Supposing Newton’s law of gravitation for gravitation forces ${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_2$ between two masses m₁ and m₂ at positions ${\overrightarrow{r}}_1$ and ${\overrightarrow{r}}_2$ ${\overrightarrow{F}}_1=-{\overrightarrow{F}}_2=-\frac{{\widehat{r}}_{12}}{{\left(r_{12}\right)}^2}G{M}_0^2{\left(\frac{m_1{m}_2}{M_0^2}\right)}^n$,where M₀ is a constant of dimension of mass, ${\overrightarrow{r}}_{12}={\overrightarrow{r}}_1-{\overrightarrow{r}}_2$ and n is a number. In such a case,

$F_1=-F_2=\frac{-r_{12}}{{r^3}_{12}} G{M^2}_0{\left(\frac{m_1{m}_2}{{M^2}_0}\right)}^n$ 

$r_{12}=r_1-{r_2}_{ }$ 

$g=\frac{G{M^2}_0{\left(m_1{m}_2\right)}^n}{{r^2}_{12}{\left(M_0\right)}^{2n}}\times \frac{1}{mass}$

Kepler's third law will  be invalid

As the force is central;    force  $\propto \frac{1}{r^2}$

first and second Kepler's  law are valid

$g=\frac{G{M^2}_0{\left(m_1{m}_2\right)}^{-n}}{{r^2}_{12}{\left(M_0\right)}^{-2n}}\times \frac{1}{mass}=\frac{G{M^{\left(1+n\right)}}_0{\left(m_1{m}_2\right)}^{-n}}{{r^2}_{12}\times mass}$

$g=\frac{G{M^2}_0}{{r^2}_{12}}{\left(\frac{{M^2}_0}{m_1{m}_2}\right)}^n\times \frac{1}{mass}$

So the bodies with less mass will experience high acceleration due to gravity.

So the bodies lighter than water will sink in water.