If a body of mass 1text{ kg} falls on the ear | vedclass

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Question

(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	e

Vedclass · Accepted Answer

$11.2	ext{ km/s}$; $6.27 	imes 10^7	ext{ J}$

Vedclass · Answer

(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 $g = 9.8	ext{ m/s}^2$ and $R = 6400	ext{ km} = 6.4 	imes 10^6	ext{ m}$.$v_e = \sqrt{2 	imes 9.8 	imes 6.4 	imes 10^6} = \sqrt{125.44 	imes 10^6} = 11.2 	imes 10^3	ext{ m/s} = 11.2	ext{ km/s}$.The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$.Substituting $m = 1	ext{ kg}$ and $v = 11.2 	imes 10^3	ext{ m/s}$:$K = \frac{1}{2} 	imes 1 	imes (11.2 	imes 10^3)^2 = 0.5 	imes 125.44 	imes 10^6 = 62.72 	imes 10^6	ext{ J} = 6.27 	imes 10^7	ext{ J}$.

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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