In 1959 Lyttleton and Bondi suggested that the expansion

Solution:

(a)

Suppose universe is a perfect sphere of radius $R$ and its constituents are Hydrogen atom are distributed uniformly.

As hydrogen atom contains one proton and one electron, charge on each hydrogen atom

$e_{H}=e_{P}+e=-(1+y) e+e=-y e=|y e|$

According to Gauss' theorem.

$\oint \mathrm{E} \cdot \mathrm{ds}=\frac{q}{\varepsilon_{0}}$

$\Rightarrow E\left(4 \pi R^{2}\right)=\frac{1}{\varepsilon_{0}}\left[\frac{4}{3}\left(\pi R^{3} N|y e|\right)\right]$

$\Rightarrow E=\frac{1}{3} \frac{N|y e| R}{s_{0}}$ $\ldots(i)$

Let, mass of each hydrogen atom $\sim m_{p}$ (mass of a proton), $G_{R}=$ gravitational field at distance $R$ on the spliere.

Then, $\quad 4 \pi R^{2} G_{R}=4 \pi G m_{P}\left(\frac{4}{3} \pi R^{3}\right) N$

$\Rightarrow G_{R}=\frac{-4}{3} \pi G m_{p} N R$ $\ldots($ ii $)$

$\therefore$ Gravitational force on this atom is $F_{G}=G_{R} \times m_{p}=\frac{-4}{3} \pi G m^{2}{ }_{p} N R \ldots$ (iii)

Coulomb force on hydrogen atom at $R$ is $F_{\mathrm{C}}=E($ ye $)=\frac{1}{3} \frac{N y^{2} e^{2} R}{\varepsilon_{0}}$ [from Eq. (i)]

Expansion of the universe will start when coulomb repulsion $F_{\mathrm{c}}>F_{\mathrm{G}}$ on the hydrogen atom.

Now, the critical value of $y$ (say $y_{2}$ to start expansion would be when, $F_{\mathrm{c}}=F_{\mathrm{G}}$

$\Rightarrow \frac{1}{3} \frac{N y^{2} e^{2} R}{\varepsilon_{0}}=\frac{4 \pi}{3} G m^{2}, N R$

$\Rightarrow \frac{1}{3} \frac{M y^{2} e^{2} R}{\varepsilon_{0}}=\frac{4 \pi}{3} G m_{p}^{2} N R$

$\Rightarrow y^{2}=\left(4 \pi \varepsilon_{0}\right) G\left(\frac{m_{F}}{e}\right)^{2}=\frac{1}{9 \times 10^{9}} \times\left(6.67 \times 10^{-11}\right)\left(\frac{\left(1.66 \times 10^{-27}\right)^{2}}{\left(1.6 \times 10^{-10}\right)^{2}}\right)=79.8 \times 10^{-i s}$

$\Rightarrow y=\sqrt{79.8 \times 10^{-33}}=8.910^{-19}$

$\Rightarrow y \simeq 10^{-13}$

(b)

Net force experience by the hydrogen atom is given by

$F_{N \sigma}=F_{c}-F_{G}=\frac{1}{3} \frac{N y^{2} e^{2} R}{\varepsilon_{0}}-\frac{4 \pi}{3} G N m_{p}^{2} R$

Due to net force, the hychrogen atom experiences an acceleration given by $m_{p} \frac{d^{2} R}{d t^{2}}$

$F=m_{p} \frac{d^{2} R}{d t^{2}}=\frac{1}{3} \frac{N Y^{2} e^{2}}{\varepsilon_{0}} R-\frac{4 \pi}{3} G m_{p}^{2} N R$

$m_{p} \frac{d^{2} R}{d t^{2}}=\left(\frac{1}{3} \frac{N Y^{2} e^{2} R}{\varepsilon_{0}}-\frac{4 \pi}{3} G m_{p}^{2} N\right) R$

$\frac{d^{2} R}{d t^{2}}=\frac{1}{m_{p}}\left[\frac{1}{3} \frac{N Y^{2} e^{2} R}{c_{0}}-\frac{4 \pi}{3} G m_{p}^{2} N\right] R=\alpha^{2} R \ldots(i v)$

Where, $\alpha^{2}=\frac{1}{m_{p}}\left[\frac{1}{3} \frac{N Y^{2} e^{2} R}{\varepsilon_{0}}-\frac{4 \pi}{3} G m_{p}^{2} N\right]$

The general solution of Eq...(iv) is given by $R=A e^{\text {at }}+B^{-\alpha t}$

Here, we are looking for expansion, here, so $B=0$ and $R=A e^{\alpha t}$

$\Rightarrow$ Velocity of expanision, $v=\frac{d R}{d t}=A e^{\text {ut }}(\alpha)=\alpha A e^{a r}=\alpha R$

Hence, $v \propto R i, e$, velocity of expansion is proportional to the distance from the centre.