From the post "Temperature",
\(\cfrac{\partial T}{\partial t}=\cfrac{L}{g_T}\cfrac{\partial^3 T}{\partial t^3}\)
But how does \(g_T\) behaves? Under normal circumstances, we would have to consider both gravity terms due to \(v^2_s\) and \(v^2_{rg}\) in the energy conservation equation,
\((mv^2_t+\cfrac{1}{2}mv^2_s+\cfrac{1}{2}mv^2_{rg})+(qv^2_{tc}+\cfrac{1}{2}qv^2_c+\cfrac{1}{2}qv^2_{rc})=c^2\)
where normal gravity is attributed to \(v^2_s\), and thermal gravity from the term \(v^2_{rg}\).
Intuitively, high temperature make space less dense and a gravity pointing towards denser/colder space pulls the heat away from the hot spot.
This gravity was approximated as \(g_T=\cfrac{H_o}{r^2}\), which is inadequate. Temperature spread out in space, thermal gravity within it is like gravity inside a solid with a material density distribution that in general thins out into space.
Inside Earth, assuming uniform density,
\(g=G.r=\cfrac{g_o}{r_e}r\)
gravity increases linearly with the radius from Earth's center.
If temperature is uniform, then thermal gravity within it is zero. We expect thermal gravity to be positive, pointing towards lower temperature, and decreases to zero at the far end.
The gravity profile we seek will be between zero and a positive monotonously decreasing bound, possibly \(\cfrac{1}{r^2}\). But, since \(T\) is propagating outward we expect the total net gravity to be outwards; \(g_T\) should be higher than the gravity of the hot body and gravity that develops within \(T\) as it spreads outwards.
