Step 1

We are provided the following data:

The longest distance between the Earth and the rocket is, ${r_C} = 16{\rm{ Mm}} \times \frac{{{{10}^6}{\rm{ m}}}}{{1{\rm{ Mm}}}} = 16 \times {10^6}{\rm{ m}}$.

The smallest distance between the Earth and the rocket is, ${r_A} = 10{\rm{ Mm}} \times \frac{{{{10}^6}{\rm{ m}}}}{{1{\rm{ Mm}}}} = 10 \times {10^6}{\rm{ m}}$.

We are required to determine the rocket’s velocity at A.

 
Step 2

The general formula of energy conservation will be,

\[\begin{array}{c} {\rm{Kinetic Energy = Potential Energy}}\\ \frac{1}{2}{m_1}{v^2} = \frac{{G{m_1}{m_2}}}{r} \end{array}\]

Here, ${m_1}{\rm{ \;and\; }}{m_2}$ represents the mass of two different objects, r represents the distance between the two objects. G represents the gravitational constant whose value is $G = 6.67 \times {10^{ - 11}}{\rm{ }}{{\rm{m}}^{\rm{3}}}{\rm{/kg}} \cdot {{\rm{s}}^2}$.

 
Step 3

On plugging the values in the above relation for point A,

\[\frac{1}{2} \times {m_r} \times v_A^2 = \frac{{G{m_r}{m_e}}}{{{r_A}}}\]

Here,${v_A}$ represents the velocity of rocket at point A, ${m_r}$ represents the mass of the rocket and ${m_e}$ represents the mass of the earth whose value is ${m_e} = 5.97 \times {10^{24}}\,{\rm{kg}}$.

 
Step 4

On plugging the values in the above relation for point C,

\[\frac{1}{2} \times {m_r} \times v_C^2 = \frac{{G{m_r}{m_e}}}{{{r_C}}}\]

Here, ${v_C}$ represents the velocity of rocket at point C.

 
Step 5

To find the formula of velocity of rocket at A, we will get the relation by equating the energy conservation of A and C,

\[\begin{array}{c} \frac{1}{2} \times {m_r} \times v_A^2 - \frac{{G{m_r}{m_e}}}{{{r_A}}} = \frac{1}{2} \times {m_r} \times v_C^2 - \frac{{G{m_r}{m_e}}}{{{r_C}}}\\ {m_r}\left( {\frac{1}{2} \times v_A^2 - \frac{{G{m_e}}}{{{r_A}}}} \right) = {m_r}\left( {\frac{1}{2} \times v_C^2 - \frac{{G{m_e}}}{{{r_C}}}} \right)\\ \left( {\frac{1}{2} \times v_A^2 - \frac{{G{m_e}}}{{{r_A}}}} \right) = \left( {\frac{1}{2} \times v_C^2 - \frac{{G{m_e}}}{{{r_C}}}} \right) \end{array}\]
 
Step 6

To find the velocity of rocket at C, we will use the conservation of angular momentum at point A and C

\[\begin{array}{c} {r_C}{v_C} = {r_A}{v_A}\\ {v_C} = \frac{{{r_A}{v_A}}}{{{r_C}}} \end{array}\]
 
Step 7

On plugging the value of ${v_C}$ in equation (1), we get

\[\begin{array}{c} \left( {\frac{1}{2} \times v_A^2 - \frac{{G{m_e}}}{{{r_A}}}} \right) = \left( {\frac{1}{2} \times {{\left( {\frac{{{r_A}{v_A}}}{{{r_C}}}} \right)}^2} - \frac{{G{m_e}}}{{{r_C}}}} \right)\\ \left( {\frac{1}{2} \times v_A^2 - \frac{{G{m_e}}}{{{r_A}}}} \right) = \left( {\frac{1}{2} \times \left( {\frac{{r_A^2v_A^2}}{{r_C^2}}} \right) - \frac{{G{m_e}}}{{{r_C}}}} \right)\\ \frac{{v_A^2}}{2} - \frac{1}{2} \times \left( {\frac{{r_A^2v_A^2}}{{r_C^2}}} \right) = \left( {\frac{{G{m_e}}}{{{r_A}}} - \frac{{G{m_e}}}{{{r_C}}}} \right)\\ \frac{{v_A^2}}{2}\left( {1 - \frac{{r_A^2}}{{r_C^2}}} \right) = \left( {G{m_e}\left( {\frac{1}{{{r_A}}} - \frac{1}{{{r_C}}}} \right)} \right)\\ \frac{{v_A^2}}{2}\left( {\frac{{r_C^2 - r_A^2}}{{r_C^2}}} \right) = G{m_e}\left( {\frac{{{r_C} - {r_A}}}{{{r_A}{r_C}}}} \right) \end{array}\]
 
Step 8

On further solving the above equation,

\[\begin{array}{c} \frac{{v_A^2}}{2} = G{m_e}\left( {\frac{{r_C^2}}{{r_C^2 - r_A^2}}} \right)\left( {\frac{{{r_C} - {r_A}}}{{{r_A}{r_C}}}} \right)\\ v_A^2 = 2G{m_e}\left( {\frac{{{r_C}}}{{\left( {{r_C} + {r_A}} \right)\left( {{r_C} - {r_A}} \right)}}} \right)\left( {\frac{{{r_C} - {r_A}}}{{{r_A}}}} \right)\\ v_A^2 = 2G{m_e}\left( {\frac{{{r_C}}}{{{r_A}\left( {{r_C} + {r_A}} \right)}}} \right)\\ {v_A} = \sqrt {2G{m_e}\left( {\frac{{{r_C}}}{{{r_A}\left( {{r_C} + {r_A}} \right)}}} \right)} \end{array}\]
 
Step 9

On plugging the values in the above relation, we get,

\[\begin{array}{l} {v_A} = \sqrt {2 \times 6.67 \times {{10}^{ - 11}}{\rm{ }}{{\rm{m}}^{\rm{3}}}{\rm{/kg}} \cdot {{\rm{s}}^2} \times 5.97 \times {{10}^{24}}\,{\rm{kg}}\left( {\frac{{16 \times {{10}^6}\,{\rm{m}}}}{{10 \times {{10}^6}\,{\rm{m}}\left( {16 \times {{10}^6}\,{\rm{m}} + 10 \times {{10}^6}\,{\rm{m}}} \right)}}} \right)} \\ {v_A} = \sqrt {2 \times 6.67 \times {{10}^{ - 11}}{\rm{ }}{{\rm{m}}^{\rm{3}}}{\rm{/}}{{\rm{s}}^2} \times 5.97 \times {{10}^{24}}\left( {0.061 \times {{10}^6} \times {{10}^{ - 12}}{\rm{ }}{{\rm{m}}^{ - 1}}\,} \right)} \\ {v_A} = 7005.74\,{\rm{ m/s}} \end{array}\]