## Essential University Physics: Volume 1 (3rd Edition)

$6.98\times10^6\ m$
We use conservation of angular momentum: $(2.91\times10^{-6}(\frac{2}{5})mR^2=(.048)(\frac{2}{5})(.6m)(xR)^2$ $x=.01$ Thus, we find: $R=.01R_{sun}=6.98\times10^6\ m$