91Ó°ÊÓ

To find the mass of the planet, we'll use the formula for the gravitational field \(g\) at the surface of the planet: \(g = \dfrac{G \cdot M}{R^{2}}\) where \(G\) is the gravitational constant (\(6.674 \times 10^{-11} \mathrm{m}^3 \cdot \mathrm{kg}^{-1} \cdot \mathrm{s}^{-2}\)), \(M\) is the mass of the planet, and \(R\) is the radius of the planet. We can solve for \(M\): \(M = \dfrac{g \cdot R^{2}}{G}\) Using the given values, \(g = 30.0 \mathrm{m} / \mathrm{s}^{2}\) and \(R = 6.00 \times 10^{7} \mathrm{m}\): \(M = \dfrac{30.0 \mathrm{m} / \mathrm{s}^{2} \cdot (6.00 \times 10^{7} \mathrm{m})^{2}}{6.674 \times 10^{-11} \mathrm{m}^3 \cdot \mathrm{kg}^{-1} \cdot \mathrm{s}^{-2}}\) After calculating, we find that the mass of the planet is \(M \approx 8.09 \times 10^{24} \mathrm{kg}\).

To escape from the gravitational pull of a planet, an object needs to have sufficient kinetic energy to overcome the gravitational potential energy. The escape speed \(v_e\) can be calculated using the formula: \(v_{e} = \sqrt{\dfrac{2 \cdot G \cdot M}{R}}\) Using the values we found for \(M\) and \(R\), as well as the value for \(G\): \(v_{e} = \sqrt{\dfrac{2 \cdot (6.674 \times 10^{-11} \mathrm{m}^3 \cdot \mathrm{kg}^{-1} \cdot \mathrm{s}^{-2}) \cdot (8.09 \times 10^{24} \mathrm{kg})}{(6.00 \times 10^{7} \mathrm{m})}}\) After calculating, we find that the escape speed is \(v_{e} \approx 1.57 \times 10^{4} \mathrm{m} / \mathrm{s}\). Thus, the escape speed from the planet is approximately \(1.57 \times 10^{4} \, \mathrm{m} / \mathrm{s}\).