tar Posted May 10, 2015 Posted May 10, 2015 pzkpfw But when you say the space guy will go off at a tangent to the circle when the wall of the ship blows out, what circle are you considering? The circle along which his head is traveling, or the circle along which his feet are traveling? I understand the principle, and I have let go of spinning ropes and devices that throw tennis balls, and I have thrown pitches and such, where my hand was traveling in an arc and the ball certainly does go off at a tangent to an arc, but it is not nescessarily a tangent to a circle centered at the shoulder, or one centered at the wrist or one centered at the 1st, 2nd or third knuckle. The directional flight of the ball, and its speed and its spin varies depending on the grip and pressure of the fingers and the order of release from the various points of contact. The momentum of the ball can be redirected onto various circular paths depending on the pressure points. Sometimes the straight line momentum of the ball is even tranferred to a rapid spin of the curve ball. I was watching an episode of St. Genius today where the host filled a long plastic daiper bag with a single breath by blowing into it from 6 inches away, lowering the pressure at the mouth of the bag and letting the air pressure of the surrounding air fill the bag. It made me think of the "lift" imparted on a plane, when the speeding air over the larger top curve of the wing, lowers the pressure on the top and lifts the plane, radially away from gravity. A tangential force applied, with the plane traveling along the curvature of the Earth, that lifts the plane radially away from the Earth. Bernolli's principle did not make sense to me, when first described. How is it possible that increasing the speed of a fluid traveling in a tube with a hole in it would lower the pressure at the hole? But it was demonstrated to me, and less fluid came out of the hole when the speed of the fluid inside the tube was increased, than when the fluid was slowed down. My speculation here, is that the momentum of an object includes both the tendency to move in a staight line, and the tendency to remain stationary and the inertia of an object is complex as the outside forces acting on it, and the inherent motion it has taken on from its history, it internal stresses and strains and elastic responses and such, conspire to not only look like a force, but to be a force that actually causes motion in a certain direction. Finger tip bowling balls have a weight block inside, with a blob of denser material stategically placed, and how you release the ball, and cause it to roll or slide along the lane with the weight block on the outside of the roll, or the inside, or with the block circulating above the equator or below, or crossing the equator at the right time, etc. can cause the ball to "flip" in a manner where it explodes into the pocket, storing inertia and releasing it, just at the right time, and in the right direction, to get really good pin action, and a higher probability of a strike. The picture of the path of a position of a spoke on a bicycle, shown earlier in the thread, illustrates this, with the rolling wheel, not causing the position on the spoke to following a circular path. So the space station inhabitant's feet and head, are traveling at different speeds, around circles with different radii. Where each would go when the wall of the ship blows out, may be tangential, but tangential to what is important. It might make difference to which direction the inhabitant flies on in, if she is walking or running or jumping or bending or stretching, this way or that, when the thing blows out. For instance, if she were to jump, just before the blow, the floor of the ship is not even touching her. How could it be applying any centripedal force? Regards, TAR
Robittybob1 Posted May 10, 2015 Posted May 10, 2015 (edited) It's lacking an incorrect analysis, which would apply to anything that cited a centrifugal force acting on the object. If an object is in orbit, the centripetal force is the gravity. There is no other force. If the object is not in orbit, the centripetal force is the difference between gravity and the normal force (which are in opposite directions) The centripetal force is less than the gravitational force. "Use" was in quotes. Don't take analogies literally. They are analogies. Since I'm only discussing forces acting on the object, I am not talking about that. It is irrelevant. When I looked at the definition of the Normal Force it was not always in the same or opposite direction as the gravitational force. So it maybe only a component of the normal force. http://en.wikipedia.org/wiki/Normal_force In mechanics, the normal force is the component, perpendicular to the surface (surface being a plane) of contact, of the contact force exerted on an object by, for example, the surface of a floor or wall, preventing the object to fall. With your definition can the weight be calculated by (m*gr), that is the residual g force (gr) not accounted for by the centripetal force? What is the reason centripetal force reduces the effectiveness of the gravitational force? Looking at the title of the thread "Centrifugal forces ' appear ' to act opposite to gravity . How is this possible?" really you seem to have it "Centripetal forces ' appear ' to act opposite to gravity . How is this possible? So it was close! Edited May 10, 2015 by Robittybob1
swansont Posted May 10, 2015 Posted May 10, 2015 When I looked at the definition of the Normal Force it was not always in the same or opposite direction as the gravitational force. So it maybe only a component of the normal force. http://en.wikipedia.org/wiki/Normal_force For the conditions described in this problem, the normal force will be opposite gravity; you don't generally place a scale on an incline. With your definition can the weight be calculated by (m*gr), that is the residual g force (gr) not accounted for by the centripetal force? Probably, i.e. if you mean what I think you mean. What is the reason centripetal force reduces the effectiveness of the gravitational force? I've answered this already. Fnet = Fcentripetal = N - mg If you don't understand the answer, you have to ask a more specific question, rather than re-stating the question in a slightly different way. Looking at the title of the thread "Centrifugal forces ' appear ' to act opposite to gravity . How is this possible?" really you seem to have it "Centripetal forces ' appear ' to act opposite to gravity . How is this possible? So it was close! But centripetal forces don't act opposite to gravity.
