Levitating Hot Body

Is it possible that a hot body levitate,  since the gravity it produces is opposite to that of Earth's.

For this construct to lift,

$Mg_T-(M+m)g>0$

We assume a linear temperature profile,

$T=T_{max}(1-\cfrac{x}{d})$

and a log squared space density profile,

${ d }_{ s }(T)=\cfrac { ({ d }_{ T_{ max } }-{ d }_{ n }) }{ ln(\cfrac { 1 }{ T_{ max }^{ 2 } } ) } ln(\cfrac { 1 }{ T^{ 2 }(x) } )+{ d }_{ n }$

${ d }_{ s }(T)=\cfrac { ({ d }_{ T_{ max } }-{ d }_{ n }) }{ ln(\cfrac { 1 }{ T_{ max }^{ 2 } } ) } ln(\cfrac { 1 }{ T^2_{max}(1-\cfrac{x}{d})^2 } )+{ d }_{ n }$

${ d }_{ s }(T)=\cfrac { ({ d }_{ T_{ max } }-{ d }_{ n }) }{ ln(T_{ max }^{ 2 }) } ln(T^{ 2 }_{ max }(1-\cfrac { x }{ d } )^{ 2 })+{ d }_{ n }$

${ d }_{ s }(T)=\cfrac { ({ d }_{ T_{ max } }-{ d }_{ n }) }{ ln(T_{ max }^{ 2 }) } (ln(T^{ 2 }_{ max })+ln((1-\cfrac { x }{ d } )^{ 2 }))+{ d }_{ n }\}$

$\cfrac { d(d_{ s }(T)) }{ dx } =\cfrac { ({ d }_{ T_{ max } }-{ d }_{ n }) }{ ln(T_{ max }^{ 2 }) } (-\cfrac { 1 }{ (1-\cfrac { x }{ d } )^{ 2 } } \cfrac { 2 }{ d } (1-\cfrac { x }{ d } ))$

$\cfrac { d(d_{ s }(T)) }{ dx } =\cfrac { ({ d }_{ T_{ max } }-{ d }_{ n }) }{ ln(T_{ max }^{ 2 }) } (-\cfrac { 2 }{ (d-x ) } )$

And we have,

$g_{ T }=\cfrac { { G }_{ T } }{ { x }_{ o }({ d }_{ n }-{ d }_{ T_{ max } }) } \cfrac { ({ d }_{ T_{ max } }-{ d }_{ n }) }{ ln(T_{ max }^{ 2 }) } (-\cfrac { 2 }{ (d-x) } )$

$g_{ T }=\cfrac { 2{ G }_{ T } }{ { x }_{ o }ln(T_{ max }^{ 2 }) }\cfrac { 1}{ (d-x) }$

Therefore,  at  $x=0$,

$M\cfrac { 2{ G }_{ T } }{ { x }_{ o }ln(T_{ max }^{ 2 }) } \cfrac { 1}{ d }-(M+m)g>0$

From this equation we see that small  $x_o$  (radius of body)  and small  $d$  provide for greater lift.  We shall now optimize  $T_{max}$  for greater lift.

It seem odd that  $g_T$  should be inversely proportional to  $T_{max}$.  In free space where  $T_{max}$  is allowed to fall over a  distance  $x=L$  to reach  $T=0$  increasing  $T_{max}$  increases  $L$  but  $d_s$  cannot fall below zero.  $\Delta d_s= d_n$  remains fixed,  as such,

$\cfrac{d(d_s(x))}{dx}=\cfrac{d_n}{L}$    when we use a linear approximation, decreases with increasing  $T_{max}$.  And so,  $g_T$  decreases.

We can fix  $L$  by placing a cold body at distance  $d$  but still  $\Delta d_s= d_n$  remains the same.

So in practice  $T_{max}$  should be increased until  $g_T$  and so lift is at a maximum and then no further.  At this point  $\Delta d_s=d_n$  and cannot be increased further.