91Ó°ÊÓ

Given data can be listed below as,

• The mass of the Mars orbiter is, 629 kg.
• The mass of the earth is,$M_E=5.97x{10}^{24}\mathrm{kg}$
• Distance of the Mars orbiter spacecraft from earth is, $R=2.87×{10}^{6}km$.
• The relative velocity of the Mars Orbiter spacecraft is, role="math" localid="1668062634224" $v\text{'}=1.20×{10}^4\mathrm{km}/\mathrm{h}\left[\frac{\left(\frac{5}{18}\right)\mathrm{m}/\mathrm{s}}{1\mathrm{km}/\mathrm{h}}\right]$.

The potential energy harbored at the height where it was before when an object started to fall (where its GPE was maximum and KE was zero); will get converted to kinetic energy fully with a minor loss of energy due to the friction of the air object.

The kinetic energy of the spacecraft can be given by,

$k=\frac{1}{2}mv{\text{'}}^2$

Here, m is the mass of the earth and v' is the relative velocity of spacecraft.

Substitute the values above,

$\begin{array}{r}k=\frac{1}{2}\left(5.97×{10}^{24}\mathrm{kg}\right){\left(3.33×{10}^3\mathrm{m}/\mathrm{s}\right)}^2\left(\frac{1\mathrm{J}/\mathrm{kg}}{1{\mathrm{m}}^2/{\mathrm{s}}^2}\right)\\ =3.49×{10}^9\mathrm{J}\end{array}$

Thus, the relative kinetic energy of spacecraft is $3.49×{10}^9\mathrm{J}$.

The gravitational potential energy of the system can be expressed as,

$U=-\frac{G{M}_{E}m}{r}$

Here, G is the constant of gravitation whose value is $G=6.673×{10}^{-11}N·m^2/k{g}^2,M_E$is the mass of earth, m is th mass of spacecraft.

Substitute values in the above equation.

$\begin{array}{r}U=−\frac{\left(6.673×{10}^{−11}\mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{kg}}^2\right)\left(5.97×{10}^{24}\mathrm{kg}\right)\left(629\mathrm{kg}\right)}{2.87×{10}^9\mathrm{m}}\left(\frac{1\mathrm{J}/\mathrm{kg}â‹\dots 1\mathrm{kg}â‹\dots \mathrm{m}/{\mathrm{s}}^2}{1{\mathrm{m}}^2/{\mathrm{s}}^{2}1\mathrm{N}}\right)\\ =−8.73×{10}^7\mathrm{J}\end{array}$

Thus, the gravitational potential energy of earth–spacecraft system is $−8.73×{10}^7\mathrm{J}$.