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#11
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help - gravity problem
In sci.space.tech message 97af933c-beda-4dba-aa2e-81364d4202ef@e20g2000
vbc.googlegroups.com, Tue, 9 Jun 2009 06:09:45, dotcom posted: I quoted the problem verbatim from my daughters text book. In that case, assuming that you actually understand and mean "verbatim", move the daughter to a different school - one that uses a better grade of textbook. What you gave cannot be verbatim from a well-written book. To go further, we need the exact words, warts and all, of the book, without any intermingled comment from yourself or your daughter. The title, author, ISBN and page reference could also be useful; some here may have access to the book and be able to see more context than it would be proper to copy here. Your daughter can copy the words, and you can certify it as a true copy. See signature below. = = = I predict that I will again receive a false message from the moderation. -- (c) John Stockton, Surrey, UK. Turnpike v6.05 MIME. Web URL:http://www.merlyn.demon.co.uk/ - FAQish topics, acronyms, & links. Proper = 4-line sig. separator as above, a line exactly "-- " (SonOfRFC1036) Do not Mail News to me. Before a reply, quote with "" or " " (SonOfRFC1036) |
#12
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help - gravity problem
Ken S. Tucker wrote:
It's ok, some specific info is lacking. How does a disabled satellite retro fire? Maybe it collides with an asteroid or an exhausted rocket stage. Then it has 0 velocity and drops vertically to 800 km. Bernhard |
#13
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help - gravity problem
Dr J R Stockton wrote:
In sci.space.tech message 97af933c-beda-4dba-aa2e-81364d4202ef@e20g2000 vbc.googlegroups.com, Tue, 9 Jun 2009 06:09:45, dotcom posted: I quoted the problem verbatim from my daughters text book. In that case, assuming that you actually understand and mean "verbatim", move the daughter to a different school - one that uses a better grade of textbook. What you gave cannot be verbatim from a well-written book. It's not so bad. Although its stated initial speed corresponds to a circular orbit at 2000km (apparently, I haven't checked), the question doesn't say that the satellite is in such an orbit. It does say that it falls to a height of 800km. So the reasonable assumption is that it's in an orbit that allows it to be at 2000km at one point in time, and 800km at another. Sylvia. |
#14
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help - gravity problem
Sylvia Else wrote:
Dr J R Stockton wrote: In sci.space.tech message 97af933c-beda-4dba-aa2e-81364d4202ef@e20g2000 vbc.googlegroups.com, Tue, 9 Jun 2009 06:09:45, dotcom posted: I quoted the problem verbatim from my daughters text book. In that case, assuming that you actually understand and mean "verbatim", move the daughter to a different school - one that uses a better grade of textbook. What you gave cannot be verbatim from a well-written book. It's not so bad. Although its stated initial speed corresponds to a circular orbit at 2000km (apparently, I haven't checked), the question doesn't say that the satellite is in such an orbit. It does say that it falls to a height of 800km. So the reasonable assumption is that it's in an orbit that allows it to be at 2000km at one point in time, and 800km at another. That would be my guess, too; it sounds asking what the speed of the satellite will be if it has a perigee of 800 km altitude and an apogee of 2000 km altitude. It's a bit glib and not terribly clear, though. I agree with the others that it's not a very useful question, especially for high school students. -- Erik Max Francis && && http://www.alcyone.com/max/ San Jose, CA, USA && 37 18 N 121 57 W && AIM/Y!M/Skype erikmaxfrancis Most men do not mature, they simply grow taller. -- Leo Rosten |
#15
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help - gravity problem
Erik Max Francis wrote:
