We’re trying to solve the system of equations left in Fascinating Frequencies 2. To start, factor out v in the second equation. We do this by multiplying both sides by ms/4πd. Then we combine the 2 equations. Next, substitute in the gravitational constant G and factor out d to find the distance. We factor out d by multiplying both sides by 72000s*d3/π and taking the fourth root of both sides. Substitute d into the first equation to solve for v. Our answer is approximately 8,032 metres per second.
\vec v=\frac{2\pi d}{72\,000s}\\[8pt] \frac{4\pi\vec vd}{ms}=\frac{Gm_1}{d^2}\\[8pt] G=\frac{6.674m^3}{10^{11}s^2kg}\\[8pt] \vec v=\frac{Gmm_1s}{4\pi d^3}=\frac{\pi d}{36\,000s}\\[8pt] \vec d\approx\sqrt[4]{\frac{2.403m^4m_1s^2}{4\times10^5\pi^2s^2kg}-1}\\[8pt] \vec v\approx\sqrt[4]{\frac{\pi^2m^4m_1s^2}{11.99 \times10^8\pi^4s^6kg}-1}\\[8pt] \vec v\approx\sqrt[4]{\frac{4.162\times10^{15}m^4}{\pi^2s^4}}=8.032\times10^3\frac ms