A planet with a radius of \(6.00 \times 10^{7} \mathrm{m}\) has a gravitational field of magnitude \(30.0 \mathrm{m} / \mathrm{s}^{2}\) at the surface. What is the escape speed from the planet?

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 \(1.57 \times 10^{4} \, \mathrm{m} / \mathrm{s}\).