Answer
$ \Delta F=\frac{4GmM_E a}{r^3}$
Work Step by Step
We know that
$\Delta F=GmM_E(\frac{1}{(r-a)^2}-\frac{1}{(r+a)^2})$
As given that for $r>>a$, $\frac{1}{(r-a)^2}-\frac{1}{(r+a)^2}\approx \frac{4a}{r^3}$
$\implies \Delta F=F_{tidal}=GmM_E(\frac{4a}{r^3})$
$\implies \Delta F=\frac{4GmM_E a}{r^3}$
