Question

The radius of an airless planet is $\mathbf{2000}\mathbf{}\mathbf{km}\left(2\times {10}^{6}m\right)$ and its mass is $\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{}\mathbf{kg}$. An object is launched straight up from just above the atmosphere of Mars (a) What initial speed is needed so that when the object is far from the planet its final speed is $\mathbf{900}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$ ? (b) What initial speed is needed so that when the object is far from the planet its final speed is $\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$ ? (This is called the escape speed.)

Short answer

(a) The initial speed when the object is far from the planet its final speed is $900\mathrm{m}/\mathrm{s}$ is$2.66\times {10}^3\mathrm{m}/\mathrm{s}$

(b) The initial speed when the object is far from the planet its final speed is $0\mathrm{m}/\mathrm{s}$ is $2·82\times {10}^3\mathrm{m}/\mathrm{s}$

Step-by-step solution

Identification of given data

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

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

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

Definition of the Escape Speed

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

(a) Calculation of the initial speed when the p is far from the planet its final speed is 900 m/s

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,

$K{E}_f+U_f=K{E}_i+U_i\cdots \cdots \text{(1)}$

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.

$K{E}_1=U_f=0$

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

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

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

Substitute the values, we get:

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

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

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

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

Substitute these values in Equation (3),

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

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

$v_i^2-810000+7920000$

$v_i^2=8730000$

$v_i=2954\mathrm{m}/\mathrm{s}$

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

(b) Calculation of the initial speed when the object is far from the planet its final speed is 0 m/s

Therefore, the Equation (3) will become,

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

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

$v_1^2=7920000$

$v_i=2814\mathrm{m}/\mathrm{s}$

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