Free fall is parabolic or elliptic?

[davey2222]

When someone tosses an object into the air, the trajectory is a parabola as I have learned in Physics.

 

But why do planets follow ellipses? I ask this since both the object and planets are being pulled by gravitational force.

[Janus]

Technically, the object tossed into the air also follows an elliptic trajectory. The reason that we treat it as a parabola is that it is that in most cases we can ignore the curvature of the Earth and its rotation. This simplifies the problem without introducing too large of an error. It also makes it easier to teach the basic concepts to start off. For long range guns and ballistic missiles, you have to use the more accurate (and complex) method.

[elfmotat]

When you're near the surface of the Earth [imath]g=GM/r[/imath] is approximately constant. (Small compared to the Earth) objects follow parabolic paths if g=constant. Over a global region, such as in planetary orbits, (small compared to the sun) objects follow ellipses.

Edited April 11, 2012 by elfmotat

[ewmon]

It's also a simplification that the Earth's surface for short trajectories is very nearly flat, thus the gravity vectors are very nearly parallel (even though they're converging toward a center ~4,000 miles away), and you can model the trajectory as a parabola and obtain almost perfect results.

Edited April 11, 2012 by ewmon

[mathematic]

In general ,the two body problem has as a solution any conic section. Planet paths are ellipses, comets may be any.

[Janus]

Just as a demonstration as to why we simplify, here's how you'd solve a problem of how far a object dropped from a height of 10m with a horizontal velocity of 10 m/s would be solved.(we'll ignore the rotation of Earth for this example)

 

Simple, less accurate method:

 

[math]D= \frac{at^2}{2}[/math]

 

Using a radius of 6378 km for the radius and 6e24 kg for the mass of the Earth, we get an "a" of 9.842452905 m/s^2

 

Solving for t gives us a fall time of 1.42548721s

 

meaning the object will travel a horizontal distance of 14.2548721m.

 

Harder, more accurate method.( I'll skip entering the numbers and just give the final answer at the end)

 

First, we find the total energy of the object by

 

[math]E = \frac{v^2m}{2}- \frac{GMm}{r}[/math]

 

We then equate this to the formula

 

[math] E = -\frac{GMm}{2A}[/math]

 

Which is another expression that gives the energy but uses the term A, which stands for the semi-major axis of an orbit.

 

By equating the equations, the "m"s drop out and we can solve for A.

 

With A and the apogee (Ra) of the orbit (10 m + the radius of the Earth),

 

we can use

 

[math]R_a = A(1+e)[/math]

 

to solve for "e" the eccentricity of the orbit.

 

now we can use the formula:

 

[math]r = \frac{A(1-e^2)}{1+e \cos q}[/math]

 

Here r is the distance from the center of the Earth of the object at an angle q from the perigee. Basically, we are looking for q when r= the radius of the Earth and the object's orbit intersects with the ground. We do this by re-arranging the above formula to solve for q.

 

If you solve for q in radians, we can then subtract this answer from pi and get the number of radians between apogee (our initial release point) and the impact point with the surface of the Earth.

 

Multiplying this answer by the radius of the Earth gives the distance traveled over the surface of the Earth by the object between release and impact.

 

If we do all of this for the stated problem we get an answer of 14.25467366 m, or an answer just about 0.2 mm different from what we got with the less accurate method. Hardly worth it for this problem.

Edited April 12, 2012 by Janus

[davey2222]

Thanks guys

[imatfaal]

Nice post Janus - really puts it into perspective to see the sums. To add to your point - this is a variation of 0.2 parts in 14000, 1 in 70,000; at present we only know G to 1 in 8300, and the radius of the earth isn't even anywhere near that good. I am not a scientist; but something that seems to run through science is an knowledge of when to approximate and when to go the whole nine yards.