Janus Posted May 10, 2015 Posted May 10, 2015 (edited) pzkpfw But when you say the space guy will go off at a tangent to the circle when the wall of the ship blows out, what circle are you considering? The circle along which his head is traveling, or the circle along which his feet are traveling? A tangent along the circle through which his center of mass is traveling. He will also spin around his center of gravity to maintain the angular momentum with respect to that COM before the ship blew out. It might make difference to which direction the inhabitant flies on in, if she is walking or running or jumping or bending or stretching, this way or that, when the thing blows out. For instance, if she were to jump, just before the blow, the floor of the ship is not even touching her. How could it be applying any centripedal force? Regards, TAR Here's what things look like for a both a person standing still and jumping as seen from both a non-rotating frame and the frame rotating with the Ship. 1a shows the person standing still when the wall gives out as seen in the non-rotating frame. The red dots are his position during the moments afterward and the blue line marks the point on the wall he was standing when it blew. 1b is the same situation as seen from the rotating ship frame. 2a shows the path of a jumping person from the non-rotating frame. He travels in a straight line that is the vector sum of his tangential velocity and his jump velocity. This will cause him to intersect the wall at some point. If the ship falls apart after the jump and before hitting the wall, he continues in a straight line. 2b show that same path as seen from the rotating ship. Even though he tries to jump directly towards the center of the ship, he will not come down at the same point of the wall. If the ship falls apart after his jump, he continues on along the path shown. If he is running walking jumping, etc., he will continue in a straight line in the direction his CoM was moving at the time. Edited May 10, 2015 by Janus
tar Posted May 10, 2015 Posted May 10, 2015 Question to the thread. If you are on a space station that is designed with radius and is rotating with a period that would provide an artifical gravity of 1G, and you were to jump one meter into the air, would you come down onto the same spot on the inside of the hull that you launched yourself from? Regards, TAR
Janus Posted May 10, 2015 Posted May 10, 2015 Question to the thread. If you are on a space station that is designed with radius and is rotating with a period that would provide an artifical gravity of 1G, and you were to jump one meter into the air, would you come down onto the same spot on the inside of the hull that you launched yourself from? Regards, TAR The post I was writing as you posted this answers that question.
tar Posted May 10, 2015 Posted May 10, 2015 Thank you Janus, Those pictures are very instructive. We cross posted, but you answered my previous question about where the jumper would land. However, I am now considering what happens to the hull when the ship falls apart. Let's construct the station in 24 peices, each 1/24th of a circumference long with one of our crew of 24 standing on the center of their designated section. At the same instant all the bolts that were holding the segments together, fail (melt, dissolve, crumble to dust, what ever, but their faliure is instantaneous, imparts no force on either section and the bolts themselves are no longer present.) The astronaut and the floor would have no particular reason to separate and both the astronaut and the segment would continue in the straight line consistent with the tangent to the circle that the center of mass of the astronaut and the segment were on at the time of the failure of the bolts. From the rotating frame the segments would be moving away from a center point at a steadily increasing radius (radial motion?) From the non-rotating frame the segments would be moving away form a center point at a steadily increasing radius (radial motion?) If the center of mass of all the separated parts is considered the center, does not the tangential motion of each of the segments, when looked at together, from either the inertial or non-inertial frame, appear to be a bunch of stuff undergoing a radially outward movement, as the circumference of the circle the masses are occupying grows? Regards, TAR
Robittybob1 Posted May 10, 2015 Posted May 10, 2015 (edited) But centripetal forces don't act opposite to gravity. If you rearrange Fcentripetal = N - mg you get Fcentripetal + mg = N Now when the mass is not moving wrt the coordinates mg = N Move mass wrt to coordinates and Fcentripetal > 0 (always greater than 0) Does mg stay constant? If so N force would increase not decrease as is the case by experiment. A.... 2b show that same path as seen from the rotating ship. Even though he tries to jump directly towards the center of the ship, he will not come down at the same point of the wall. If the ship falls apart after his jump, he continues on along the path shown. .... (Just within the ship) So do you think he will land ahead of where he jumped from? Edited May 10, 2015 by Robittybob1
