91Ó°ÊÓ

The given data can be listed below as,

• The masses of spacecraft,$\mathrm{m}=5000\text{ k²µ}$.
• The distance of the first orbit from mars surface is,${\mathrm{h}}_1=2000\text{ k³¾}$.
• The distance of the second orbit from mars surface is,${\mathrm{h}}_2=4000\text{ k³¾}$.

Energy (the capacity to do work) cannot be formed or destroyed; it can only be altered from one form to another.

The conservation of the energy is expressed as,

$\begin{array}{c}{K}_1+U_1+W_0=K_2+U_2\\ W_0=(K_2−K_1)+(U_2−U_1)\end{array}$ ...(i)

Here $W_o$is the work required, $U_2$is the gravitational potential energy for the orbital second, $U_1$is the gravitational potential energy for orbital first orbital, $K_2$is the kinetic energy for the orbital second,$K_1$is the kinetic energy for orbital first orbital.

The gravitational potential energy for orbital first is expressed as,

role="math" localid="1655787246995" $U_1=−\frac{Gm{M}_m}{r_1}$ ...(ii)

Here$U_1$is the potential energy,$G$is the gravitational constant,$M_m$is the mass of the mar,$m$is the mass of the spacecraft, and$r_1$is the radius of the first orbit.

The radius of the first orbit is expressed as,

role="math" localid="1655787243442" $r_1=h_1+R_m$

Here$R_m$is the radius of mars.

Substitute $2000×{10}^3\text{″¾}$ for$h_1$and$3.39×{10}^6\text{″¾}$for$R_m$in the above equation.

$\begin{array}{c}{r}_1=2000×{10}^3\text{″¾}+3.39×{10}^6\text{″¾}\\ \text{=5.39}×{\text{10}}^6\text{″¾}\end{array}$

Substitute $6.67×{10}^{−11}{\text{ N³¾}}^{\text{2}}/{\text{kg}}^{\text{2}}$for $G$, $6.42×{10}^{23}\text{ k²µ}$for$M_m$,$5000\text{ k²µ}$for$m$and$5.39×{10}^6\text{″¾}$ for$r_1$in the equation.(ii).

$\begin{array}{c}{U}_1=−\frac{6.67×{10}^{−11}{\text{ N³¾}}^{\text{2}}/{\text{kg}}^{\text{2}}×5000\text{ k²µ}×6.42×{10}^{23}\text{ k²µ}}{5.39×{10}^6\text{″¾}}\\ =−3.9723×{10}^{10}\text{ J}\end{array}$

The gravitational potential energy for orbital second is expressed as,

$U_2=−\frac{Gm{M}_m}{r_2}$ ...(iii)

Here $U_2$is the potential energy, $G$is the gravitational constant, $M_m$is the mass of the mar, $m$is the mass of spacecraft, and $r_2$is the radius of the second orbit.

The radius of the second orbit is expressed as,

$r_2=h_2+R_m$

Here$R_m$is the radius of mars.

Substitute$4000×{10}^3\text{″¾}$for$h_2$and$3.39×{10}^6\text{″¾}$for$R_m$in the above equation.

$\begin{array}{c}{r}_2=4000×{10}^3\text{″¾}+3.39×{10}^6\text{″¾}\\ \text{=7.39}×{\text{10}}^6\text{″¾}\end{array}$

Substitute$6.67×{10}^{−11}{\text{ N³¾}}^{\text{2}}/{\text{kg}}^{\text{2}}$for $G$, $6.42×{10}^{23}\text{ k²µ}$for$M_m$,$5000\text{ k²µ}$for$m$and$7.39×{10}^6\text{″¾}$for$r_1$in the equation (iii).

$\begin{array}{c}{U}_2=−\frac{6.67×{10}^{−11}{\text{ N³¾}}^{\text{2}}/{\text{kg}}^{\text{2}}×5000\text{ k²µ}×6.42×{10}^{23}\text{ k²µ}}{7.39×{10}^6\text{″¾}}\\ =−2.89725×{10}^{10}\text{ J}\end{array}$

The spacecraft move in circular orbit, and then the speed of spacecraft is expressed as,

$v=\sqrt{\frac{G{M}_m}{r}}$

Here$v$ is the speed of spacecraft,$G$ is the gravitational constant,$M_m$ is the mass of the mars and$r$ is the radius of spacecraft.

The relation of kinetic energy is expressed as,

$K=\frac{1}{2}m{v}^2$ ...(iv)

Here K is the kinetic energy.

Substitute the value of$v$ in equation (iv)

$\begin{array}{c}K=\frac{1}{2}m{\left(\sqrt{\frac{G{M}_E}{r}}\right)}^2\\ =\frac{Gm{M}_E}{2r}\end{array}$

The value of $−\frac{Gm{M}_m}{r}$is equal to the gravitational potential energy and then the kinetic energy is expressed as,

$K=−\frac{1}{2}U$

The kinetic energy for first orbit is expressed as,

$K_1=−\frac{1}{2}{U}_1$

Substitute$−3.9723×{10}^{10}\text{ J}$ for$U_1$ , in the above equation.

$\begin{array}{c}{K}_1=−\frac{1}{2}×(−3.9723×{10}^{10}\text{ J})\\ =1.98615×{10}^{10}\text{ J}\end{array}$

The kinetic energy for second orbit is expressed as,

$K_2=−\frac{1}{2}{U}_2$

Substitute$−2.89725×{10}^{10}\text{ J}$ for$U_2$ , in the above equation.

$\begin{array}{c}{K}_2=−\frac{1}{2}×(−2.89725×{10}^{10}\text{ J})\\ =1.448625×{10}^{10}\text{ J}\end{array}$

Substitute the value of$U_1$ , $U_2$, $K_1$, and$K_2$ , in the equation (i).

$\begin{array}{c}{W}_0=(1.448625×{10}^{10}\text{ J}−1.98615×{10}^{10}\text{ J})+(−2.89725×{10}^{10}\text{ J}+3.9723×{10}^{10}\text{ J})\\ =5.37×{10}^{10}\text{ J}\end{array}$

Hence the work required is,$5.37×{10}^{10}\text{ J}$ .