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The given data can be listed below as,

• The mass of the spacecraft is, m.
• Theradius of the earth, $R_E$.
• The mass of the earth is, $m_E$.

According to the law of conservation of energy, the total energy of an isolated system is constant.

The energy of the satellite on earth surface is expressed as,

$E_1=-G\frac{m{m}_E}{R_E}$

Here, G is the gravitational constant.

The energy of the satellite in orbit i.e.$r=R_E$ (Here is the distance between earth surface to satellite) is expressed as,

$E_2=G\frac{m{m}_E}{R}$

Here$R=r+R_E=R_E+R_E$

$\begin{array}{l}{E}_2=G\frac{m{m}_E}{R_E+R_E}\\ {}_{}\\ E_2=G\frac{m{m}_E}{2R_E}\end{array}$

The work required to launch a spacecraft in orbit is expressed as,

$W=E_2-E_1$

Here, W is the work.

Substitute all the value in the above equation.

$\begin{array}{l}w=G\frac{m{m}_E}{2R_E}+G\frac{m{m}_E}{R_E}\\ =G\frac{m{m}_E}{2R_E}\end{array}$

Hence the work required to launch a spacecraft in orbit is, $G\frac{m{m}_E}{2R_E}$.

Since the total energy of satellites far from Earth is zero, the additional work required is expressed as,

$W=E_T-E_2$

Here $E_T$is the total energy. Substitute all the value in the above equation.

$\begin{array}{l}w=0-\left(-G\frac{m{m}_E}{2R_E}\right)\\ \\ =G\frac{m{m}_E}{R_E}\end{array}$

Here, G is the gravitational acceleration.

Hence the minimum amount of addition work required is, $G\frac{m{m}_E}{R_E}$.

The work required to bring a satellite into orbit was the same as that required to bring the satellite from orbit to the edge of space.