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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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 \(1 \gg \frac{Gm}{rc^2}\)