Answer
$3.08\times10^{-38}$
Work Step by Step
To estimate the ratio of the gravitational potential energy to the nuclear potential energy for two neutrons separated by 1.0 fm, we’ll calculate both values separately and then find their ratio.
From the potential-energy diagram:
At a separation of $ r = 1.0 $ fm (from $r=1$ fm to $r=2$ fm), the nuclear potential energy $ U_{\text{nuclear}} $ is approximately $-38\, \text{MeV}$.
$$U_{\text{nuclear}}=-38\;\rm MeV $$
Hence,
$$U_{\text{nuclear}}=(-38)(10^6)(1.6\times 10^{-19})= \bf -6.08\times 10^{-12}\;\rm J$$
The gravitational potential energy $ U_{\text{grav}} $ between two masses $ m $ separated by a distance $ r $ is given by:
$$
U_{\text{grav}} = -\frac{G m_1 m_2}{r}=-\dfrac{Gm^2}{r}
$$
Substituting the known:
$$
U_{\text{grav}} =-\dfrac{( 6.674 \times 10^{-11} )(1.675 \times 10^{-27})^2}{(1.0 \times 10^{-15})}$$
$$=\bf -1.87\times 10^{-49}\;\rm J
$$
Thus the needed ratio is then
$$
\frac{U_{\text{grav}}}{U_{\text{nuclear}}} = \frac{-1.87\times 10^{-49} }{-6.08\times 10^{-12}}
$$
$$
\frac{U_{\text{grav}}}{U_{\text{nuclear}}} \approx \color{red}{\bf 3.08\times10^{-38}}
$$
