John Cuthber Posted July 26, 2017 Posted July 26, 2017 15 hours ago, ivylove said: "The ISS is in a state of continuous free-fall." What is This? I this statement a joke? (since someone brought up the centripetal force) I could use a laugh. If you don't think it's falling, then something must be keeping it from doing so. What do you think is holding it up? Skyhooks? 1
ivylove Posted July 28, 2017 Author Posted July 28, 2017 (edited) Using a space station orbital radius of 4.04 x 10^5 m and 2 km/month orbital decay, the free fall rate of the international space station would be approximately 0.00027 mph. Also, it should be: "If it was the centripetal force that was causing the weightlessness of the 50 kg astronaut then the angular velocity would be constant" Edited July 28, 2017 by ivylove -1
imatfaal Posted July 28, 2017 Posted July 28, 2017 56 minutes ago, ivylove said: Using a space station orbital radius of 4.04 x 10^5 m and 2 km/month orbital decay, the free fall rate of the international space station would be approximately 0.00027 mph. Also, it should be: "If it was the centripetal force that was causing the weightlessness of the 50 kg astronaut then the angular velocity would be constant" Orbiting at a radius of 404km? - that's a neat trick around a planet with a radius of 6300km. You mean orbital height That measure you give would be the orbital decay - which is caused by friction from the very rarefied atmosphere in the lower orbits. If anything it is a measure of the tiny amount by which the ISS is not quite in free fall - but the size of the adjustments required mean that we are safe in saying that ISS is in freefall. Alternatively: Can you not just look up freefall on wikipedia, read that it is the state when the only force acting on an object is gravity, then stop arguing? 1
studiot Posted July 28, 2017 Posted July 28, 2017 6 minutes ago, imatfaal said: Orbiting at a radius of 404km? - that's a neat trick around a planet with a radius of 6300km. You mean orbital height That measure you give would be the orbital decay - which is caused by friction from the very rarefied atmosphere in the lower orbits. If anything it is a measure of the tiny amount by which the ISS is not quite in free fall - but the size of the adjustments required mean that we are safe in saying that ISS is in freefall. Alternatively: Can you not just look up freefall on wikipedia, read that it is the state when the only force acting on an object is gravity, then stop arguing? Well spotted. I particularly like your alternative. +1
imatfaal Posted July 28, 2017 Posted July 28, 2017 (edited) 9 minutes ago, studiot said: Well spotted. I particularly like your alternative. +1 Without getting too mutually congratulatory - I really liked your write up of the Cavendish Experiment and it was the fact that this had fallen on deaf ears that made me a little waspish Edited July 28, 2017 by imatfaal gramma erra
studiot Posted July 28, 2017 Posted July 28, 2017 Thanks. I had considered the Wiki definitive statement a while back but had basically given up on the OP. I did hold back, jut now on a really waspish (but very funny) comment in the Huygens thread. I try to tell myself that maybe the OP is one of these people who has trouble communicating and deserves special consideration.
Strange Posted July 28, 2017 Posted July 28, 2017 1 hour ago, ivylove said: Also, it should be: "If it was the centripetal force that was causing the weightlessness of the 50 kg astronaut then the angular velocity would be constant" Only for a circular orbit: http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c5
ivylove Posted July 28, 2017 Author Posted July 28, 2017 Using a space station orbital radius of 6.771 x 10^6 m and 2 km/month orbital decay, the free fall rate of the international space station would be approximately 0.00027 mph. Also, it should be: "If it was the centripetal force that was causing the weightlessness of the 50 kg astronaut then the angular velocity would be constant" -1
Strange Posted July 28, 2017 Posted July 28, 2017 12 minutes ago, ivylove said: Also, it should be: "If it was the centripetal force that was causing the weightlessness of the 50 kg astronaut then the angular velocity would be constant" I guess if you are just going to repeat the same nonsense, I can only repeat the same reply: Only for a circular orbit: http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c5
ivylove Posted July 28, 2017 Author Posted July 28, 2017 (edited) "If it was the centripetal force that was causing the weightlessness of the 50 kg astronaut then the angular velocity would be constant" is a correction to the post----> Posted Tuesday at 05:18 PM (edited) · Report post Also, respectfully,was the free fall argument a motion to concede the centripetal force argument? Edited July 28, 2017 by ivylove
swansont Posted July 28, 2017 Posted July 28, 2017 40 minutes ago, ivylove said: Using a space station orbital radius of 6.771 x 10^6 m and 2 km/month orbital decay, the free fall rate of the international space station would be approximately 0.00027 mph. Also, it should be: "If it was the centripetal force that was causing the weightlessness of the 50 kg astronaut then the angular velocity would be constant" Freefall is not from orbital decay.
