dgary@ecsvax.UUCP (D Gary Grady) (12/20/85)
I recently posted a puzzlement: If you're in the space shuttle orbiting 300 miles up, the acceleration of gravity very little less than it is at the surface. If weight = mass X acceleration, how can you be "weightless"? As I noted, the answer to this requires a somewhat more subtle definition than you normally see in general physics texts. In article <96@decwrl.DEC.COM> osman@sprite.DEC (Eric, DIGITAL, Burlington Ma. 617 273-7484) writes: >I believe the key to the answer is the word "orbit". If you're in orbit, >you're whizzing around the earth just fast enough so that your centrifical >force OUTWARD matches gravity's force INWARD. Centrifugal "force" is a fictitious force (measured in figmentive newtons, normally called "fig newtons," but that's beside the point). >Hence the total acceleration for you is zero, and hence you are indeed >weightless. But you must be accelerated, otherwise you would go in a straight line! >p.s. Does someone remember the details of the "thought" experiment > involving being in a windowless elevator, and trying to determine > whether you are accelerating or just plain heavy ? Well, the result of that thought experiment is called the General Theory of Relativity which starts by saying (speaking loosely) that you can't tell a difference because both possibilities amount to the same thing. Anyway, I'm glad you brought up the elevator (don't say a word, Henry). If you're in a falling elevator you're weightless too, aren't you? But the gravitational field is still the same, isn't it? Let me offer a hint: We need a better definition of weight, and possibly a better notion of acceleration, to solve the problem. But no deep dark physics is needed, just careful thought. -- D Gary Grady Duke U Comp Center, Durham, NC 27706 (919) 684-3695 USENET: {seismo,decvax,ihnp4,akgua,etc.}!mcnc!ecsvax!dgary
dgary@ecsvax.UUCP (D Gary Grady) (12/20/85)
I recently posted a puzzlement: If you're in the space shuttle orbiting 300 miles up, the acceleration of gravity is very little less than it is at the surface. If weight = mass X acceleration, how can you be "weightless"? As I noted, the answer to this requires a somewhat more subtle definition than you normally see in general physics texts. In article <96@decwrl.DEC.COM> osman@sprite.DEC (Eric, DIGITAL, Burlington Ma. 617 273-7484) writes: >I believe the key to the answer is the word "orbit". If you're in orbit, >you're whizzing around the earth just fast enough so that your centrifical >force OUTWARD matches gravity's force INWARD. Centrifugal "force" is a fictitious force (measured in figmentive newtons, normally called "fig newtons," but that's beside the point). >Hence the total acceleration for you is zero, and hence you are indeed >weightless. But you must be accelerated, otherwise you would go in a straight line! >p.s. Does someone remember the details of the "thought" experiment > involving being in a windowless elevator, and trying to determine > whether you are accelerating or just plain heavy ? Well, the result of that thought experiment is called the General Theory of Relativity which starts by saying (speaking loosely) that you can't tell a difference because both possibilities amount to the same thing. Anyway, I'm glad you brought up the elevator (don't say a word, Henry). If you're in a falling elevator you're weightless too, aren't you? But the gravitational field is still the same, isn't it? Let me offer a hint: We need a better definition of weight, and possibly a better notion of acceleration, to solve the problem. But no deep dark physics is needed, just careful thought. -- D Gary Grady Duke U Comp Center, Durham, NC 27706 (919) 684-3695 USENET: {seismo,decvax,ihnp4,akgua,etc.}!mcnc!ecsvax!dgary
bet@ecsvax.UUCP (Bennett E. Todd III) (12/20/85)
In article <970@ecsvax.UUCP> dgary@ecsvax.UUCP (D Gary Grady) writes: >If you're in the space shuttle orbiting 300 miles up, the acceleration >of gravity is very little less than it is at the surface. If weight = >mass X acceleration, how can you be "weightless"? First off, let us define acceleration as the second derivative of position with respect to time. So what's position? You know, spatial coordinates. There's where you get into trouble in this one -- an object in free fall is weightless *in it's local reference frame* -- a falling rock wouldn't normally be called weightless; the taken-for-granted reference frame for falling objects is the "outside" world, against which they are falling. Space ships, by finding a ballistic trajectory which doesn't intersect the surface of the Earth, can keep falling for a good while, and with their size, make an interesting reference frame in their own right. Thus "weightlessness". On the other hand, the gravitation attraction of the Earth is indeed still important; if it weren't then the craft wouldn't orbit, it would head elsewhere to frolic. So in the reference frame of the Earth the satellite is not in fact weightless; it weighs something less than what it does on the surface of the Earth, and its weight is its mass times the acceleration it is suffering to keep it in orbit. -Bennett -- "Hypocrisy is the vaseline of social intercourse." (Who said that?) Bennett Todd -- Duke Computation Center, Durham, NC 27706-7756; (919) 684-3695 UUCP: ...{decvax,seismo,philabs,ihnp4,akgua}!mcnc!ecsvax!duccpc!bet