51 | | To get the mass of the shell we take the difference between the masses of the original uniform sphere and the shell at some later time t. |

| 51 | To get the mass of the shell we take the difference between the masses of the original uniform sphere and the shell at some later time t. |

| 52 | |

| 53 | Here is a graphic of the situation: |

| 54 | |

| 55 | |

| 56 | Since for uniform collapse, we can numerically solve for R(t), and the uniform rho(t) of the collapsing sphere, we can calculate the mass of the shell outside of the BE sphere, |

| 57 | |

| 58 | |

| 59 | [[latex($M(t)= \frac {4}{3} \pi (R^3(t) - R_{BE}^3) \rho(t)$)]] |

| 60 | |

| 61 | Then the total mass that has fallen "onto" the BE sphere is given by, |

| 62 | |

| 63 | [[latex($M_{shell} = M_{init} - M(t)$)]] |

| 64 | |

| 65 | where |

| 66 | |

| 67 | [[latex($M_{init} = \frac{4}{3}\pi R_0^3 \rho_0$)]] |