Mike Smith Cosmos Posted May 10, 2015 Author Posted May 10, 2015 (edited) Thank you Janus, Those pictures are very instructive. We cross posted, but you answered my previous question about where the jumper would land. However, I am now considering what happens to the hull when the ship falls apart. Let's construct the station in 24 peices, each 1/24th of a circumference long with one of our crew of 24 standing on the center of their designated section. At the same instant all the bolts that were holding the segments together, fail (melt, dissolve, crumble to dust, what ever, but their faliure is instantaneous, imparts no force on either section and the bolts themselves are no longer present.) The astronaut and the floor would have no particular reason to separate and both the astronaut and the segment would continue in the straight line consistent with the tangent to the circle that the center of mass of the astronaut and the segment were on at the time of the failure of the bolts. From the rotating frame the segments would be moving away from a center point at a steadily increasing radius (radial motion?) From the non-rotating frame the segments would be moving away form a center point at a steadily increasing radius (radial motion?) If the center of mass of all the separated parts is considered the center, does not the tangential motion of each of the segments, when looked at together, from either the inertial or non-inertial frame, appear to be a bunch of stuff undergoing a radially outward movement, as the circumference of the circle the masses are occupying grows? Regards, TAR Tar , I know your model is the plastercine and the pick sticks with blobs . But am I to understand, with your encouraged diagrams from Janus . That :- If one sets up a ROTATIONAL ACCELERATING FRAME, rotating about a central point , where at a radius of approx 6100 Kms with a rotational speed at that 6100 km orbit of 17,700 mph . That , working on a (reflected , reactional ,fictitious ? ) value of centrifugal force , to gravity. In other words , centrifugal force , be it apparent, fictitious , or not that :- the device , if it were in a tube, reaching from the centre of the earth , all the way to 100, Kms above the Earths surface , namely 6100 Kms. That The mass would rise up the tube , from the centre of the Earth to orbital height ? ( 6100 Kms from earth centre. ) Be it that it is not up a straight up a radius ( as one normally visualises a radius ) . But rather a drawn out radius, as per Janus's diagram? This must only be an earth size version ,of what goes on in a medical, chemical , test tube centrifugal separator( filter) . I am pretty certain a tube model with a steel ball starting at the centre , rotating , would see the steel ball , rising up the tube . Mike Edited May 10, 2015 by Mike Smith Cosmos
tar Posted May 10, 2015 Posted May 10, 2015 Thread, Segment number 18 (segments numbered 1-24 counter clockwise) will fly off on a tangent that will get closer and closer to the radial line drawn, at the time of the bolt failure, through segment 24. The segment (18) will get farther and farther from the radial line drawn through segment 18 at the time of the bolt failure (as the radial line drawn through segment 18 at breakup is orthogonal to the tangent line of 18 at time of break up, but parallel the radial line of segment 24. 18's tangent line will never reach 24's radial line but it will cross 19's, 20's, and 21's, 22's and 23's radial lines in pretty quick order (rough drawing looks like it would cross 23's radial line be it traveled a circumference (pi D) away in the radial direction. I will not ever cross 24's but it will look more and more like it is traveling in 24's radial direction since the direction is identical (the tangent of 18 and the radial of 24 are parallel, and only 1 r apart. So, initially at breakup the 24 segments are moving tangentially, but after about a period of previous rotation, the segments are going radially outward along the radial line of the segment that was 90 degrees clockwise at the time of breakup. After this initial period, anybody looking at the event would see only segments traveling radially outward from the center, where the space station was at time of breakup. Unless they knew which was number 18 and which was 24 it would not matter, it would look like they were all moving in a radially outward direction. There would be no rotation, either real or apparent, from either the inertial or non-inertial frame, after about a period...or so says my minds eye. Well I guess if you maintained the rotating non--inertial frame virtually, the segments would look like they were headed out in an ever widening spiral pattern, but after all the station broke up, so the rotating reference frame is no more. Regards, TAR Mike, No, I don't think it would rise up the tube you suggest. I think it would rise up the tube 90 degrees away in the direction of rotation. Which is, after all the tangential, straight line direction we are being told it should go in. But this would only be an opposite gravity direction on the equator, and it would only be close to parallel to radial, after a period of rotation or so, according to the 24 segment thought experiment. Regards, TAR
