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In 1959 Lyttleton and Bondi suggested that the expansion of the Universe could be explained il matter carried a net charge. Suppose that the Universe is made up of hydrogen atoms with a number density N, which is maintained a constant. Let the charge on the proton be : e_(p) =-(1+y)ewhere e is the electronic charge. (a) Find the critical value of y such that expansion may start. (b) Show that the velocity of expansion is proportional to the distance from the centre. |
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Answer» Solution :(a) Let the Universe have a radius R. Assume that the hydrogen atoms are uniformly distributed. The expansion of the universe will start if the COULOMB REPULSION on a hydrogen atom at R is larger that the gravitational attraction. The hydrogen atom CONTAINS one proton and one electron, charge on each hydrogen atom. `e_(1p) = e_(p) + e =(-1 +y)e +e` `=-e + ye + e` = ye Let E be electric field intensity at distance R, on the surface of the sphere, then according to Gauss. theorem, `oint vecE.dvecS = q/epsilon_(0)` `therefore E(4piR^(2)) = 4/3 (piR^(3)N|ye)/epsilon_(0)`......(1) Let us suppose the mass of each hydrogen atom = `m_p` = mass of a proton and `G_R` = gravitational field at distance R on the sphere. Then `=-4piR^(2)G_(R) = 4piGm_(p)(4/3piR^(3))N` `therefore G_(B) =-4/3piGm_(p).NR`........(2) Gravitational force on this atom is, `F_(G) =m_(p) xx G_(R) = (-4pi)/3 Gm_(p)^(2)NR`......(3) Coulomb force on hydrogen atom at R is, `F_( C) = (ye)E = 1/3(y^(2)e^(2)NR)/epsilon_(0)` [From eqns (1)] Now to expansion `F_(C) gt F_(G)` and critical value of Y to start expansion would be when `F_( C) = F_(G)` `therefore 1/3 (Ny^(2)e^(2)R)/epsilon_(0) = (4pi)/3 Gm_(p)^(2)NR` `therefore y^(2) = (4piepsilon_(0))G xx (m_(p)/e)^(2)` `therefore y = sqrt(79.8 xx 10^(-38)) = 8.9 xx 10^(-19) = 10^(-18)` Hence, `10^(-18)`is the required critical value of y corresponding to which expansion of universe would start, (b) Net force experience by the hydrogen atom is given by, `F = F_( C) - F_(G) = 1/3 (Ny^(2)e^(2)R)/epsilon_(0) -(4pi)/3 Gm_(p)^(2)NR` Because of this net force, the hydrogen atom experience an acceleration such that, `m_(p) xx (d^(2)R)/(dt^(2)) = F = 1/3 (Ny^(2) e^(2)R)/epsilon_(0) - (4pi)/3 Gm_(p)^(2)NR` `therefore (d^(2)R)/(dt^(2)) = alpha^(2)R`......(4) where, `alpha^(2) =1/m_(p) (1/3 (Ny^(2)e^(2))/epsilon_(0) -(4pi)/3 Gm_(p)^(2)N)` The solution of equation (4) is given by,`R = Ae^(alphat) + Be^(-alphat)`. We are looking for expansion, here so B = 0 and `R = Ae^(alphat)` Velocity of expansion, `v= (dR)/(dt) = d/(dT) (Ae^(alphat))` `therefore v prop R` Velocity of expansion is proportional to the distance from the CENTRE. |
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