Question

The each revolves round the sun due to gravit ational attraction. Suppose that the sun and the earth are point particles with their existing mas ses and that Bohr's quantization rule for angular momentum is vaild case of gravitation.

(a) Calculate the minimum radius the earth can have for its orbit.

(b) What is value of the principal quantum num ber n for the present radius ? Mass of the earth = $6.0\times {10}^{24}$ kg, Mass of the sun = $2.0\times {10}^{30}$ kg, earth-sun distance = $1.5\times {10}^{11}$ m.

Answer 1

Mass of Earth = $M_e=6.0\times {10}^{24}$ kg

Mass of Sun = $M_8=2.0\times {10}^{30}$ kg

Earth- Sun distance = $1.5\times {10}^{11}$ m

Since mvr = $\frac{nh}{2\pi }$

or, $m^2{v}^2{r}^2=\frac{n^2{h}^2}{2{\pi }^2}.....\left(1\right)$

$\frac{G{M}_e{M}_8}{r^2}=\frac{M_e{v}^2}{r}$

or $v^2=\frac{G{M}_8}{r}$ ...............(2)

Dividing (1) and (2)

We get $M_e^{2}r=\frac{n^2{h}^2}{4{\pi }^{2}G{M}_8}$

for n = 1,

r = $\sqrt{\frac{h^2}{{40}^{2}G{M}_8{M}_e^2}}$

= $\sqrt{\frac{(6.6\times {10}^{-24}{)}^2}{4\times (3.14{)}^2\times 6.67\times {10}^{-11}\times (6\times {10}^{24})\times 2\times {10}^{30}}}$

= $2.29\times {10}^{-138}m$

= $2.3\times {10}^{138}m$

(b) $n^2$ = $\frac{M_e^2\times r\times 4\times {\pi }^2\times G\times M_8}{h^2}$

$\frac{(6\times {10}^{24}{)}^2\times 1.5\times {10}^{11}\times 4\times (3.14{)}^2\times 6.67\times {10}^{-11}\times 2\times {10}^{30}}{(6.6\times {10}^{-34}{)}^2}$

= $2.5\times {10}^{74}$