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I was hoping to get help with both a specific example and a general idea of how to do the problem.

Here's my current understanding, which may or may not be correct: When objects are thrown from a moving object, they have the speed of the object they are moving at, as well as the speed they were thrown with, which can be combined with vectors. Assuming no other forces besides gravity act upon the object, this means that an object thrown straight up will land exactly where it was thrown from.

Here's the parts I don't understand: This idea seems to break down at higher altitudes. At a sufficiently high altitude, well into space, an object's orbital circumference would be much greater than the circumference than Earth. Going by the logic that the object was previously on Earth, and therefore had a velocity that traveled one times the circumference of the Earth per day (let's use 1670 km/h, since that's what it is at the equator), it would then stand to reason that when this object was thrown directly up, it would be still traveling 1670 km/h (if we're ignoring upwards and downwards motion). Therefore, if we say it comes down to Earth one day after it was thrown, it would have failed to complete one full orbit around the Earth (with it being at a larger circle (the orbit) (and with that, a larger circumference) yet traveling at 1670 km/h.), yet the Earth would have completed one full rotation. With the object not completing a full revolution around the Earth, and the Earth completed a full revolution around it's axis, it seems like the object would land in a different place than it was thrown.

For example, let's say that an object was thrown up, and for simplicity's sake, teleports to an orbit exactly twice the circumference of Earth and will come directly down after one day. Moving at the speed of Earth's rotation, it should complete half its orbit, since the orbit has twice the circumference of Earth. In the same time, the Earth completes a full rotation. It seems like when the object comes down, it will come down at the exact opposite side of the world from where it's launched.

This only gets more confusing when we add the projectile's variable height in: objects can't teleport as in the above example, so if an object traveled normally to an orbit twice the circumference of Earth and immediately started falling after it reached that altitude (all governed by the formula $ h(t) = -1/2 g * t + vt + h$, assuming no other propulsion and that h = 0), constantly changing heights would mean anyone trying to calculate the path that the object was taking relative to Earth would have to calculate circumferences for every height between the upper and lower bounds, as well as how much the object traveled those circumferences, which seems like quite the task.

The general question was for a way to solve the problem above: how can one figure out where a projectile will land given the upwards velocity launched from Earth, when the projectile will fly substantial distances from the Earth? Drag and other affecting forces besides gravity need not be accounted for.

The more specific question is this: a projectile is being fired at a variable direction at a speed of eight miles per second. The projectile needs to land five miles from the point of launch (with precision to +- one mile), and come down at least 48 hours (but preferably not more than a month) later. What angle should the projectile be fired at? Is 8 miles a second fast enough to leave the projectile suspended for 48 hours, and if not, what speed would be applicable? (Note: I was informed that an object fired at 8 miles a second will leave Earth's orbit. What speed, then, would be applicable to accomplish this objective?) Like above, drag and any force besides gravity does not need be be added.

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  • $\begingroup$ If your projectile is launched at 8 miles per second, it's not coming back down. Ever. $\endgroup$ – Mark H May 16 '17 at 7:05
  • $\begingroup$ Ah, thanks for that. I edited the question to reflect that. $\endgroup$ – lbs21 May 16 '17 at 7:08
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Even supposing a constant downwards gravitation acceleration if you do the math of your problem you get a four degree polynomial (parabola-circumference intersection): Suppose you are on a latitude with the radius $r$ to the rotation axis, and $\omega$ earth angular speed. You throw the corp vertically with speed $v$. We consider an inertial system with center on earth axis and same latitude, y axis in the vertical direction the corp is thrown and x toward the tangent speed $r\omega$. The corp motion law in this system is $y(t)=-\frac 12gt^2+vt+r$, $ x(t)=\omega rt$ so if you want to know where it lands you have to impose that it is on the circumference: $(-\frac 12gt^2+vt+r)^2+ \omega^2r^2=r^2$, which is a 4 grade equation. Suppose $\lambda_0$ to be the root we are interested in, then you can find the landing position relative to the starting one in radiant $\Delta\theta=\tan^{-1}\frac{y(\lambda_0)}{x(\lambda_0)}-\omega\lambda_0$. If you consider very long distance you should even consider the change in the direction of the gravitational force, so even more complex (supposing this can be done without escaping earth gravitational field).

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