Angular speed of earth around sun is\[ \dot{\theta }=\frac{2\pi }{\left ( 364\right ) \left ( 24\right ) \left ( 60\right ) \left ( 60\right ) }\] Force on earth is therefore \(mr\dot{\theta }^{2}\). Equating this to \(F=G\frac{m_{e}m_{s}}{r^{2}}\) and solving for \(r\) gives\[ r\dot{\theta }^{2}=G\frac{m_{s}}{r^{2}}\] One equation with one unknown \(r\). Solving gives (taking the positive root)\[ r=149.26\times 10^{9}\text{ meter}\]
\[ y\left ( x\right ) =8\sin \left ( \pi x\right ) \] We want to solve for \(v\) in \[ \frac{v^{2}}{\rho }=g \] But \(\rho =\frac{\left ( 1+y^{\prime }\left ( x\right ) ^{2}\right ) ^{\frac{3}{2}}}{\left \vert y^{\prime \prime }\left ( x\right ) \right \vert }\). To find what \(x\) to use, since at top of hill, then we want \(\sin \left ( \pi x\right ) =1\) or \(x=\frac{1}{2}\). Plugging this into \(\rho \) gives \[ \rho =0.0126651 \] Hence \[ \frac{v^{2}}{0.0126651}=9.81 \] Or\[ v=0.3525\text{ m/s}\]