3 ms·
Only tangentially related to your comment but it was recently pointed out to me that the trajectory of thrown objects is really an ellipse, not a parabola, sinc
by ahelwer 4y ago
Only tangentially related to your comment but it was recently pointed out to me that the trajectory of thrown objects is really an ellipse, not a parabola, since they're in a (very short!) orbit while airborne. Of course parabolas are a very, very good approximation over short distances and much easier to deal with mathematically than ellipses. Still, wish they'd mentioned this as an aside in physics classes.
- injidup 4y agoA parabolic trajectory would occur if the earth was flat.
- guerby 4y agoContext is 17th century so ball is not going very far and parabola will be perfect with the measuring technology of the time I think :)
- chongli 4y agoYes and since the earth is large enough to be locally flat a parabola is a great approximation of what happens when we throw a ball.
- ahelwer 4y agoI think it would also require that gravity doesn't obey the inverse-square law, so stays constant in magnitude throughout the entire vertical component of the trajectory. An infinitely-flat earth would also mean the force vector points straight down, instead of shifting slightly to point to the earth's center of mass across the horizontal component of the trajectory.
- WastingMyTime89 4y agoParabolas and ellipses are different conic sections. Objects thrown are actually following a parabolic trajectory and not an elliptic one. It’s not an approximation. They are not in orbit. Objects are actually in orbit when their trajectory shift from a parabola to an ellipse. If you go even faster, you get another conic section as a solution - an hyperbola - and the object escape.
- ahelwer 4y agoI don't buy this. A thrown object is in orbit around the earth's center of mass if one were to treat it as a point mass. Since the earth is (roughly) spherical its gravitational pull is equivalent to being a point mass. The thrown object doesn't "know" that it will impact the surface of the earth and thus switch between parabola/ellipse. You can read some of the answers here for a better idea: https://physics.stackexchange.com/q/373250 https://physics.stackexchange.com/q/373250
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- albrewer 4y agoAre you contesting the fact that a thrown ball's distance above the ground (y) at a particular time (t) relative to its initial velocity (v) is equal to y = vt + 0.5 * g * t ^ 2, (where g is the direction of acceleration due to gravity) which can very clearly a modeled by a parabola? I get that the "error" is pretty low, but there are tons of equations in physics used every day that rely on the fact that the difference in some value is negligible (often things like the sine of an angle being approximately equal to the angle itself).
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- WastingMyTime89 4y agoNo but I get what they mean. If you view the Earth as a point system, you can use the general expression of gravity between two objects and you will indeed find an extremely elongated ellipse as a solution. Considering a parabola can be viewed as an ellipse with a focal point at infinity, I’m guessing the difference between the two solutions will be so small as to be negligible. They both depend on intentional inexactitudes (g is not a fixed vector, the earth is not a point system) which don’t matter at the scale. I wouldn’t personally view the elliptical solution as truer but I see how you could.