Answer
$r_{\circ}=11.1Mm$, $\Delta v=814m/s$
Work Step by Step
We can determine the $r_{\circ}$ and the change in velocity as follows:
$r_{\circ}=\frac{2GM_M}{v^2_A}$
We plug in the known values to obtain:
$r_{\circ}=\frac{2\times 66.73\times 10^{-12}\times 0.1074\times 5.976\times 10^{24}}{(10\times \frac{10^6}{3600})^2}$
This simplifies to:
$r_{\circ}=11.1\times 10^6=11.1Mm$
Now $v_{Ac}=\sqrt{\frac{66.73\times 10^{-12}(0.1074)\times 5.976\times 10^{24}}{11.1\times 10^6}}$
$\implies v_{Ac}=1964.19m/s$
$\Delta v=v_{Ac}-v_A$
$\implies \Delta v=1964.19-(10\times\frac{10^6}{3600})$
This simplifies to:
$\Delta v=814m/s$
