91Ó°ÊÓ

The radius of the planet is $2000\mathrm{km}$

The mass is $1.2×{10}^{23}\mathrm{kg}$

The final speed is $900\mathrm{m}/\mathrm{s}$

The Escape Speed is defined as the minimum speed required to break free an object from the gravitational pull of a planet.

The principle of the conservation of energy is the addition of final kinetic and potential energy is equal to the addition of initial kinetic and potential energy, The expression will be,

$K{E}_f+U_f=K{E}_i+U_i⋯⋯\text{(1)}$

Where,

$K{E}_f=$Final Kinetic energy

$U_f=$final potential energy

$K{E}_i=$initial Kinetic energy

$U_i=$initial potential energy

Here, the final kinetic energy and potential energy are zero because the final speed is zero when the escape speed is used.

$K{E}_1=U_f=0$

$K{E}_i=\frac{1}{2}m{v}_e^2……\text{(2)}$

Here, $m=$mass, $v_e=$escape speed

$U_i=-\frac{GMm}{R}$

Substitute the values, we get:

$\frac{1}{2}m{v}_i^2+0=\frac{1}{2}m{v}_i^2+\frac{GMm}{R}$

$\frac{1}{2}{v}_i^2=\frac{1}{2}{v}_f^2+\frac{GM}{R}⋯⋯\left(3\right)$

Where, $G=6·6×{10}^{-11}\mathrm{m}/{\mathrm{s}}^2$(gravitational acceleration of planet) $M=1.2×{10}^{23}\mathrm{kg}$(mass of planet)

$R=2×{10}^6\mathrm{m}$(radius of planet)

Substitute these values in Equation (3),

$\frac{1}{2}{v}_i^2-\frac{1}{2}(900{)}^2+\frac{\left(6·6×{10}^{-11}\right)×\left(1.2×{10}^{23}\right)}{2×{10}^6}$

$v_i^2=(900{)}^2+\frac{\left(6.6×{10}^{11}\right)×\left(1.2×{10}^{23}\right)}{{10}^6}$

$v_i^2-810000+7920000$

$v_i^2=8730000$

Hence, the initial speed when the object is far from Mars its final speed is $900\mathrm{m}/\mathrm{s}$, is $2.9×{10}^3\mathrm{m}/\mathrm{s}$

Therefore, the Equation (3) will become,

$\frac{1}{2}{v}_i^2=0+\frac{GM}{R}$

$\frac{1}{2}{v}_1^2=\frac{\left(6.6×{10}^{-11}\right)×\left(1.2×{10}^{23}\right)}{2×{10}^6}$

$v_1^2=7920000$

Hence, the initial speed when the object is far from Mars its final speed is $0\mathrm{m}/\mathrm{s}$ is $2·8×{10}^3\mathrm{m}/\mathrm{s}$