If a body of mass \(1\) 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 \(6400\) km and \(g = 9.8\) m/s\(^2\))

By conservation of mechanical energy, the total energy of the body at infinity is zero.

When the body reaches the surface of the earth, its potential energy is \(-\dfrac{GMm}{R}\) and kinetic energy is \(k = \dfrac{1}{2}mv^2\).

\(0 = k - \dfrac{GMm}{R}\)

\(k = \dfrac{GMm}{R} = mgR\)

Substituting \(m = 1\) kg, \(g = 9.8\) m/s\(^2\), and \(R = 6400 \times 10^3\) m:

\(k = 1 \times 9.8 \times 6400 \times 10^3 = 6.272 \times 10^7\) J

The velocity \(v\) is given by \(\dfrac{1}{2}mv^2 = k\).

\(v = \sqrt{2gR} = \sqrt{2 \times 9.8 \times 6400 \times 10^3}\)

\(v = \sqrt{125.44 \times 10^6} = 11.2 \times 10^3\) m/s \(= 11.2\) km/s

Answer: \(11.2\) km/s; \(6.27 \times 10^7\) J