posted
A question has been bothering me for a while. My mind can have some pretty random thoughts when I space out, and on the trip to Bob and Dana's wedding, this one popped in.
Pretending that earth's atmosphere doesn't exist, so there is no wind resistance, if I shot a 10 pound bowling ball off the top of Mount Everest, how fast would it have to go to travel all the way around the earth and hit the same spot?
As an added bonus, factor it air/wind resistance now and tell me how far I could shoot the bowling ball (given that you can have any amount of speed) without it breaking the Earth's atmosphere?
As a side note: I only took one year of Physics and am in no way intelligent in this area.
Posts: 9754 | Registered: Jul 2002
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posted
Neglecting wind resistance, projectiles follow a parabolic path with respect to a frame of reference assumed to be flat. I'm not a physicist, I'm an engineer, so I would simply take the distance required to circumvent the globe in a given direction from wherever Everest is, and treat the projectile's trajectory as though it were going that distance over a flat surface. Simple, but probably not precisely correct. A good engineering approximation. Posts: 10886 | Registered: Feb 2000
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quote:Pretending that earth's atmosphere doesn't exist, so there is no wind resistance, if I shot a 10 pound bowling ball off the top of Mount Everest, how fast would it have to go to travel all the way around the earth and hit the same spot?
I don't think this is possible. Gravity is going to be pulling it down no matter how fast its going, at the rate of 9.8 meters/sec^2. Even a bullet falls at that rate, no matter how fast its shot. You'd have to angle the ball above 180 degrees. So at the very least, angle needs to be considered. Then you'd need to factor in escape velocity from earth's gravity well.
posted
But if the bowling ball travels fast enough, the arc of it's flight will be over the horizon. This is basic in orbital mechanics, no matter how low.
Posts: 410 | Registered: Apr 2005
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posted
Phil, remember that the earth is round. Therefore, if the cannonball is travelling fast enough, by the time it falls, say, an inch, it will have traveled far enough horizontally that the curvature of the earth will have fallen an inch.
posted
Hmmm... might not be so low after all. According to Wikipedia, satellites in low earth orbit (350km to 1400 km) travel at about 27400 km/hr, which is about 17025.5 mi/hr.
Posts: 4534 | Registered: Jan 2003
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posted
I've started doing that again since leaving the defense industry, now that no one will think I'm a communist for using SI units.
Posts: 4534 | Registered: Jan 2003
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Figuring it out with air resistance becomes almost impossible, even assuming constant air resistance, because the path would no longer be spherical, or even elliptical due to constant decrease in velocity.
But if we assume the object traveling through the air is perfectly laminar, then I can figure it out.
posted
Well, I hate to burst any bubbles but I have to disagree with you scholars on this one.
Websters defines atmosphere as:
the gaseous envelope of a celestial body (as a planet) b : the whole mass of air surrounding the earth
SO, if there were no air (so T_Smith would not encounter any resistence, wind etc.) then T-Smith could not throw the bowling ball since we'd all be dead.
Last time I checked I don't breath too well without air, although some nights under the bridge it come awful close...
posted
I dont think T_Smith can throw a bowling ball 17716.9 mi/hr with or without an atmosphere. Or spacesuit.
Posts: 2756 | Registered: Jul 2002
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posted
One of the orbital mechanics problems that Newton solved was for an isolated non-spinning astronomical body with no atmosphere. An object launched for a ballistic flight -- ie no artificial acceleration or deceleration during the flight itself -- at any velocity above minimum orbital velocity and below escape velocity will cause the object to hit the launchpoint after one orbit.
For such an Earth-mass"planet", it would take ~87minutes30seconds to make the flight in the minimum "ground-skimming" orbit. And if the launch is fast enough to reach the altitude of Moon-orbit, it would take ~27days8hours.
posted
aspectre, any speed in that range works? I'll have to think about that. Saxon75, there's a much simpler calculation for skimming orbit speed. According to Wiki, orbital speed would be the escape velocity divided by sqr(2), or 28500 km/hr, at sea level. This neglects rotation of the earth and the initial height of release. And no gutter balls.
With the height, maybe 28490.
quote: In the hypothetical case of uniform density, the velocity that an object would achieve when dropped in a hypothetical vacuum hole from the surface of the earth to the center of the earth is the escape velocity divided by √2, i.e. the speed in a circular orbit at a low height.
posted
There's technically no "range" of velocities that would work, only one will do it exactly, but if it's close then it will either lose or gain just a small amount of distance each rotation. Just as it's possible for satellites to have a degenerating orbit that ends up in them crashing to the earth, it's possible for them to be going to fast and end up spinning away from the planet.
quote: An object launched for a ballistic flight -- ie no artificial acceleration or deceleration during the flight itself -- at any speed between minimum orbital velocity and escape velocity will cause the object to hit the launchpoint after one orbit.
From this description, it sounds like the object would travel in an orbit, at minimum, or a cardioid shape, gaining altitude for half its journey and then losing altitude until it hits it's launch point.
But part of the description is missing. Are we assuming the object is fired tangentially to the earth? Must be. Then perhaps rather than a cardioid shape, the minimun orbit would be circular, while adding velocity would extend the orbit to make it elliptical, with the launch point at perigee.
