The escape velocity is given by the formula: \[ v_{escape} = \sqrt{\frac{2GM}{R}} \]
Step 2: Given Mass and Radius RelationshipsWe are given the following mass and radius relationships: \[ \frac{M_{planet}}{M_{earth}} = \frac{1}{8}, \quad \frac{R_{planet}}{R_{earth}} = \frac{1}{2} \]
Step 3: Calculate the Ratio of Escape VelocitiesThe ratio of the escape velocities is: \[ \frac{v_{escape, planet}}{v_{escape, earth}} = \sqrt{\frac{M_{planet} R_{earth}}{M_{earth} R_{planet}}} = \frac{1}{2} \]
Step 4: Calculate the Escape Velocity from the PlanetTherefore, the escape velocity from the planet is: \[ v_{escape, planet} = \frac{1}{2} \times 11.2 = 5.6 \, \text{km/s} \]
Final Answer: \[ v_{escape, planet} = 5.6 \, \text{km/s} \]A block of mass 2 kg is tied to one end of a 2 m long metal wire of 1.0 mm² area of cross-section and rotated in a vertical circle such that the tension in the wire is zero at the highest point. If the maximum elongation in the wire is 2 mm, the Young's modulus of the metal is:
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: