravity, you get certain conic sections. You get ellipses, circles (in very exact cases) and hyperbolas. Why not any parabolas? Or are there parabolas, but they're just as rare as exactly circular orbits?

Staff Emeritus
Gold Member
The ball's motion is actually an ellipse, but it can be approximated very well with a parabola. You can get a true parabola in the case where an object is moving at exactly escape velocity.

syhprum
I am very pleased to see someone else making this point, I wince whenever I read about projectiles moving in a parabolic path.
I was taught this erorroneus idea at school in the fourties and found it confusing when the age of artificial satellites arrived and take every opurtunity to correct it when I see it in print but it normally falls on deaf ears.

Staff Emeritus
Gold Member
I am very pleased to see someone else making this point, I wince whenever I read about projectiles moving in a parabolic path.
I was taught this erorroneus idea at school in the fourties and found it confusing when the age of artificial satellites arrived and take every opurtunity to correct it when I see it in print but it normally falls on deaf ears.

Hmm, well for motion in a uniform gravitational field, the trajectory is parabolic. That follows directly from the equations of motion. So, the question you have to ask is, to what degree is the field non-uniform in the context of the problems done in grade school, and does it matter?

In response to the original post, yes, parabolas and hyperbolas are conic sections as well, and yes there are examples of parabolic and hyperbolic "orbits." I think a good example might be a "flyby" of a planet done by a spacecraft to get a gravitational boost. Of course, these are not closed curves, meaning that, during this interaction, the object does not become gravitationally *bound* to the planet, but rather is on an *unbound* trajectory, eventually escaping "to infinity." I'm not 100% sure, but I think that some comets that get knocked into the inner solar system come in on unbound trajectories and eventually move back out again, never to return.

Which conic section you end up with depends on the amount of angular momentum in the system.

cephron
Is the movement of a projectile so close to being parabolic because the earth's gravity field is so close to being uniform?
Or, in other words, in a planar world where gravity is uniformly -10m/s^2 in the y direction, would projectiles then have a truly parabolic trajectory?

Staff Emeritus
Gold Member
Is the movement of a projectile so close to being parabolic because the earth's gravity field is so close to being uniform?
Or, in other words, in a planar world where gravity is uniformly -10m/s^2 in the y direction, would projectiles then have a truly parabolic trajectory?

Yes. You can assume that the gravitational force is constant in magnitude, since your distance from the centre of the Earth doesn't change appreciably at the surface of the Earth compared to a few tens of metres above the surface of the Earth (or even higher still). Mathematically, this is equivalent to setting GM/R2 = g = const. in Newton's law of gravitation where M is the mass of the Earth and R is its radius. It should really be GM/r2, where r = R + h is the distance between the object and the centre of the Earth, with h being that object's height above the surface. So we're basically saying that h << R, and so we can ignore it. Therefore, we're saying that the gravitational field strength (i.e. acceleration) at height h is basically the same as the gravitational field strength at the surface (i.e. they are both g).

That's why we can pretend the force is constant in magnitude. But why can we pretend it is constant in direction (even though it is always radial, and the "radially inward" direction changes as you move along the surface of a sphere)? The answer is because, close to the surface of the Earth, the curvature is so slight that even if you move a distance along the surface, the "radial" direction is not much different from what it was it was where you started. So we can just pretend it hasn't changed direction at all, and set the radial direction to be constant and equal to the "y" direction. This is the same as pretending that Earth is flat.

Newton's second law then gives:

$$\textbf{F} = m\textbf{a}$$

$$-mg\hat{\textbf{y}} = m\textbf{a}$$

$$-g\hat{\textbf{y}} = \frac{d^2 \textbf{r}}{dt^2}$$

This is the equation of motion of the particle. Here, r(t) = (x(t), y(t)) is the position vector of the particle. It varies with time, describing a trajectory. This equation can be separated into two differential equations (one for each vector component):

$$-g = \frac{d^2 y}{dt^2}$$

$$0 = \frac{d^2 x}{dt^2}$$

Their solutions are:

$$y(t) = y_0 + v_{0y}t - \frac{1}{2}gt^2$$

$$x(t) = x_0 + v_{0x}t$$

Here, v0x and v0y are the initial (t=0) speeds of the particle in the x and y directions respectively. Similarly, (x0,y0) is the initial position of the particle (at t = 0). If you use the second equation above to solve for t in terms of x, and then substitute that into the first equation above, you will obtain an expression for y as a function of x that is quadratic. As I'm sure you can understand, the curve of y as a function of x is the physical shape of the particle's trajectory.

syhprum
If one carefully measured the trajectory and found it to be parabolic it would be proof that the Earth was both flat and infinite in extent.