Strange Posted July 28, 2017 Posted July 28, 2017 (edited) 1 hour ago, ivylove said: "If it was the centripetal force that was causing the weightlessness of the 50 kg astronaut then the angular velocity would be constant" is a correction to the post----> It is still wrong. Quote Also, respectfully,was the free fall argument a motion to concede the centripetal force argument? I think it was just a futile attempt to explain some basic schoolboy physics to you. But you are unwilling/unable to learn so we are still going round in circles. Edited July 28, 2017 by Strange
imatfaal Posted July 28, 2017 Posted July 28, 2017 7 minutes ago, Strange said: It is still wrong. I think it was just a futile attempt to explain some basic schoolboy physics to you. But you are unwilling/unable to learn so we are still going round in circles. If we were going round in circles then it would be a constant angular velocity - but as we are going round in ellipses then it isn't.
Strange Posted July 28, 2017 Posted July 28, 2017 11 minutes ago, imatfaal said: If we were going round in circles then it would be a constant angular velocity - but as we are going round in ellipses then it isn't. And somewhat eccentric ellipses at that.
ivylove Posted July 28, 2017 Author Posted July 28, 2017 (edited) Let say that it is approximately a constant angular velocity for a one hour time interval. Now explain the effect (weightlessness) of the astronaut propagating in the opposite direction of the angular velocity and why a person that is propelled at a velocity of five mph, from the space station, towards Uranus propagates in the direction of Uranus in a straight line since the centripetal force would prevent such an occurrence. Maybe some sort of mistake such as mistaking Jupiter for the moon? Also, in your free fall argument if it is even a argument or maybe a joke, what free fall are you speaking of? (Maybe the fall during the elliptical orbit.) since if it wasn't for the decay of the orbit and space station would remain in orbit forever and ever. Is this not irreversible and definitive proof that women are smarter than men? "Freefall is not from orbital decay. " What then? (give me some scientific facts) Edited July 28, 2017 by ivylove -2
J.C.MacSwell Posted July 28, 2017 Posted July 28, 2017 23 minutes ago, ivylove said: Let say that it is approximately a constant angular velocity for a one hour time interval. Now explain the effect (weightlessness) of the astronaut propagating in the opposite direction of the angular velocity and why a person that is propelled at a velocity of five mph, from the space station, towards Uranus propagates in the direction of Uranus in a straight line since the centripetal force would prevent such an occurrence. Maybe some sort of mistake such as mistaking Jupiter for the moon? Well now. That's at least a hint of understanding.
Strange Posted July 29, 2017 Posted July 29, 2017 (edited) It has already been a cplainedvthat freefall means that the only force is that of gravity. With no sideways motion, an object in freefall will fall (freely) to the Earth. With sufficient sideways force it can enter a stable orbit. Because the orbital speed does not depend on the mass of the orbiting object, all objects at the same distance fall at the same speed. Therefore (with no force applied) there will be no relative motion. Hence the apparent weightlessness. Is your incoherent rambling about Uranus, perhaps, a reference to the Coriolis force? Edited July 29, 2017 by Strange
Janus Posted July 29, 2017 Posted July 29, 2017 2 hours ago, ivylove said: Let say that it is approximately a constant angular velocity for a one hour time interval. Now explain the effect (weightlessness) of the astronaut propagating in the opposite direction of the angular velocity and why a person that is propelled at a velocity of five mph, from the space station, towards Uranus propagates in the direction of Uranus in a straight line since the centripetal force would prevent such an occurrence. Maybe some sort of mistake such as mistaking Jupiter for the moon? Also, in your free fall argument if it is even a argument or maybe a joke, what free fall are you speaking of? (Maybe the fall during the elliptical orbit.) since if it wasn't for the decay of the orbit and space station would remain in orbit forever and ever. Is this not irreversible and definitive proof that women are smarter than men? "Freefall is not from orbital decay. " What then? (give me some scientific facts) A person moving at 5 mph relative to the space station, would not travel toward Uranus or anywhere else in a straight line they would just enter a different type of orbit around the Earth than the space station. If he were free to do so, (He wasn't inside the space station when he started.) this new orbit