(A) The velocity attained by a body falling from infinity to the earth's surface is equal to the escape velocity, $v_e = \sqrt{2gR}$.Given $g = 9.8\te

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$11.2\text{ km/s}$; $6.27 \times 10^7\text{ J}$

(A) The velocity attained by a body falling from infinity to the earth's surface is equal to the escape velocity, $v_e = \sqrt{2gR}$.Given $g = 9.8\text{ m/s}^2$ and $R = 6400\text{ km} = 6.4 \times 10^6\text{ m}$.$v_e = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} = \sqrt{125.44 \times 10^6} = 11.2 \times 10^3\text{ m/s} = 11.2\text{ km/s}$.The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$.Substituting $m = 1\text{ kg}$ and $v = 11.2 \times 10^3\text{ m/s}$:$K = \frac{1}{2} \times 1 \times (11.2 \times 10^3)^2 = 0.5 \times 125.44 \times 10^6 = 62.72 \times 10^6\text{ J} = 6.27 \times 10^7\text{ J}$.

Class 11 Physics Gravitation Escape Velocity and Escape Energy

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If a body of mass $1\text{ kg}$ falls on the earth from infinity, it attains velocity $(v)$ and kinetic energy $(k)$ on reaching the surface of earth. The values of $v$ and $k$ respectively are . . . . . . . (Take radius of earth to be $6400\text{ km}$ and $g = 9.8\text{ m/s}^2$)

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A
$11.2\text{ km/s}$; $6.27 \times 10^7\text{ J}$
B
$11.2\text{ km/s}$; $12.54 \times 10^7\text{ J}$
C
$8.8\text{ km/s}$; $6.27 \times 10^7\text{ J}$
D
$8.8\text{ km/s}$; $12.54 \times 10^7\text{ J}$

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If a body of mass $1\text{ kg}$ falls on the earth from infinity, it attains velocity $(v)$ and kinetic energy $(k)$ on reaching the surface of earth. The values of $v$ and $k$ respectively are . . . . . . . (Take radius of earth to be $6400\text{ km}$ and $g = 9.8\text{ m/s}^2$)

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