Earth's rotation having effect on sub-orbital trajectory

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  • #1
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If you throw a ball obscenely high (ignoring air resistance, etc), and the Earth is rotating, the ball will land in a different spot. Relative to an observer on the Earth, the ball has a sub-orbital trajectory across the surface. If we attach thrusters onto that ball, and have it burn at its top point until it achieves orbit, will its expended fuel be less than if the Earth wasn't rotating?

Taking this to the extreme, if the Earth was rotating very quickly, would shooting something out of the atmosphere be sufficient for it to be in orbit? Or would it follow a spiral-like trajectory across the surface? Is the reference frame of an Earth observer useless because it is non-inertial?
 

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  • #2
Bandersnatch
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If we attach thrusters onto that ball, and have it burn at its top point until it achieves orbit, will its expended fuel be less than if the Earth wasn't rotating?
That depends entirely on which direction you're going to make the thrusters fire - against or with the rotation.

Taking this to the extreme, if the Earth was rotating very quickly, would shooting something out of the atmosphere be sufficient for it to be in orbit?
There's no need for extreme changes in rotation. Providing you raise the object high enough, it can get in orbit with only the tangential velocity of the rotation.
Try calculating the radius at which centripetal acceleration provided by Earth's gravity is just right to curve the path of an object moving at the speed of Earth's surface at the equator.
 
  • #3
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Try calculating the radius at which centripetal acceleration provided by Earth's gravity is just right to curve the path of an object moving at the speed of Earth's surface at the equator.
As you wish: 1.84 * 108 km, given a resulting circular orbit. I suspect the height for an elliptical-but-stable orbit is less, but the calculation is harder.
 
  • #4
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As you wish: 1.84 * 108 km, given a resulting circular orbit. I suspect the height for an elliptical-but-stable orbit is less, but the calculation is harder.
I get 1.84 * 106 km ([itex]GM/(R\omega)^2[/itex])... 1.84 * 108 seems like a large number... 25% further than the distance between the earth and sun.
 
  • #5
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I had a conversation with my professor, and came to an understanding - within the reference frame of 'space', what's happening is the object being given tangential velocity equal to the Earth's rotation. This means that you're essentially throwing it at a vector relative to the Earth's surface. You can create as big a parabolic trajectory as you want (even reach escape velocity), but you can't get it into orbit from a single throw.
 
  • #6
A.T.
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within the reference frame of 'space', what's happening is the object being given tangential velocity equal to the Earth's rotation. This means that you're essentially throwing it at a vector relative to the Earth's surface.
What is "throwing at a vector"? Relative to the Earth's surface it is thrown vertically, with no tangential velocity. Relative to an inertial frame it has a tangential velocity equal to the Earth's rotation.
 
  • #7
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By 'vector' I mean the combined vector of the vertical and tangential velocity. Throwing something upward on a spinning globe is equivalent to throwing something at an angle on a non-spinning globe.
 
  • #8
jbriggs444
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It will not be a "parabolic" trajectory unless you happen to achieve escape velocity exactly. It could be...

A circular trajectory, if it were not for the fact that the launch angle has a vertical component, so it can't be that.

An elliptical trajectory. Which would intersect with the surface of the earth. So eventually your projectile will crash. That's the key problem with getting to orbit in a single throw.

A parabolic trajectory. If launched exactly at escape velocity.

A hyperbolic trajectory. If launched in excess of escape velocity.

A straight line (degenerate ellipse, parabola or hyperbola) if it were not for the fact that the [inertial-frame-relative] launch angle has a horizontal component, so it can't be that.
 

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