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$$^2$$)

The body starts from rest at infinity, so its initial mechanical energy is zero (both kinetic and potential energies are zero at infinity).

When it reaches the Earth’s surface, let its speed be $$v$$. Its kinetic energy is $$\dfrac12\,mv^{2}$$ and its gravitational potential energy is $$-\dfrac{GMm}{R}$$, where $$G$$ is the universal gravitational constant, $$M$$ is Earth’s mass, $$R$$ is Earth’s radius.

Using conservation of mechanical energy (initial energy = final energy): $$0 = \dfrac12\,mv^{2} - \dfrac{GMm}{R}$$ $$\Rightarrow \dfrac12\,mv^{2} = \dfrac{GMm}{R}$$ $$\Rightarrow v^{2} = \dfrac{2GM}{R} \qquad -(1)$$

Recall that $$g = \dfrac{GM}{R^{2}}$$. Multiplying both sides by $$R$$ gives $$gR = \dfrac{GM}{R}$$. Substitute this in $$(1)$$:

$$v^{2} = 2gR \qquad -(2)$$

Numerical substitution (convert radius to metres): $$R = 6400\text{ km} = 6400 \times 10^{3}\text{ m} = 6.4 \times 10^{6}\text{ m}$$

Using $$g = 9.8\text{ m s}^{-2}$$, from $$(2)$$ we get $$v = \sqrt{2 \times 9.8 \times 6.4 \times 10^{6}}$$ $$v = \sqrt{1.2544 \times 10^{8}}$$ $$v = 1.12 \times 10^{4}\text{ m s}^{-1}$$ $$v = 11.2\text{ km s}^{-1}$$

The kinetic energy on arrival is $$k = \dfrac12\,mv^{2}$$ For $$m = 1\text{ kg}$$, $$k = \dfrac12 \times 1 \times (1.12 \times 10^{4})^{2}$$ $$k = \dfrac12 \times 1.2544 \times 10^{8}$$ $$k = 6.272 \times 10^{7}\text{ J}$$ $$k \approx 6.27 \times 10^{7}\text{ J}$$

Thus, $$v = 11.2\text{ km s}^{-1}$$ and $$k = 6.27 \times 10^{7}\text{ J}$$.

Option A which is: 11.2 km/s; 6.27 × 107 J