Mike Smith Cosmos Posted May 10, 2015 Author Posted May 10, 2015 (edited) Mike, No, I don't think it would rise up the tube you suggest. I think it would rise up the tube 90 degrees away in the direction of rotation. Which is, after all the tangential, straight line direction we are being told it should go in. But this would only be an opposite gravity direction on the equator, and it would only be close to parallel to radial, after a period of rotation or so, according to the 24 segment thought experiment. Regards, TAR But it would still rise up the tube though, whether it was tangential to radius or not, surely ! . ( which I sort of understand . ) but I think it is sort of oscillating into and out of 90 degrees as it ( is tangential, in orbit, is climbing non tangential, tangential, in orbit , non tangential , climbing . Etc ) though I must say not sure ? Mike Edited May 10, 2015 by Mike Smith Cosmos
Janus Posted May 10, 2015 Posted May 10, 2015 (edited) Thank you Janus, Those pictures are very instructive. We cross posted, but you answered my previous question about where the jumper would land. However, I am now considering what happens to the hull when the ship falls apart. Let's construct the station in 24 peices, each 1/24th of a circumference long with one of our crew of 24 standing on the center of their designated section. At the same instant all the bolts that were holding the segments together, fail (melt, dissolve, crumble to dust, what ever, but their faliure is instantaneous, imparts no force on either section and the bolts themselves are no longer present.) The astronaut and the floor would have no particular reason to separate and both the astronaut and the segment would continue in the straight line consistent with the tangent to the circle that the center of mass of the astronaut and the segment were on at the time of the failure of the bolts. Yes they would. The segment would travel along the tangent of its CoM and the Astronaut along the the tangent of his CoM Unless these CoMs are the same they will have different distances from the center of the ship and different tangential velocities and the astronaut and segment will separate. From the rotating frame the segments would be moving away from a center point at a steadily increasing radius (radial motion?) From the non-rotating frame the segments would be moving away form a center point at a steadily increasing radius (radial motion?) If the center of mass of all the separated parts is considered the center, does not the tangential motion of each of the segments, when looked at together, from either the inertial or non-inertial frame, appear to be a bunch of stuff undergoing a radially outward movement, as the circumference of the circle the masses are occupying grows? Regards, TAR No they do not. Here's what it will look like from both the non-rotating and rotating frames: The right image is the non- rotating frame. The CoM of each segment continues along a straight line while each segment maintains its original angular momentum with respect to its own Com. The left image shows the same thing from the rotating frame. The red lines in the right image shows the radial lines between where the center of one segment would be if the structure did not break up compared to where the segment ends up after the break up. As the segment increases its distance form the center the angular momentum of its CoM with respect to the center of the original structure must be conserved. Thus its angular velocity with respect to that center decreases and it travels a smaller angle than it would if it had maintained its original distance from the center. The angular speed and momentum around its own CoM remains unchanged and thus its orientation with respect to the original structure's center changes. Relative to the frame in the right image, neither the structure nor segments rotate. However the segments follow a curved path which causes the segments to change orientation as seen from the center of the structure. In neither case can you say that the segments appear to exhibit simple radial motion away from the center. Edited May 10, 2015 by Janus
swansont Posted May 10, 2015 Posted May 10, 2015 Move mass wrt to coordinates and Fcentripetal > 0 (always greater than 0) Sorry, why always greater than zero? The centripetal force is inthe direction of the net force, i.e. in the same direction as gravity. In this case, Fc is negative
tar Posted May 10, 2015 Posted May 10, 2015 Actually It will not get "closer and closer" to the radial. The tangent segment 18 goes on is exactly parallel to number 24's radial at the start and onward. They start out separated by one radius and since they are parallel lines, they will stay separated by one radius, no matter how long after the breakup, you look at it.
swansont Posted May 11, 2015 Posted May 11, 2015 Be it that it is not up a straight up a radius ( as one normally visualises a radius ) . But rather a drawn out radius, as per Janus's diagram? A spiral, perhaps? I am pretty certain a tube model with a steel ball starting at the centre , rotating , would see the steel ball , rising up the tube . Yes. As has been described several times in this thread. But absent an outward pushing force.
