An astronaut whose height is $1.50 \text{m}$ floats "feet down" in an orbiting space shuttle at a distance $\text{r}=\sqrt[3]{6.67}\times {10}^6 \text{m}$ away from the centre of Earth. The magnitude of difference between the gravitational acceleration at her feet and at her head is found to be $\text{N}\times {10}^{-6} \mathrm{m} {\mathrm{s}}^{-2}$. What is the value of $\mathrm{N}$?   
[${\text{M}}_{\text{E}}=6\times {10}^{24} \text{kg}$ and $\mathrm{G}=6.67\times {10}^{-11} \mathrm{N} {\mathrm{m}}^2 {\mathrm{kg}}^{-2}$]