Sylvia Else wrote: Dr J R Stockton wrote: In sci.space.tech message 97af933c-beda-4dba-aa2e-81364d4202ef@e20g2000 vbc.googlegroups.com, Tue, 9 Jun 2009 06:09:45, dotcom posted: I quoted the problem verbatim from my daughters text book. In that case, assuming that you actually understand and mean "verbatim", move the daughter to a different school - one that uses a better grade of textbook. What you gave cannot be verbatim from a well-written book. It's not so bad. Although its stated initial speed corresponds to a circular orbit at 2000km (apparently, I haven't checked), the question doesn't say that the satellite is in such an orbit. It does say that it falls to a height of 800km. So the reasonable assumption is that it's in an orbit that allows it to be at 2000km at one point in time, and 800km at another. That would be my guess, too; it sounds asking what the speed of the satellite will be if it has a perigee of 800 km altitude and an apogee of 2000 km altitude. It's a bit glib and not terribly clear, though. I agree with the others that it's not a very useful question, especially for high school students. It couldn't have an apogee at 2000km, because it's going at the wrong speed. But the question doesn't require any assumptions about the actual orbit. I don't really see the objection to the question. It requires the ability to calculate the potential energy of an object in a gravitational field, an understanding of kinetic energy, and the application of the law of conservation of energy. Is this not highschool maths/physics? Sylvia. |
#16
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help - gravity problem
In article
, dotcom wrote: I thought I understood basic gravity problems but the following high school physics problem from my daughter has me stumped ( I think) Q. a disabled ( meaning of disable not defined) Presumably it doesn't have any propulsion systems so it is in free-fall. satellite of mass 2400kg is in orbit at a ht of 2000 km above the earth at a speed of 6900 m/s. ( my calc show that is exaclty the speed required for a circular orbit at that ht). I haven't done the calculation, but assuming that you did it correctly: If the orbital speed for a circular orbit is 6900 m/s at 2000 km height, that does not mean that an object with that height and that speed is in a circular orbit. Suppose you climb a 2000 km tall ladder (from Antarctica so you can ignore Earth's rotation and are initially motionless), and then fire a rifle that has a muzzle speed of 6900 m/s. Every bullet you fire will have a 6900 m/s speed at 2000 km height. If you fire it horizontally, then you will have a circular orbit. If you fire it straight up then it will have a long, skinny orbit that intersects Earth before it gets all the way around. If you fire it straight down it will have the same orbit, only it will complete even less of an orbit. If you fire it at an angle, then it will have a perigee below 2000 km and a apogee above that height. It will, of course, travel faster at perigee and slower at apogee. If the apogee is below 800 km, then you can complete the question. It turns out to have the same answer for all orbits with an apogee below 800 km (including the shoot-straight-down case). it then says the satelite falls to a ht of 800 km calculate what the new speed at the lower ht.. well I simply calculated the gain in potential energy ( PE = delta GMm/r) and equated this to the gain in kinetic energy ( =0.5 mv^2) as the satellite must speed up. and added this to the original speed of 6900 m/s to get 10870 m/s , but I am not sure that this correct. it certainly doesnt give me the answer in the school text book of 7900 m/s ( I used G=6.67E-11, M =5.98E24 kg and r= 6.38E6 m. using this the loss in potential energy = 1.9E10 J You can't take E = 1.9e10 J - 3970 m/s and add that to 6900 m/s to get the new speed. You have to take Ekinetic_start = 5.7e10 J + Epotential_change = 1.9e10 J = Ekinetic_end = 7.6e10 J And 7.6e10 J - the 7900 m/s that's the textbook answer. Basically, you have to add the energies, not the velocities. Which means that you have to add the square of the velocity and the square of the KE-equivalent velocity and take the square root of the sum. (The phrase used is 'add in quadrature') I suspect I am going wrong somewhere in not accounting for the fact velocity is a vector quantitiy. Surely it must depend on the direction the satellite is heading initally. is this really a solvable problem? Yes, the stated problem is a conservation of energy problem, and solvable. -- David M. Palmer (formerly @clark.net, @ematic.com) |
#17
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help - gravity problem
Sylvia Else wrote:
