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: \[ v_{escape, planet} = 5.6 \, \text{km/s} \]The velocity (v) - time (t) plot of the motion of a body is shown below :
The acceleration (a) - time(t) graph that best suits this motion is :
A wheel of a bullock cart is rolling on a level road, as shown in the figure below. If its linear speed is v in the direction shown, which one of the following options is correct (P and Q are any highest and lowest points on the wheel, respectively) ?
A body of mass 1000 kg is moving horizontally with a velocity of 6 m/s. If 200 kg extra mass is added, the final velocity (in m/s) is:
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).