would take him just a bit further from the Earth than the space station before returning to the same distance from the Earth one orbit later. This orbit will also take a bit longer than the space station orbit. In order to permanently leave the vicinity of the Earth, he would have to increase his speed by another ~41 percent that of the space station itself (for the ISS this is 7.67 km/sec, so he would have to increase his speed by nearly 3.2 km/ sec. Even then, he wouldn't travel in a straight line, but on a parabolic trajectory, And once he got far enough away from the Earth that its gravitational effect was negligible, it would still be constrained to an orbit around the Sun, This drift with respect to the ISS would be hard to notice for an object inside the station. Imagine you are at one end of a 10 meter long section of the station and you toss an object from one end to the other at 5mph. The "outward" drift of the this object in the time it took to travel the length of the 10 meters works out to be just 2 inches. This is a smaller distance than likely to be caused just by the inaccuracy of the thrower. For a person traveling from one end to the other at typical speeds, his tendency to drift in towards the Earth or away from it is so small, you would need very sensitive measurements to notice it. It is a bit silly to argue that Newton's equation is wrong when there are countless examples that show its accuracy. Every satellite and space probe arrived where it is by virtue of precise calculations that are derived from the equation, If it was in error, these calculation would give the wrong answers and those craft would have ended up way off course instead of exactly where we planned to put them. 1
beecee Posted July 29, 2017 Posted July 29, 2017 15 minutes ago, Janus said: If it was in error, these calculation would give the wrong answers and those craft would have ended up way off course instead of exactly where we planned to put them. As long as we use the correct system of measurements and not confuse metric with Imperial.
John Cuthber Posted July 29, 2017 Posted July 29, 2017 19 hours ago, ivylove said: Using a space station orbital radius of 6.771 x 10^6 m and 2 km/month orbital decay, the free fall rate of the international space station would be approximately 0.00027 mph. ... The station isn't very much further from the centre of the earth than we are. (6771 km compared to 6,371) (about 6% further our) So the local free fall acceleration should be only slightly smaller (about 12 % less) than the acceleration near ground level- about 9.8 m/s/s So, why is the orbital decay rate not something like 8.7 m/s/s? What's holding it up?
John Cuthber Posted July 29, 2017 Posted July 29, 2017 16 hours ago, ivylove said: . Is this not irreversible and definitive proof that women are smarter than men? No, It shows that one person, who tacitly reports them self as female has made a monumental mistake. It says absolutely nothing about anyone else. 1
ivylove Posted July 31, 2017 Author Posted July 31, 2017 (edited) Using a radius of 6.771 x 10^6 m and a period of 5400 sec (1.5 hours), the centripetal force would be 458.4 N and the Newton's gravitational force is 438.4 N for a 50 kg astronaut which produces a 20 N force pointing away from the earth. Hi. My name is Ivy would you like a physics lesson? Edited July 31, 2017 by ivylove -1
swansont Posted July 31, 2017 Posted July 31, 2017 2 hours ago, ivylove said: Using a radius of 6.771 x 10^6 m and a period of 5400 sec (1.5 hours), the centripetal force would be 458.4 N and the Newton's gravitational force is 438.4 N for a 50 kg astronaut which produces a 20 N force pointing away from the earth. How did you determine the period? If you chose it arbitrarily, then it won't be a circular orbit. The conditions being discussed have the gravitational force and centripetal force being equal.
Klaynos Posted July 31, 2017 Posted July 31, 2017 12 minutes ago, swansont said: How did you determine the period? If you chose it arbitrarily, then it won't be a circular orbit. The conditions being discussed have the gravitational force and centripetal force being equal. The height and period are consistent with the iss. For the rest of the post I've not calculated it so wouldn't like to comment...
Klaynos Posted July 31, 2017 Posted July 31, 2017 6 minutes ago, Klaynos said: The height and period are consistent with the iss. For the rest of the post I've not calculated it so wouldn't like to comment... Sorry. I'm going to correct myself. On reflection given the dependence here on the orbital time you need to be quite accurate. Therefore 5559 seconds is what you need. This give 432 N. Using a more precise orbital height will deal with the rest of the difference between that and Newtonian gravity.
Recommended Posts