## Physics: Principles with Applications (7th Edition)

$1.21\times10^6m$
Find the 'new' Earth radius by setting the acceleration due to gravity at the Sun's surface equal to the acceleration due to gravity at the 'new' Earth's surface. $$g_{Earth new}=g_{Sun}$$ $$\frac{GM_{Earth}}{r^2_{Earthnew}}=\frac{GM_{Sun}}{r^2_{Sun}}$$ $$r_{Earthnew}=r_{Sun}\sqrt{\frac{M_{Earth}}{M_{Sun}}}=(6.96\times10^8m)\sqrt{\frac{5.98\times10^{24}kg}{1.99\times10^{30}kg}}=1.21\times10^6m$$