Janus Posted July 31, 2017 Share Posted July 31, 2017 3 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. Hi. My name is Ivy would you like a physics lesson? No. The given radius gives an orbital speed of 7673.556779 m/sec. At that speed and radius, the centripetal force would be 7673.556779^2 x 50/7=6,771e6 = 434.8211... N The force of gravity would be 3.987e14 x 50/6,771e6^2 = 434.8211.. N They are equal to the same degree of accuracy we have for the given parameters The 3.9987e14 is the "gravitational parameter" for the Earth which is equal to GM,where M is the mass of the Earth. This gives a more accurate answer as we know it to a better accuracy than we know either the gravitational constant or the mass of the Earth separately. 2 Link to comment Share on other sites More sharing options...
John Cuthber Posted July 31, 2017 Share Posted July 31, 2017 4 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. Hi. My name is Ivy would you like a physics lesson? Hi My name is Barack Obama. Would you like a lesson on establishing identities on the internet? 1 Link to comment Share on other sites More sharing options...
Bender Posted July 31, 2017 Share Posted July 31, 2017 4 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. To elaborate on Janus's point. Instead of making the calculations with accurate input, like he did, you could also make yours with inaccurate input, but then you have to take the significant figures into account: For a period of 5x10³ s (about 1.5 hours) and a mass of 5x10 kg (about 50 kg), you get a force of 4x10² N. The gravitational force with the same level of accuracy is 4x10² N. As you can see, once you remove all the insignificant figures from your calculation, the result is spot on. To put it differently, the error bar on your result of 20 N is about 100 N, so the observed 0 N fits nicely within that range. The null hypothesis (physics works) cannot be rejected. Link to comment Share on other sites More sharing options...
swansont Posted July 31, 2017 Share Posted July 31, 2017 2 hours 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... So 20 N might just be roundoff. Significant digits were not presented appropriately in the problem, I see. edit: and already mentioned by Bender and Janus Link to comment Share on other sites More sharing options...
ivylove Posted August 1, 2017 Author Share Posted August 1, 2017 (edited) So now can you NOW tell me how Cavendish measured a 2 microgram weight in 1797. Girl Power rules. Edited August 1, 2017 by ivylove -1 Link to comment Share on other sites More sharing options...
imatfaal Posted August 1, 2017 Share Posted August 1, 2017 1 hour ago, ivylove said: So now can you NOW tell me how Cavendish measured a 2 microgram weight in 1797.... Please try not to be so rude when you are asking questions. Especially questions which have been well covered by Studiot and Strange above. I am not sure anyone is going to go to the trouble of reprising all of Cavendish's calculations - but Strange's post explains what was done and Studiot's post shows how G could be determined. Link to comment Share on other sites More sharing options...
swansont Posted August 1, 2017 Share Posted August 1, 2017 2 hours ago, ivylove said: So now can you NOW tell me how Cavendish measured a 2 microgram weight in 1797. Girl Power rules. You were asked for the reference where this appears, and I don't see where you have done so. Link to comment Share on other sites More sharing options...
ivylove Posted August 1, 2017 Author Share Posted August 1, 2017 https://en.wikipedia.org/wiki/Cavendish_experiment Link to comment Share on other sites More sharing options...
ivylove Posted August 1, 2017 Author Share Posted August 1, 2017 (edited) In Einstein's paper, "The Foundation of the Generalised Theory of Relativity" (1916), Einstein represents gravity with Maxwell's electromagnetic field using Maxwell's equations. dh/dt + rot e = 0.. div h = 0... rot h - de'/dt = i.... div e' = p" (Einstein5, § 20). Einstein is representing gravity with Maxwell's electromagnetic field that is based on Faraday's induction effect but a small stone that is affected by gravity but not attracted to a magnet which is experimental proof gravity is not an electromagnetic phenomenon. Girl power definitely superior. Edited August 1, 2017 by ivylove Link to comment Share on other sites More sharing options...
imatfaal Posted August 2, 2017 Share Posted August 2, 2017 16 hours ago, ivylove said: In Einstein's paper, "The Foundation of the Generalised Theory of Relativity" (1916), Einstein represents gravity with Maxwell's electromagnetic field using Maxwell's equations. dh/dt + rot e = 0.. div h = 0... rot h - de'/dt = i.... div e' = p" (Einstein5, § 20). Einstein is representing gravity with Maxwell's electromagnetic field that is based on Faraday's induction effect but a small stone that is affected by gravity but not attracted to a magnet which is experimental proof gravity is not an electromagnetic phenomenon. Girl power definitely superior. ! Moderator Note Dumping irrelevant copynpasta from a usenet sci-forum is not acceptable. If you continue to troll you will be banned. Thread Locked. Link to comment Share on other sites More sharing options...
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