4 ms·

Parabolas and ellipses are different conic sections. Objects thrown are actually following a parabolic trajectory and not an elliptic one. It’s not an approxima

Parabolas 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 ago
I 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 ago
  Are 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 ago
    No 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.
    - albrewer 4y ago
      According to their link, the error at a human scale is about 1e-6 units on Earth. You need to be doing the math at velocities such that the object's apogee is over 100k feet / meters from the surface to be "off" by more than 1 unit.
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ahelwer 4y ago

I 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

deleted 4y ago
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albrewer 4y ago
Are 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 ago
  No 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.
  - albrewer 4y ago
    According to their link, the error at a human scale is about 1e-6 units on Earth. You need to be doing the math at velocities such that the object's apogee is over 100k feet / meters from the surface to be "off" by more than 1 unit.
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