Answer: B. At $r=2.0\times10^7\text{ m}$ the graph gives $g=5.0\text{ N kg}^{-1}$ Since $g=GM/r^2$ $v_{\rm esc}=\sqrt{2GM/r}=\sqrt{2gr}=\sqrt{2(5.0)(2.0\times10^7)}=1.4\times10^4\text{ m s}^{-1}$

A: uses the circular orbital speed $\sqrt{gr}$ omitting the factor 2 under the square root.
C: assumes escape speed is twice the circular orbital speed, rather than $\sqrt{2}$ times it.
D: reads $g=20\text{ N kg}^{-1}$ from the graph at $r=1.0\times10^7\text{ m}$ not the value at the satellite’s radius.

Remember: escape speed is $\sqrt{2}$ times circular orbital speed at the same radius.