And Newton Rolls Over In His Grave

Just you might be happy with the previous derivations,

If we start from,

$T(x)=\cfrac{A}{\sqrt{x}}$

A body experiences gravitational effect wholly due the hot mass behind it,  the net effect of all hot mass above the sphere containing the hot mass is zero.

We have the total  $T(x)$  behind the body at point  $x$ from the center of  $T$.

Total  $T(x)$  in a sphere of   $T(x)$   of radius  $x$,

$M_T=   \int _{ 0 }^{ x }{ 4\pi x^{ 2 }T(x) } dx=\int _{ 0 }^{ x }{ 4\pi x^{ 2 }\cfrac { A }{ \sqrt { x }  }  } dx=\cfrac { 8A\pi  }{ 5 } { x }^{ \cfrac { 5 }{ 2 }  }$

Then we consider a hot body of total equivalent mass  $M_T$,  what is its gravity inward?   If we consider spheres of area

$4\pi x^2$

and the total gravitational flux through them from a unit mass of  $T$,  assuming space is empty,

$g_{o}.4\pi x^2_o=g.4\pi x^2$

and that at  $x=x_o$,  $G=g_{o}.4\pi x^2_o$

$g=\cfrac{G}{4\pi x^2}$    per unit mass

$G$  increases with equivalent mass by  $\cfrac { 8A\pi  }{ 5 } { x }^{ \cfrac { 5 }{ 2 }  }$  as  $x$  increases.

$g_T=\cfrac{G}{4\pi x^2}\cfrac { 8A\pi  }{ 5 } { x }^{ \cfrac { 5 }{ 2 }  }=G_{o}\sqrt{x}$

where  $G_{o}=\cfrac{2AG}{5}$

This is the gravity due to the mass of  $T$  which is here assumed to be proportional to  $T$.  It is assumed here that gravity is directly proportional to its mass.  $g_T$  increase in the square root of $x$ until  $T=0$  at some distance  $x=L$  from the hot body.