Velocity and Kinetic Energy of Body Falling from Infinity to Earth Surface

To find the velocity ($v$) and kinetic energy ($k$) of a body falling from infinity to the surface of the Earth, we apply the Law of Conservation of Mechanical Energy.

1. Given Data:

• Mass of the body, $m = 1\text{ kg}$
• Radius of the Earth, $R = 6400\text{ km} = 6.4 \times 10^6\text{ m}$
• Acceleration due to gravity at Earth's surface, $g = 9.8\text{ m/s}^2$

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2. Conservation of Energy:

At infinity ($r = \infty$):

• Initial Potential Energy, $U_i = 0$
• Initial Kinetic Energy, $K_i = 0$
• Total Initial Mechanical Energy, $E_i = U_i + K_i = 0$

At the surface of the Earth ($r = R$):

• Final Potential Energy, $U_f = -\frac{GMm}{R}$
• Final Kinetic Energy, $K_f = k = \frac{1}{2}mv^2$

Since $g = \frac{GM}{R^2}$, we can rewrite $GM = gR^2$. Thus, the potential energy at the surface is: $U_f = -\frac{(gR^2)m}{R} = -mgR$

By the conservation of energy: $E_i = E_f$ $0 = K_f + U_f$ $k - mgR = 0 \implies k = mgR$

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3. Calculation of Kinetic Energy ($k$):

Substitute the given values into the equation for kinetic energy: $k = mgR$ $k = 1\text{ kg} \times 9.8\text{ m/s}^2 \times (6.4 \times 10^6\text{ m})$ $k = 62.72 \times 10^6\text{ J} = 6.27 \times 10^7\text{ J}$

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4. Calculation of Velocity ($v$):

Using the expression for kinetic energy: $k = \frac{1}{2}mv^2 = mgR$ $v = \sqrt{2gR}$

Substitute the given values: $v = \sqrt{2 \times 9.8\text{ m/s}^2 \times 6.4 \times 10^6\text{ m}}$ $v = \sqrt{125.44 \times 10^6\text{ m}^2/\text{s}^2}$ $v = 11.2 \times 10^3\text{ m/s} = 11.2\text{ km/s}$

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Conclusion:

• $v = 11.2\text{ km/s}$
• $k = 6.27 \times 10^7\text{ J}$

Hence, the correct option is A.