## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(a) The radius of the planet is $1.80\times 10^7~m$ (b) The rocket needs a minimum speed of 9.41 km/s to escape from the planet.
(a) We can find the radius $R$ of the planet. Let $M$ be the mass of the planet which is twice the mass of the earth. We know that the free-fall acceleration on the earth is $g = 9.80~m/s^2$. $\frac{G~M}{R^2} = \frac{1}{4}~g$ $R^2 = \frac{4~G~M}{g}$ $R = \sqrt{\frac{4~G~M}{g}}$ $R =\sqrt{\frac{(4)(6.67\times 10^{-11}~m^3/kg~s^2)(2)(5.98\times 10^{24}~kg)}{9.80~m/s^2}}$ $R = 1.80\times 10^7~m$ The radius of the planet is $1.80\times 10^7~m$ (b) We can use the equation for escape speed to find the escape speed from this planet. $v_{esc} = \sqrt{\frac{2~G~M}{R}}$ $v_{esc}=\sqrt{\frac{(2)(6.67\times 10^{-11}~m^3/kg~s^2)(2)(5.98\times 10^{24}~kg)}{1.80\times 10^{7}~m}}$ $v_{esc} = 9.41\times 10^3~m/s$ $v_{esc} = 9.41~km/s$ The rocket needs a minimum speed of 9.41 km/s to escape from the planet.