The escape speed ($v_e$) from the surface of a planet is given by the formula:
$ v_e = \sqrt{\frac{2GM}{R}} $
where $G$ is the universal gravitational constant, $M$ is the mass of the planet, and $R$ is its radius.
Identify Given Information:
Write Escape Speed Formulas:
Calculate the Ratio $\frac{v_2}{v_1}$:
Divide the expression for $v_2$ by the expression for $v_1$:
$ \frac{v_2}{v_1} = \frac{\sqrt{\frac{2GM}{R_2}}}{\sqrt{\frac{2GM}{R_1}}} $
Simplify the ratio:
$ \frac{v_2}{v_1} = \sqrt{\frac{\frac{2GM}{R_2}}{\frac{2GM}{R_1}}} = \sqrt{\frac{R_1}{R_2}} $
Substitute the Radius Relationship:
Substitute $R_2 = \frac{R_1}{2}$ into the ratio:
$ \frac{v_2}{v_1} = \sqrt{\frac{R_1}{\frac{R_1}{2}}} = \sqrt{\frac{2 R_1}{R_1}} = \sqrt{2} $
The ratio $\frac{v_2}{v_1}$ is $\sqrt{2}$.
The power of a crane, which lifts a mass of $1000 \text{ kg}$ to a height of $20 \text{ m}$ in $10 \text{ s}$ is :
($g = 9.8 \text{ m/s}^2$)
A thin wire of length 'L' and linear mass density 'm' is bent into a circular ring (in x-y plane) with centre 'C' as shown in figure. The moment of inertia of the ring about an axis yy' (tangent in the plane) will be :
A frictionless circular wire of unit radius is fixed on the horizontal plane. Two point particles of unit mass start moving simultaneously from point $A \left( \theta = \frac{\pi}{2} \right)$ with identical uniform angular speeds in opposite directions, and meet again at point $B \left( \theta = -\frac{\pi}{2} \right)$. During this time, which of the following figures schematically represent the magnitude of the total linear momentum $\vec{P}$ of the system, as a function of $\theta$?

Suppose there is a uniform circular disc of mass $M \text{ kg}$ and radius $r \text{ m}$ shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis $A$ of the disc is given by $\frac{x}{256} M r^2$. The value of $x$ is ________.