Erik Max Francis wrote: Sylvia Else wrote: So the reasonable assumption is that it's in an orbit that allows it to be at 2000km at one point in time, and 800km at another. That would be my guess, too; it sounds asking what the speed of the satellite will be if it has a perigee of 800 km altitude and an apogee of 2000 km altitude. It's a bit glib and not terribly clear, though. I agree with the others that it's not a very useful question, especially for high school students. It couldn't have an apogee at 2000km, because it's going at the wrong speed. That seems in direct contradiction to what you just said above It starts with an altitude of 2000 km, is hit by something, and ends up at 800 km altitude. There's nothing about it ending up in a 800 km _circular_ orbit. -- Erik Max Francis && && http://www.alcyone.com/max/ San Jose, CA, USA && 37 18 N 121 57 W && AIM/Y!M/Skype erikmaxfrancis Only the ephemeral is of lasting value. -- Ionesco |
#18
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help - gravity problem
Erik Max Francis wrote:
Sylvia Else wrote: Erik Max Francis wrote: Sylvia Else wrote: So the reasonable assumption is that it's in an orbit that allows it to be at 2000km at one point in time, and 800km at another. That would be my guess, too; it sounds asking what the speed of the satellite will be if it has a perigee of 800 km altitude and an apogee of 2000 km altitude. It's a bit glib and not terribly clear, though. I agree with the others that it's not a very useful question, especially for high school students. It couldn't have an apogee at 2000km, because it's going at the wrong speed. That seems in direct contradiction to what you just said above It starts with an altitude of 2000 km, is hit by something, and ends up at 800 km altitude. There's nothing about it ending up in a 800 km _circular_ orbit. I haven't said it's hit by something. Indeed, if it is, then the question is unanswerable. My view is that it's in some orbit, being an orbit that rises above 2000km, and falls below 800km. Sylvia. |
#19
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help - gravity problem
Erik Max Francis wrote:
Sylvia Else wrote: Erik Max Francis wrote: Sylvia Else wrote: So the reasonable assumption is that it's in an orbit that allows it to be at 2000km at one point in time, and 800km at another. That would be my guess, too; it sounds asking what the speed of the satellite will be if it has a perigee of 800 km altitude and an apogee of 2000 km altitude. It's a bit glib and not terribly clear, though. I agree with the others that it's not a very useful question, especially for high school students. It couldn't have an apogee at 2000km, because it's going at the wrong speed. That seems in direct contradiction to what you just said above It starts with an altitude of 2000 km, is hit by something, and ends up at 800 km altitude. There's nothing about it ending up in a 800 km _circular_ orbit. Also, look at David M Palmers response to the OP. Sylvia. |
#20
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help - gravity problem
In article
, dotcom wrote: I thought I understood basic gravity problems but the following high school physics problem from my daughter has me stumped ( I think) Q. a disabled ( meaning of disable not defined) satellite of mass 2400kg is in orbit at a ht of 2000 km above the earth at a speed of 6900 m/s. ( my calc show that is exaclty the speed required for a circular orbit at that ht). it then says the satelite falls to a ht of 800 km calculate what the new speed at the lower ht.. well I simply calculated the gain in potential energy ( PE = delta GMm/r) and equated this to the gain in kinetic energy ( =0.5 mv^2) as the satellite must speed up. and added this to the original speed of 6900 m/s to get 10870 m/s , but I am not sure that this correct. it certainly doesnt give me the answer in the school text book of 7900 m/s ( I used G=6.67E-11, M =5.98E24 kg and r= 6.38E6 m. using this the loss in potential energy = 1.9E10 J I suspect I am going wrong somewhere in not accounting for the fact velocity is a vector quantitiy. Surely it must depend on the direction the satellite is heading initally. is this really a solvable problem? Sadly I dont have the maths to work this out, but since this is only high school work, perhaps you just need to calculate the speed at 800km You guys are very cluey, but you wouldn't expect a school student to do all other stuff you are considering Perhaps someone could just calculate the speed required to keep it in orbit at 800km David who is trying to learn maths, but... |
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