Chapter 8, Problem 105A

To determine

The Roche limit for the earth-moon system.

Explanation of Solution

Given:

Radius of the Earth, ( $R$ ) = 6371000 m

Volumetric density of earth, ${\rho }_M=5510{\text{kgm}}^{\text{-3}}$

Volumetric density of moon, ${\rho }_m=3340{\text{kgm}}^{\text{-3}}$

Formula used:

  $d=2.423\times R\times {\left(\frac{{\rho }_M}{{\rho }_m}\right)}^{\frac{1}{3}}$

Where, $d$ is Roche limit.

  $R$ = radius of the planet

  ${\rho }_M$ = volumetric density of planet

  ${\rho }_m$ = volumetric density of satellite

Calculation:

On putting the values

  $\begin{array}{l}d=2.423\times 6371000\times {\left(\frac{5510}{3340}\right)}^{\frac{1}{3}}=18240191.63\text{m}\\ d=1.8\times {10}^7\text{m}\end{array}$

Hence Roche limit of Earth-Moon system is $1.8\times {10}^7\text{m}$ .

Conclusion:

Thus, Roche limit of Earth-Moon system is $1.8\times {10}^7\text{m}$ .