Gravitation Inside Earth — Question 10

Question 10

A density jump changes the slope of gravity without making gravity jump

A spherical body has a finite density discontinuity at radius $c$, but no infinitesimally thin massive surface layer. Let $g(r)=GM(r)/r^2$ be its inward field magnitude.

1. Derive $dg/dr$ in terms of local density and $g$.

2. Explain the continuity of $g$ and calculate the outward jump in slope if density falls from $\rho_1$ to $\rho_2$.

Question 10 - Solution

A density jump changes the slope of gravity without making gravity jump

(a) Derivation and calculation

Differentiate the inward field magnitude: $$g(r)=\frac{GM(r)}{r^2},\quad \frac{dg}{dr}=\frac G{r^2}\frac{dM}{dr}-\frac{2GM(r)}{r^3}.$$ Since $M'(r)=4\pi r^2\rho(r)$, $$\boxed{g'(r)=4\pi G\rho(r)-\frac{2g(r)}r}.$$

(b) Result and interpretation

Without a massive surface sheet, the enclosed mass is continuous: $$M(c^+)-M(c^-)=\lim_{\delta\to0}\int_{c-\delta}^{c+\delta}4\pi r^2\rho(r)\,dr=0.$$ Hence $g(c^+)=g(c^-)$ and the $-2g/r$ terms cancel in the slope difference: $$\boxed{g'(c^+)-g'(c^-)=4\pi G(\rho_2-\rho_1)<0}.$$

Physical check

A finite downward density jump lowers the slope, but need not make the field itself decrease immediately.