The earth revolves round the sun due to gravitatinal attraction. Suppose that the sun and the earth are point particle with their existing masses and that Bhor's quantization rule for angular monentum is valid in the case of gravitation (a) Calculate the minimum radius the earth can have for its orbit. (b) What is the value of the principle quantum number n for the present radius ? Mass of the earth = `6.0 xx 10^(24)` kg, mass of the sun = `2.0 xx 10^(30)` kg, earth-sun distance = `1.5 xx 10^(11)m`.

Answer

A, B, C, D

Explanation

Mass of Earth`=M_e = 6.0 xx 10^24 kg`
Mass of Sun `=M_s = 2.0 xx 10^30 kg`
Earth-Sun distance `=1.5 xx 10^11 m`
Since `mvr = (nh)/(2pi)`
`or, `m^2v^2r^2 = (n^2h^2)/(2pi^2)` …(1)
`(GM_eM_s)/r^2 = (M_ev^2)/r`
or `v^2 = (GM_s)/r` ...(2)
`Dividing (1) and (2)
We get `M_e^2r = (n^2h^2)/(4pi^2GM_s)`
for `n=1`,
`r=sqrt((h^2)/(40^2GM_sM_e^2))`
`=sqrt(((6.6 xx 10^-24)^2)/(4 xx (3.14)^2 xx 6.67 xx 10^-11 xx (6 xx 10^24) xx 2 xx 10^30))`
`=2.29 xx 10^-138 m`
`=2.3 xx 10^-138 m`
(b) `n^2 = (M_e^2 xx r xx 4 xx pi^2 xx G xx M_s)/(h^2)`
`((6xx10^24) xx 1.5 xx 10^11 xx 4 xx (3.14)^2 xx 6.67 xx 10^-11 xx 2 xx 10^30)/((6.6 xx 10^-34)^2)`
`=2.5 xx 10^74`