\(\mathbf{22.4 \text{ km/s}}\)
Step 1: Understanding the escape velocity formula The escape velocity is given by the formula: \[ v_e = \sqrt{\frac{2 G M}{R}} \] where: - \( v_e \) is the escape velocity, - \( G \) is the universal gravitational constant, - \( M \) is the mass of the planet, - \( R \) is the radius of the planet.
Step 2: Expressing in terms of density Since density \( \rho \) is given by: \[ \rho = \frac{M}{\frac{4}{3} \pi R^3} \] we can express the mass in terms of density and radius: \[ M = \rho \times \frac{4}{3} \pi R^3 \] Substituting in the escape velocity formula: \[ v_e = \sqrt{\frac{2 G \rho \times \frac{4}{3} \pi R^3}{R}} \] Since \( \rho \) is constant for both Earth and the given planet, the escape velocity depends on: \[ v_e \propto \sqrt{\frac{M}{R}} \] Step 3: Finding the relation between escape velocities For Earth: \[ v_{e,E} = \sqrt{\frac{2 G M_E}{R_E}} \] For the given planet (with mass \( M_P = 8 M_E \) and same density): \[ \frac{M_P}{M_E} = 8 \] Since density is the same: \[ \frac{M_P}{M_E} = \frac{R_P^3}{R_E^3} \] Taking cube root: \[ \frac{R_P}{R_E} = \sqrt[3]{8} = 2 \] Now, the escape velocity ratio is: \[ \frac{v_{e,P}}{v_{e,E}} = \sqrt{\frac{M_P}{R_P}} \] \[ = \sqrt{\frac{8 M_E}{2 R_E}} = \sqrt{4} = 2 \] Thus, \[ v_{e,P} = 2 v_{e,E} \]
Step 4: Calculating the final value Given that: \[ v_{e,E} = 11.2 \text{ km/s} \] \[ v_{e,P} = 2 \times 11.2 = 22.4 \text{ km/s} \]
Step 5: Verifying the correct option Comparing with given options, the correct answer is: \[ \mathbf{22.4 \text{ km/s}} \]
A small point of mass \(m\) is placed at a distance \(2R\) from the center \(O\) of a big uniform solid sphere of mass \(M\) and radius \(R\). The gravitational force on \(m\) due to \(M\) is \(F_1\). A spherical part of radius \(R/3\) is removed from the big sphere as shown in the figure, and the gravitational force on \(m\) due to the remaining part of \(M\) is found to be \(F_2\). The value of the ratio \( F_1 : F_2 \) is: