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Universitas Scripta is licensed under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License.
Your cookies may be used by Google and mathjax. See Google's privacy policy.
Friday, 21 January 2011
Doctor Who - The origin of the Time Lord
You may know 'the end of time', which has been prevented by the Doctor from the Time Lords. However, what if the end of time resumes, and the Time Lord becomes the only life form in the universe even out of physical form? Then what would happen? Maybe in some universe, some parallel universe, it should happen. Falling, the whole disaster falling over the earth...
Wednesday, 8 December 2010
Problem of the moral theory - Partial answer
Surely, for the person, even for wrong decisions, knowing things is much preferred than not knowing though not knowing gives right logic to the wrong decision. May be given: should human flesh be provided, sold, and eaten, shall the customers know the truth or not? The truth must be revealed is the preferred answer in the theory.
Tuesday, 7 December 2010
Problem of the moral theory
A friend of mine suggested the problem of conscientiousness: intended distortion of information and not.
Monday, 15 November 2010
Korean Galaxy S upgraded into froyo
Today, galaxy s in Korea is upgraded into android froyo. You may know the conflicts between Samsung and SK Telecom. The final version is little bit worse than the last firmware. It means, the last firm was the very complete version of Samsung on the ecclair. The base memory use is about 160MB out of 340MB, while the last version uses 112MB out of 310 MB. You need to initialise the phone to make speedy after the installation. Still lags in some point. You cannot delete Google account in froyo, and cannot even change the password, and I think it is platform problem, so cannot be wished to be fixed until the next android.Also, you do not have to enter annoying T store as the market includes it.
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Thursday, 11 November 2010
Problem: Disappearance of gravitation
Region: Newton Mechanics, Classical Gravitation.
Consider a static, uniform spherical mass (star) and an object, particle mass, which is approaching to the star by only the gravitation of the star during \(t=-\infty\) through \(t=0\), where no collision occurs. At \(t=0\), sudden explosion of the star occurs, and the mass of the star is spread uniformly over the infinite universe, thus which is equivalent to the star disappearing, within negligible time interval. (a) What does happen to the object? Does any physical quantity change between the before and the after of the explosion? Assume the mass of the star can penetrate anything so that any effect of contact (collision) is not but is only of the distant force field. (b) Obtain the total mechanical energy of the object before and after the explosion. Does it make sense? (c) Calculate the total energy of the system before and after the explosion. Does it make sense? (d) How much of energy should be supplied or subtracted for occurence of the explosion? (e) At the explosion, energy E is born with loss of mass \(\frac{E}{c^2}\) of the star. How much the mass is lost? (f) On what criteria does this kind of explosion never happen? Compare to the Schwarzschild radius \(r_s=\frac{2GM}{c^2}\).(C) Albertus Liberius
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Solution
(a) As no force field acts, the object will conserve the momentum (inertial motion)
(b) Just before the explosion, \(E = \frac{1}{2} mv^2 - \frac{GMm}{r}\). Just after the explosion, \(E = \frac{1}{2} mv^2\). It does not make sense.
(c) The gravitational bonding energy (internal energy) is \(-\int_0^{M(R)} \frac{GM(r) dM(r)}{r} = -\int_0^R \frac{G}{r} \rho \frac{4}{3} \pi r^3 \rho 4\pi r^2 dr = -\frac{16}{15}\pi^2 R^5 G \rho^2 = -\frac{3}{5}\frac{GM^2}{R}\). Thus, before: \(E = \frac{1}{2} mv^2 - \frac{GMm}{r} - \frac{3}{5}\frac{GM^2}{R}\); and after: \(E = \frac{1}{2} mv^2\). It does not make sense.
(d) \(\Delta E = \frac{GMm}{r} + \frac{3}{5}\frac{GM^2}{R} \) is required to be supplied for occurrence of the explosion.
(e) \(-\Delta M = \frac{GMm}{rc^2} + \frac{3}{5}\frac{GM^2}{Rc^2} \) should be lost to supply the energy.
(f) \( \frac{GMm}{rc^2} + \frac{3}{5}\frac{GM^2}{Rc^2} > M\) i.e. \(R \leq \frac{\frac{3}{5}\frac{GM}{c^2}}{1-\frac{Gm}{rc^2}} \approx\frac{3}{10}r_s\) as \(1 \gg \frac{Gm}{rc^2}\)
(b) Just before the explosion, \(E = \frac{1}{2} mv^2 - \frac{GMm}{r}\). Just after the explosion, \(E = \frac{1}{2} mv^2\). It does not make sense.
(c) The gravitational bonding energy (internal energy) is \(-\int_0^{M(R)} \frac{GM(r) dM(r)}{r} = -\int_0^R \frac{G}{r} \rho \frac{4}{3} \pi r^3 \rho 4\pi r^2 dr = -\frac{16}{15}\pi^2 R^5 G \rho^2 = -\frac{3}{5}\frac{GM^2}{R}\). Thus, before: \(E = \frac{1}{2} mv^2 - \frac{GMm}{r} - \frac{3}{5}\frac{GM^2}{R}\); and after: \(E = \frac{1}{2} mv^2\). It does not make sense.
(d) \(\Delta E = \frac{GMm}{r} + \frac{3}{5}\frac{GM^2}{R} \) is required to be supplied for occurrence of the explosion.
(e) \(-\Delta M = \frac{GMm}{rc^2} + \frac{3}{5}\frac{GM^2}{Rc^2} \) should be lost to supply the energy.
(f) \( \frac{GMm}{rc^2} + \frac{3}{5}\frac{GM^2}{Rc^2} > M\) i.e. \(R \leq \frac{\frac{3}{5}\frac{GM}{c^2}}{1-\frac{Gm}{rc^2}} \approx\frac{3}{10}r_s\) as \(1 \gg \frac{Gm}{rc^2}\)
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