Can heritability be deduced from a correlation coefficient?
No. But a strong correlation may nevertheless be an indicator for heritability (it makes it more likely).

The reason is, that the correlation in the phenotype (birth weight) is not only due a correlation in the genotype but also a correlation in the environment of the child and its mother (culture, economical situation, genes, etc of their respective mothers). 

If you were able to remove the association of the environments and of the genotypes and the environments by an intervention (eg. shuffling the zygotes (ideally of parents who were shuffled as zygotes as well)) then the correlation coefficient of the phenotype between one of two parents and the child would be half the heritability. $corr=\frac{1}{2} H^2$



Phenotype ($P$), Genotype ($G$), Environment ($E$)

$P_{parent}=G_{parent}+E_{parent}$

$P_{child}=G_{child}+E_{child}$

$
\begin{eqnarray*}
corr(P_{parent},P_{child}) &amp;=&amp; \frac{COV(P_{parent},P_{child})}{\sqrt{VAR(P_{parent})}\cdot \sqrt{VAR(P_{child})}} \\
&amp;=&amp; \frac{COV(P_{parent},P_{child})}{ VAR(P)}\\
&amp;=&amp; \frac{COV(G_{par}+E_{par},G_{chi}+E_{chi})}{ VAR(P)}\\
&amp;=&amp; \frac{COV(G_{p},G_{c})+ COV(G_{p},E_{c})+ COV(E_{p},G_{c})+ COV(E_{p},E_{c})}{VAR(P)}\\
&amp;=&amp; \frac{\frac{1}{2}VAR(G)+ COV(G_{p},E_{c})+ COV(E_{p},G_{c})+ COV(E_{p},E_{c})}{VAR(P)}\\
&amp;\overset{(1)}{=}&amp; \frac{\frac{1}{2}VAR(G)}{VAR(P)}\\
&amp;\overset{(2)}{=}&amp;\frac{1}{2}H^2
\end{eqnarray*}
$

$(1): \,\,\,$if you managed to get $COV(G_{p},E_{c}), COV(E_{p},G_{c}),COV(E_{p},E_{c}) = 0 $

$(2): \,\,\,H^2=\frac{VAR(G)}{VAR(P)}$ - the definition  of heritability