The escape speed from Earth is \(11.2 \mathrm{km} / \mathrm{s},\) but that is only the minimum speed needed to escape Earth's gravitational pull; it does not give the object enough energy to leave the solar system. What is the minimum speed for an object near the Earth's surface so that the object escapes both the Earth's and the Sun's gravitational pulls? Ignore drag due to the atmosphere and the gravitational forces due to the Moon and the other planets. Also ignore the rotation and the orbital motion of the Earth.

First, we need to find the initial potential energy of the object due to both Earth's and Sun's gravitational pull. The gravitational potential energy is given by the formula: \(U = - \frac{G * m * M}{r}\). For Earth's gravitational pull, we have: - \(G\): The gravitational constant (\(6.674 \times 10^{-11} \,\mathrm{N\cdot m^2 / kg^2}\)) - \(m\): The mass of the object (unknown but will cancel out later) - \(M_E\): The mass of Earth (\(5.97 \times 10^{24} \mathrm{kg}\)) - \(r_E\): The radius of Earth (\(6.37 \times 10^6 \mathrm{m}\)) We can now find the potential energy due to Earth's gravitational pull as: \(U_E = - \frac{G * m * M_E}{r_E}\). For the Sun's gravitational pull, we have: - \(M_S\): The mass of the Sun (\(1.989 \times 10^{30} \mathrm{kg}\)) - \(r_S\): The distance of the Earth from the Sun (\(1.496 \times 10^{11} \,\mathrm{m}\)) We can now find the potential energy due to Sun's gravitational pull as: \(U_S = - \frac{G * m * M_S}{r_S}\). The total initial potential energy of the object is the sum of these values: \(U_{total} = U_E + U_S\).

The escape velocity from Earth's gravitational pull is given as \(v_E = 11.2 \frac{\mathrm{km}}{\mathrm{s}} = 11200 \frac{\mathrm{m}}{\mathrm{s}}\). We can now find the initial kinetic energy of the object due to Earth's gravitational pull as: \(K_E = \frac{1}{2} * m * v_E^2\).

The object is now located at a distance \(r_S\) from the Sun and has already escaped Earth's gravitational pull. The escape velocity from the Sun's gravitational pull can be calculated using the formula: \(v_S^2 = 2 * G * M_S / r_S\). We now find the initial kinetic energy of the object due to the Sun's gravitational pull as: \(K_S = \frac{1}{2} * m * v_S^2\).

The total initial kinetic energy of the object is the sum of the kinetic energy due to Earth's and Sun's gravitational pulls: \(K_{total} = K_E + K_S\).

The object will escape both Earth and the Sun if the sum of its kinetic and potential energy is greater than or equal to zero, i.e., \(K_{total} + U_{total} \ge 0\). Substitute the expressions found in the previous steps, and we'll get a single equation with the only unknown, the mass of the object (\(m\)). However, it's worth noting that mass cancels out, and our final expression for the minimum velocity to escape both Earth's and Sun's gravitational pulls will not have \(m\) in it. The result will provide the value for the minimum velocity for an object near the Earth's surface to escape both Earth's and the Sun's gravitational pulls.