Step 1

We are given the circular orbit at altitude of $h = 20\,{\rm{Mm}}$.

We are asked to determine the minimum increment speed it must have in order to escape the earth’s gravitational field.

 
Step 2

Find the radius of the rocket orbit using the following relation.

\[{r_O} = h + {R_e}\]

Here, ${R_e}$ is the radius of the Earth and its value is $6378 \times {10^3}\;{\rm{m}}$.

On substituting the known values in the above equation we get,

\[\begin{array}{c} {r_O} = \left( {20\,{\rm{Mm}} \times \frac{{{{10}^6}\,{\rm{m}}}}{{1\,{\rm{Mm}}}}} \right) + \left( {6378 \times {{10}^3}\,{\rm{m}}} \right)\\ = \left( {20 \times {{10}^6}\,{\rm{m}}} \right) + \left( {6378 \times {{10}^3}\,{\rm{m}}} \right)\\ = 26378 \times {10^3}{\rm{m}} \end{array}\]
 
Step 3

Find the speed of the rocket at circular orbit using the following relation.

\[{v_s} = \sqrt {\frac{{G{M_e}}}{{{r_O}}}} \]

Here, ${M_e}$ is the mass of the Earth and its value is $5.976 \times {10^{24}}\;{\rm{kg}}$ and $G$ is the gravitational constant and its value is $\left( {66.73 \times {{10}^{ - 12}}\,{{\rm{m}}^3}/{\rm{kg}} \cdot {{\rm{s}}^2}} \right)$.

On substituting the known values in the above equation we get,

\[\begin{array}{c} {v_s} = \sqrt {\frac{{\left( {66.73 \times {{10}^{ - 12}}\,{{\rm{m}}^3}/{\rm{kg}} \cdot {{\rm{s}}^2}} \right)\left( {5.976 \times {{10}^{24}}\,{\rm{kg}}} \right)}}{{26378 \times {{10}^3}{\rm{m}}}}} \\ = \sqrt {\frac{{\left( {398.7785 \times {{10}^{12}}{{\rm{m}}^{\rm{3}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}} \right)}}{{26378 \times {{10}^3}\,{\rm{m}}}}} \\ = \left( {3888.17\,{\rm{m/s}} \times \frac{{{{10}^{ - 3}}\,{\rm{km/s}}}}{{1\,{\rm{m/s}}}}} \right)\\ = 3.8882\,{\rm{km/s}} \end{array}\]
 
Step 4

Find the speed of the rocket at parabolic trajectory to escape from the earth’s gravitational force using the following relation.

\[{v_p} = \sqrt {\frac{{2G{M_e}}}{{{r_O}}}} \]

On substituting the known values in the above equation we get,

\[\begin{array}{c} {v_p} = \sqrt {\frac{{2\left( {66.73 \times {{10}^{ - 12}}\,{{\rm{m}}^3}/{\rm{kg}} \cdot {{\rm{s}}^2}} \right)\left( {5.976 \times {{10}^{24}}\,{\rm{kg}}} \right)}}{{\left( {26378 \times {{10}^3}{\rm{m}}} \right)}}} \\ = \sqrt {\frac{{\left( {797.557 \times {{10}^{12}}{{\rm{m}}^{\rm{3}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}} \right)}}{{\left( {26378 \times {{10}^3}\,{\rm{m}}} \right)}}} \\ = \left( {5498.7\,{\rm{m/s}} \times \frac{{{{10}^{ - 3}}\,{\rm{km/s}}}}{{1\,{\rm{m/s}}}}} \right)\\ = 5.4987\,{\rm{km/s}} \end{array}\]
 
Step 5

Find the minimum increment speed it must have in order to escape the earth’s gravitational field using the following relation.

\[\Delta v = {v_p} - {v_s}\]

On substituting the known values in the above equation we get,

\[\begin{array}{c} \Delta v = \left( {5.4987\,{\rm{km/s}}} \right) - \left( {3.8882\,{\rm{km/s}}} \right)\\ = 1.611\,{\rm{km/s}} \end{array}\]