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- Thread starter nhmllr
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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.

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cepheid

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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.

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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?

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cepheid

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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/R

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:

[tex] \textbf{F} = m\textbf{a} [/tex]

[tex] -mg\hat{\textbf{y}} = m\textbf{a} [/tex]

[tex] -g\hat{\textbf{y}} = \frac{d^2 \textbf{r}}{dt^2} [/tex]

This is the

[tex] -g = \frac{d^2 y}{dt^2} [/tex]

[tex] 0 = \frac{d^2 x}{dt^2} [/tex]

Their solutions are:

[tex] y(t) = y_0 + v_{0y}t - \frac{1}{2}gt^2 [/tex]

[tex] x(t) = x_0 + v_{0x}t [/tex]

Here, v

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