Gravitation Near Earth's Surface — Question 9

Question 9

Vertical free-fall separation grows because the lower body accelerates more

Two small freely falling objects are initially separated vertically by $\ell\ll R$, close to the surface of a nonrotating spherical Earth. Their initial relative velocity is zero. Use $g_0=10\,\mathrm{m/s^2}$, $R=6.4\times10^6\,\mathrm m$ and $\ell=2\,\mathrm m$.

1. Find their initial relative acceleration, keeping first order in $\ell/R$.

2. Estimate the increase in separation after $1\,\mathrm s$, over which field changes can be neglected.

Question 9 - Solution

Vertical free-fall separation grows because the lower body accelerates more

(a) Derivation and calculation

Take outward radial coordinate positive, so $a_r=-GM/r^2$. For separation $\ell$, $$\ddot\ell\simeq\frac{da_r}{dr}\ell =\frac{2GM}{R^3}\ell=\frac{2g_0\ell}{R}.$$ $$\boxed{\ddot\ell=\frac{2(10)(2)}{6.4\times10^6} =6.25\times10^{-6}\ \mathrm{m/s^2}}.$$

(b) Result and interpretation

With zero initial relative speed and approximately constant tidal acceleration, $$\Delta\ell\simeq\frac12\ddot\ell t^2 =\frac12(6.25\times10^{-6})(1)^2.$$ $$\boxed{\Delta\ell=3.125\times10^{-6}\ \mathrm m=3.125\ \mu\mathrm m}.$$ The lower particle falls faster, increasing their vertical separation.

Physical check

The fractional separation change is $1.5625\times10^{-6}$, consistent with the short-time linearization.