But the original description, launching the object from Everest, means that a circular orbit would be at the altitude of Everest, while a velocity lower than what is needed for a circular orbit would create an elliptical orbit with the starting point at apogee.
Thinking out loud. Does this make sense, anybody?
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posted
I'm pretty sure that any velocity less than that required for a circular orbit will result in an unstable and therefore incomplete orbit.
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posted
Isn't the top of Everest already orbiting the earth's core? Can we just leave our bowling ball on the mountain and come back in 24 hours?
Posts: 2655 | Registered: Feb 2004
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posted
Another thinkg to consider is that an orbit has to travel through two antipodal points. So an object launched from everest would have to travel across the equator, where sea level is higher than it is at Everest's latitude.
So an object launched in a circular orbit from north or south of the equator would need to be launched high enough to clear the equatorial bulge.
Just another point.
And thinking, this all means that the L.E.M was essentially in orbit right up until just about the point where it landed, right?
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Note that the point farthest from the center of the Earth is not Everest, but Chimborazo in the Chilean Andes. This is due to the bulge at the equator.
Posts: 26071 | Registered: Oct 2003
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posted
Yeah, Rivka. That's a bookmark, for sure. Very cool.
However, don't try typing in a velocity. It looks like it accepts it, but it doesn't change the path unless you use the slide bar.
Posts: 3735 | Registered: Mar 2002
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posted
OK, if assumptions play into the theory of velocity etc., etc., etc., then we have to assume that the inexplicable disappearance of atmosphere from the earth did not equate to the disappearance of gravity from the earth. Assuming this is true, then we have to deduce that T-Smith cannot throw the bowling ball in any manner that would place it in orbit.
It is not air that ties matter to the earth, it is gravity.
So, assume that away.
I'll be on the bridge.
Posts: 65 | Registered: Mar 2005
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quote: I'm pretty sure that any velocity less than that required for a circular orbit will result in an unstable and therefore incomplete orbit
It shouldn't matter. If a ball were shot from the just above ground from the antipodal point from everest, at a speed that created an elliptical orbit that just skimmed everest, (that's an orbital velocity greater than circular) it would have the same effect as launching another ball from everest at the same speed at which the first ball skimmed by. This velocity would be lower than what is necessary for a circular orbit at everest's altitude.
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posted
Neglected to mention that it's also assuming that the initial trajectory -- velocity is speed&direction -- won't result in a collision with the planet/etc from which the object is launched before the orbit is completed.
However, Newton's results are actually even cuter, Morbo. Assuming a non-rotating, non-moving uniform spherical body in a vacuum, any object dropped from the surface into a straight tunnel that passes through the planet/etc takes the same amount of time to resurface at the other end of the tunnel, then return to the original drop point as the amount of time it takes for the minimum ballistic "ground-skimming" orbit. eg Eliminating friction, an object accelerated&decelerated by gravity alone sliding in a straight tunnel between NewYorkCity and Boston then back to NewYorkCity would take the same amount of time as an object sliding between NewYorkCity and Sydney,Australia then back to NewYorkCity : the same ~87minutes30seconds as that minimum ballistic orbit.
So if an object could pass through a planet/etc without interaction other than purely gravitational -- ie through a trajectory-matching tunnel -- even an object with a trajectory which intersects with the planet/etc would have the same orbital time as an object on a non-intersecting trajectory which reaches the same maximum height above the surface.
posted
"Neglected to mention that it's also assuming that the initial trajectory -- velocity is speed&direction -- won't result in a collision with the planet/etc from which the object is launched before the orbit is completed"
Hence Mount Everest. I figured that would negate that problem, but I could be wrong.
Posts: 9754 | Registered: Jul 2002
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posted
Hum... I think throw a bowling ball from the up of Everest is dangerous : It can occure an avalanche, or threaten somebody down to you. This is not a good attitude In addition to that, you want to retire the atmosphear !! You want to kill us ?
Nonono ! Stop it now ! And give me your ball ! That is confiscate !
Posts: 1189 | Registered: Dec 2004
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posted
BTW: That java cannon-launch program is deceptive if ya don't know what their assumptions are. Haven't done the calculations myself, but just eyeballing the picture, it appears that they assume the mountain is ~1100kilometres/660miles high. And the speed gain from that drop from mountain height to the surface has to be factored in with the initial speed, as well as the lower gravity at that mountain height. In this case, the minimum non-intersecting initial orbital velocity of ~15,524mph/~25,000kph from the mountain is about ~1500mph/~2400kph less than the initial minimum non-intersecting orbital velocity of an Mt.Everest launch. And the escape velocity would be comparably lower from that mountain.
posted
Sorry, Choobak, we have to remove the atmosphere. For Science. Otherwise the 10pound/4.5kilogram ball would explode upon hitting the air at 27,428kph/7.67kps with the force of ~17kilograms/37.4pounds of TNT. And if the bowling ball exploded upon contact with air, it would ruin the experiment. Which is why the directional component of velocity's speed&direction is important, T_Smith. Even assuming a vacuum at the top of Mt.Everest, if mackillian threw the bowling ball at the ground, the resulting impact explosion would damage the scenery.
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