Zero, a hypothetical planet, has amass of$\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{}\mathbf{\text{kg}}$, a radius of$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathbf{\text{m}}$ ,and no atmosphere. A$\mathbf{10}\mathbf{}\mathbf{\text{kg}}$ space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial energy of $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{}\mathbf{\text{J}}$, what will be its kinetic energy when it is$\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathit{m}$ from the center of Zero? (b) If the probe is to achieve a maximum distance of $\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathit{m}$from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?

Mass of Zero plant is, $\text{M}=5\times {10}^{23}\text{kg}$

The radius of Zero planet is, $\text{R}=3\times {10}^6\text{m}$

Mass of probe is, $\text{m}=10\text{kg}$

Initial kinetic energy is, ${\text{KE}}_{\text{1}}=5\times {10}^7\text{J}$

To find kinetic energy at$\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathit{m}$, first, calculate potential energy at that distance, followed by the calculation of potential energy onthesurface of Zero planet. Then, use conservation of energy to find kinetic energy.

To findtheinitial kinetic energy when the probe is at a maximum height$\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathit{m}$, find potential energy. At maximum height, the velocity is zero. So, kinetic energy is also zero. Hence, use the conservation of energy to find initial kinetic energy.

Formulae are as follows:

$$\begin{gathered} \mathit{K}{\mathit{E}}_{\mathbf{1}}\mathbf{+}{\mathit{U}}_{\mathbf{1}}\mathbf{=}\mathit{K}{\mathit{E}}_{\mathbf{2}}\mathbf{+}{\mathit{U}}_{\mathbf{2}} \\ \mathit{U}\mathbf{=}\frac{\mathbf{G}\mathbf{M}\mathbf{m}}{\mathbf{r}} \end{gathered}$$

where, M, and m are masses, r is the radius, G is the gravitational constant, KE is kinetic energy and U is potential energy.

Firstly, the initial potential energy on the surface of zero planets can be found as,

$\begin{array}{rcl}{U}_1& =& -\frac{GMm}{R}\\ & =& -\frac{6.67\times {10}^{-11}N·m^2·k{g}^{-2}\times 5\times {10}^{23}kg\times 10kg}{3\times {10}^{6}m}\\ & =& -1.12\times {10}^{8}J\\ & & \end{array}$

Now, potential energy at $\text{r}=4\times {10}^6\text{m}$ is,

$\begin{array}{rcl}{U}_2& =& -\frac{GMm}{r}\\ & =& -\frac{6.67\times {10}^{-11}N·m^2·k{g}^{-2}\times 5\times {10}^{23}kg\times 10kg}{4\times {10}^{6}m}\\ & =& -0.84\times {10}^{8}J\\ & & \end{array}$

Now, to find the kinetic energy at $4\times {10}^{6}m$ from center of the Zero planet, use the formula for conservation of energy as follows:

$$\begin{gathered} K{E}_1+U_1=K{E}_2+U_2 \\ K{E}_2=K{E}_1+U_1-U_2 \\ K{E}_2=5\times {10}^{7}J-11.2\times {10}^{7}J+8.4\times {10}^{7}J \\ K{E}_2=2.2\times {10}^{7}J \end{gathered}$$

So, kinetic energy at a distance of $4\times {10}^{\text{3}}m$ is $2.2\times {10}^{7}J$.

Kinetic energy at maximum height is zero. So,

$K{E}_2=0J$

The potential energy at distance $8\times \${10}^{6}m$ is as follows:

$\begin{array}{rcl}{U}_3& =& -\frac{GMm}{r}\\ & =& -\frac{6.67\times {10}^{-11}N·m^2·k{g}^{-2}\times 5\times {10}^{23}kg\times 10kg}{8\times {10}^{6}m}\\ & =& -4.2\times {10}^{7}J\\ & & \end{array}$

Potential energy on the surface of zero planets is calculated in part a.

So,

$U_1=-11.2\times {10}^{7}J$

Now,the initial kinetic energy is as follows:

$$\begin{gathered} K{E}_1+U_1=K{E}_2+U_2 \\ K{E}_1=K{E}_2+U_2-U_1 \\ K{E}_1=0-4.2\times {10}^{7}J+11.2\times {10}^{7}J \\ K{E}_1=7.0\times {10}^{7}J \end{gathered}$$

Hence, initial kinetic energy when the maximum distance of the probe is $8\times {10}^{6}m$is$7.0\times {10}^{7}J$

Therefore, using the concept of conservation of energy,thatis,the total energy (potential and kinetic energy) of the system remains constant, and the kinetic energy of the planet can be found.