A particle falls on the earth from: (1) infinity, (2) the height of 10 times the radius of the earth. What will be the ratio of velocity gain on reaching the surface of the earth?
A) $\sqrt {11} :\sqrt {10} $
B) $\sqrt {10} :\sqrt {11} $
C) 10:11
D) 11:10

Hint: There is no external force acting in the system of the earth particle system. Therefore, Energy of the system would be conserved, hence, apply this idea for initial and final configurations of the earth and body system for the two cases.

Formula Used:
Potential Energy due to gravitational pull of earth: $P.E = \dfrac{{ - Gm{M_e}}}{r}$
Where $G$ is the gravitational constant, $m$ is the mass of earth and $M$ is the mass of Earth.
Kinetic energy of a particle moving with velocity $v$: \[K.E = \dfrac{1}{2}m{v^2}\]
Where $m$ is the mass of the particle and $v$ is velocity.
Total Energy of the system is conserved: $K.E + P.E = k$
Where $k$ is constant.

Complete step by step answer:
Given initial radii in two cases: $r = \infty $ in case particles coming from infinity and $r = 10{R_{earth}}$ in other case.
Let final velocity be ${v_1}$ and ${v_2}$ for two cases respectively
The following diagram is representing the two cases.

Step 1:
Case 1: The total energy of the combined system earth and particle:
${(T.E)_{initial}} = P.E + K.E$
Initially particle stationery, so $K.E. = 0$
${(T.E)_{initial}} = \dfrac{{ - G{M_e}m}}{\infty } + 0 = 0$ …… (1)
Similarly, in Case 2:
${(T.E)_{initial}} = P.E + K.E$
Initially particle stationery, so $K.E. = 0$
${(T.E)_{initial}} = \dfrac{{ - G{M_e}m}}{{11{R_e}}} + 0$ …… (2)
Step 2: The total energy of the combined system when reaching the earth’s surface:
Case 1:
${(T.E)_{final}} = P.{E_{surface}} + K.{E_{final}}$
${(T.E)_{final}} = \dfrac{{ - G{M_e}m}}{{{R_e}}} + \dfrac{1}{2}m{v_1}^2$ …… (3)
Similarly, in Case 2:
${(T.E)_{final}} = \dfrac{{ - G{M_e}m}}{{{R_e}}} + \dfrac{1}{2}m{v_2}^2$ …… (4)

Step 3: Using conservation of energy condition in each case:
${(T.E)_{initial}} = {(T.E)_{final}}$ …… (5)
Case 1:
Equating equation (1) and (3) we get-
$0 = \dfrac{{ - G{M_e}m}}{{{R_e}}} + \dfrac{1}{2}m{v_1}^2$
$ \Rightarrow - \dfrac{1}{2}m{v_1}^2 = \dfrac{{ - G{M_e}m}}{{{R_e}}}$
$ \Rightarrow {v_1} = \sqrt {\dfrac{{2G{M_e}}}{{{R_e}}}} $ …… (6)
Similarly, in case 2:
Equating equation (2) and (4) we get-
$\dfrac{{ - G{M_e}m}}{{11{R_e}}} + 0 = \dfrac{{ - G{M_e}m}}{{{R_e}}} + \dfrac{1}{2}m{v_2}^2$
$ \Rightarrow \dfrac{{ - G{M_e}}}{{11{R_e}}} + \dfrac{{G{M_e}}}{{{R_e}}} = \dfrac{1}{2}{v_2}^2$
$ \Rightarrow \dfrac{1}{2}{v_2}^2 = \dfrac{{ - G{M_e} + 11G{M_e}}}{{11{R_e}}} = \dfrac{{10G{M_e}}}{{11{R_e}}}$
$ \Rightarrow {v_2} = \sqrt {\dfrac{2}{1}\dfrac{{10G{M_e}}}{{11{R_e}}}} $ …… (7)

Step 4: Taking ratio of two final velocity from equation (6) and (7) we get:
$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{\sqrt {\dfrac{{2G{M_e}}}{{{R_e}}}} }}{{\sqrt {\dfrac{2}{1}\dfrac{{10G{M_e}}}{{11{R_e}}}} }} = \dfrac{{\sqrt {11} }}{{\sqrt {10} }}$

Therefore, the correct option is (A) $\sqrt {11} :\sqrt {10} $.

Note:
The total energy of a closed system is always conserved. If there is no energy transformation going on. It means that energy is not dissipating through friction or generating heat or sound or light etc. That is why it is said as a closed system, the total energy of the Universe is constant.