Question

A spacecraft is coasting toward Mars. The mass of Mars is $\mathbf{6}\mathbf{.}\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{kg}$ and its radius is $\mathbf{3400}\mathbf{}\mathbf{km}\left(3.4\times {10}^{6}m\right)$. When the spacecraft is $\mathbf{7000}\mathbf{}\mathbf{km}\left(7\times {10}^{6}m\right)$ from the center of Mars, the spacecraft's speed is $\mathbf{3000}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$ . Later, when the spacecraft is $(4000\mathrm{km}\left(4\times {10}^{6}m\right)$ from the center of Mars, what is its speed? Assume that the effects of Mar's two tiny moons, the other planets, and the Sun are negligible. Precision is required to land on Mars, so make an accurate calculation, not a rough, approximate calculation.

Short answer

The speed of a spacecraft is$3000\mathrm{m}/\mathrm{s}$

Step-by-step solution

Identification of given data

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

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

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

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

- 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 $m$

Concept of Principle of Energy Conservation

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, $P_i+K_i=P_f+K_t\cdots \cdots \left(1\right)$

Determination of the speed of a satellite

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

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

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

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

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

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

From Equation (2),

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

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

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

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

$=3000\mathrm{m}/\mathrm{s}$

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