tar Posted May 11, 2015 Posted May 11, 2015 Janus, Thank you for the pictures and descriptions. It does differ from what my minds eye was envisioning, but I do agree that your model is correct and mine had the segment's maintaining their orientation to the the center. You showed about a quarter of a period. My speculation about where the motion would start to look more and more like radial motion was after a quarter period. Can you show us both frames during two complete periods. It is here that I am thinking the motion will look like rays from the center, and not spirals in the inertial frame. The rotating frame is pretty much gone, as soon at the bolts fail. Regards, TAR Janus, My thought being that each of the segments are pulling each other in toward the center, around in the circle, and once the bolts dissolve the circular motion is instantly gone. All that is left is tangential motion in a straight line. No material will follow a curved path in the inertial frame and all 24 observers are now in the same inertial frame and none are undergoing any acceleration due to a change in direction. The segments may each take on a rotation of their own, as seen by the other 23, and the other 23 thought of by the one as reference points might as well indicate this rotation to the one, but the CoM of each segment will be traveling in a straight line, not a curved path. All these straight lines have an origin at the original location of the spinning space station. What might look like a spiral during the first quarter period, should look more like a star after a couple periods. There are no forces causing anything to curve. Regards, TAR
Robittybob1 Posted May 11, 2015 Posted May 11, 2015 (edited) Sorry, why always greater than zero? The centripetal force is inthe direction of the net force, i.e. in the same direction as gravity. In this case, Fc is negative Is gravity a negative force too then? Why I said centripetal force is positive was that the formula Fc = m*v^2/r yields a positive value. Mass would normally be positive, v^2 would give a positive number whether the velocity was a negative or positive figure, and radius would the a positive value too. So how do you get a negative value to centripetal force? Reference to: http://www.scienceforums.net/topic/88420-centrifugal-forces-appear-to-act-opposite-to-gravity-how-is-this-possible/page-13#entry866983 A spiral, perhaps? Yes. As has been described several times in this thread. But absent an outward pushing force. Do you think if you blocked the end of the tube with a scale, and spin it up that the scale would not register any outward force? I'm picking there would be an outward force while the ball is in the tube. Edited May 11, 2015 by Robittybob1 -1
swansont Posted May 11, 2015 Posted May 11, 2015 Is gravity a negative force too then? Why I said centripetal force is positive was that the formula Fc = m*v^2/r yields a positive value. Mass would normally be positive, v^2 would give a positive number whether the velocity was a negative or positive figure, and radius would the a positive value too. So how do you get a negative value to centripetal force? Forces are vectors. In one dimension, you use the convention of + being in one direction and - being the other. The sign of the result tells you the direction. Typically + is up but the choice is arbitrary, and once decided upon must be applied consistently. Do you think if you blocked the end of the tube with a scale, and spin it up that the scale would not register any outward force? I'm picking there would be an outward force while the ball is in the tube. The scale would give a positive number, meaning it is pushing inward on the ball. The force on the ball is inward. For the nth time, the force exerted by the ball is irrelevant to the discussion. 3
Mike Smith Cosmos Posted May 11, 2015 Author Posted May 11, 2015 (edited) It is interesting , last night an horizon program BBC 4 , was interviewing major , popular scientists, One of which was Michael Kaku . As well as Mark Tegmark , etc they were discussing singularities , black holes , and the beginning of the Universe. All of which little if nothing is known about , apart from their effect. Gravity was an issue , as Einstein was said to be the first to mention black holes from his General Theory of Gravity / relativity. Michael Kaku was saying about gravity being distortions in space. "Remember " Kaku said " GRAVITY is NOT PULLING DOWN on a mass , towards the Earth, like on a string . But SPACE IS PUSHING DOWN! towards the centre of the earth . If this is the case then the reactive ( equal and opposite force ) is pushing up on the mass surely ? Mike Edited May 11, 2015 by Mike Smith Cosmos -1
swansont Posted May 11, 2015 Posted May 11, 2015 It is interesting , last night an horizon program BBC 4 , was interviewing major , popular scientists, One of which was Michael Kaku . As well as Mark Tegmark , etc they were discussing singularities , black holes , and the beginning of the Universe. All of which little if nothing is known about , apart from their effect. Gravity was an issue , as Einstein it sId was the first to mention black holes from his General Theory of Gravity / relativity. Michael Kaku was saying about gravity being distortions in space. Remember he said ..Gravity is NOT PULLING DOWN on a mass , towards the Earth, like on a string . Bur SPACE IS PUSHING DOWN! towards the centre of the earth . If this is the case then the reactive ( equal and opposite force ) is pushing up on the mass surely ? Mike Completely irrelevant to the discussion. Action/reaction forces and circular motion are introductory physics. How about trying to understand the basics before gallivanting off into concepts of advanced relativity?
