91Ó°ÊÓ

- The mass of Mars is $M=6.4×{10}^3\mathrm{kg}$

- The radius of Mars is $3.4×{10}^6\mathrm{m}$

- The initial distance of spacecraft is $d_i=7×{10}^{\mathrm{b}}\mathrm{m}$

- The final distance of spacecraft is $d_f=4×{10}^6\mathrm{m}$

- The spacecraft's initial speed is $v_i=3000\mathrm{m}/\mathrm{s}$

- The spacecraft's initial speed is $v_f$

- The mass of a satellite is $m$

The principle of energy conservation states that, the addition of initial kinetic and potential energy is equal to the addition of final potential and kinetic energy. The expression will be, $P_i+K_i=P_f+K_t⋯⋯\left(1\right)$

The potential energy and kinetic energy is conserved for the system, From Equation (1), we can get the expression for speed.

$-\frac{GMm}{d_i}+\frac{1}{2}m{v}_i^2=-\frac{GMm}{d_f}+\frac{1}{2}m{v}_f^2$

$-\frac{GM}{d_i}+\frac{1}{2}{v}_i^2=-\frac{GM}{d_f}+\frac{1}{2}{v}_f^2$

$v_f=\sqrt{2\left(-\frac{GM}{d_i}+\frac{1}{2}{v}_i^2+\frac{GM}{d_f}\right)}⋯⋯\left(2\right)$

( where, $G=$Gravitational constant $=6.67×{10}^{-11}{\mathrm{m}}^3/\mathrm{kg}·{\mathrm{s}}^2$)

Substitute the values of $M,d_i,d_f,v_i$from the given data.

From Equation (2),

$v_f=\sqrt{2\left(-\frac{\left(6.67×{10}^{-11}{\mathrm{m}}^3/\mathrm{kg}·{\mathrm{s}}^2\right)×6.4×{10}^3\mathrm{kg}}{7×{10}^6\mathrm{m}}+\frac{1}{2}×{3000}^2\frac{{\mathrm{m}}^2}{{\mathrm{s}}^2}+\frac{\left(6.67×{10}^{-11}{\mathrm{m}}^3/\mathrm{kg}·{\mathrm{s}}^2\right)×6.4×{10}^3\mathrm{kg}}{4×{10}^6\mathrm{m}}\right)}$

$=\sqrt{2\left(-\frac{\left(6.67×{10}^{-11}\right)×6.4×{10}^3}{7×{10}^6}+\frac{1}{2}×{3000}^2+\frac{\left(6.67×{10}^{-11}\right)×6.4×{10}^3}{4×{10}^6}\right)\left(\frac{1{\mathrm{m}}^3/\mathrm{kg}·{\mathrm{s}}^2}{1\mathrm{m}}+\frac{1{\mathrm{m}}^2}{1{\mathrm{s}}^2}+\frac{1{\mathrm{m}}^3/\mathrm{kg}}{1\mathrm{m}}\right.}$

$=\sqrt{2\left(-\frac{\left(6.67×{10}^{11}\right)×6.4×{10}^3}{7×{10}^6}+\frac{1}{2}×{3000}^2+\frac{\left(6.67×{10}^{11}\right)×6.4×{10}^3}{4×{10}^6}\right)\left(\frac{1{\mathrm{m}}^3·}{1\mathrm{kg}·{\mathrm{s}}^2·\mathrm{m}}+\frac{1{\mathrm{m}}^2}{1{\mathrm{s}}^2}+\frac{1{\mathrm{m}}^3}{1\mathrm{kg}·{\mathrm{s}}^2}.\right.}$

$=\sqrt{2\left(-\frac{\left(6.67×{10}^{-11}\right)×6.4×{10}^3}{7×{10}^6}+\frac{1}{2}×{3000}^2+\frac{\left(6.67×{10}^{-11}\right)×6.4×{10}^3}{4×{10}^6}\right)\left(\frac{1{\mathrm{m}}^2}{1{\mathrm{s}}^2}+\frac{1{\mathrm{m}}^2}{1{\mathrm{s}}^2}+\frac{1{\mathrm{m}}^2}{1{\mathrm{s}}^2}\right)}$

Hence, the speed of a satellite in a circular orbit near the Earth is $3000\mathrm{m}/\mathrm{s}$