Robittybob1 Posted May 11, 2015 Posted May 11, 2015 (edited) Forces are vectors. In one dimension, you use the convention of + being in one direction and - being the other. The sign of the result tells you the direction. Typically + is up but the choice is arbitrary, and once decided upon must be applied consistently. The scale would give a positive number, meaning it is pushing inward on the ball. The force on the ball is inward. For the nth time, the force exerted by the ball is irrelevant to the discussion. Forces are vectors. In one dimension, you use the convention of + being in one direction and - being the other. The sign of the result tells you the direction. Typically + is up but the choice is arbitrary, and once decided upon must be applied consistently. The scale would give a positive number, meaning it is pushing inward on the ball. The force on the ball is inward. For the nth time, the force exerted by the ball is irrelevant to the discussion. When the scales I weigh myself I have to press on then first to get a reading. The spring inside gets stretched and then it reads my weight. Only as the ball pushes on the scale will the scale push back on the ball. That would be a rather easy experiment to check the physics. Are you saying that a centripetal force changes sign depending on which sector it is in. That is news to me. That does not sound right. Can you show me a link that confirms that please? Edited May 11, 2015 by Robittybob1
swansont Posted May 11, 2015 Posted May 11, 2015 When the scales I weigh myself I have to press on then first to get a reading. Fine. Irrelevant. The spring inside gets stretched and then it reads my weight. Or compressed. Only as the ball pushes on the scale will the scale push back on the ball. Yes. The scale pushes on the ball. Got that? The scale pushes on the ball. Inward. That would be a rather easy experiment to check the physics. Yes, indeed. Feel free. Are you saying that a centripetal force changes sign depending on which sector it is in. That is news to me. That does not sound right. Can you show me a link that confirms that please? Changes sign? That depends on your coordinate system. In a polar coordinate system, the sign would always be negative, as it would point to the center (assuming you've defined out as being positive, which is the norm). In a cartesian system, it would be toward the origin, so negative when the corresponding coordinate is positive. Any elementary physics textbook that explains coordinate systems will confirm this concept. Though some of them may assume that you already know this from the prerequisite math class. http://en.wikipedia.org/wiki/Polar_coordinate_system
Mike Smith Cosmos Posted May 11, 2015 Author Posted May 11, 2015 (edited) A spiral, perhaps? Yes. As has been described several times in this thread. But absent an outward pushing force.. Yes, but if the tube extended into the sky , above the surface of the earth , the metal ball would continue to rise up the tube " as if by a fictitious centrifugal force ( or whatever else you want to call it ) ""Looking like"" it was being, pushing in a skyward mode " , is that not so ? ( as far as looking at, in , up and around the tube ) the ball in the tube would 'look like ' an ascending elevator . Edited May 11, 2015 by Mike Smith Cosmos
swansont Posted May 11, 2015 Posted May 11, 2015 . Yes, but if the tube extended into the sky , above the surface of the earth , the metal ball would continue to rise up the tube " as if by a fictitious centrifugal force ( or whatever else you want to call it ) ""Looking like"" it was being, pushing in a skyward mode " , is that not so ? ( as far as looking at, in , up and around the tube ) the ball in the tube would 'look like ' an ascending elevator . image.jpg Do we observe things ascending like that? Can you figure out why?
Mike Smith Cosmos Posted May 11, 2015 Author Posted May 11, 2015 (edited) Do we observe things ascending like that? Can you figure out why?.. Yes I figure out its possible, but not observed as everyday happening, because either the tube or the earths surface would need to be travelling at a circumferential speed of approx .17,700 mph . Which is not an easy , everyday occurrence . But in theory it is possible and with a bit of 'gigery pokery ' it could be arranged practically! Currently ,this is the sort of trajectory NASA uses to launch satellites. Except they use a, loose rocket ,rather than a vast tube ! Mike Edited May 11, 2015 by Mike Smith Cosmos
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