Question: The mysterious visitor that appears in the enchanting story The Little Prince was said to come from a planet that “was scarcely any larger than a house!” Assume that the mass per unit volume of the planet is about that of Earth and that the planet does not appreciably spin. Approximate (a) the free-fall acceleration on the planet’s surface and (b) the escape speed from the planet.

The mass of the earth is$M_E=6.0\times {10}^{24}kg$

The radius of the earth is$R_E=6.4\times {10}^{6}m$

When an object falls through a vacuum, it is solely subject to one external force: gravitational force, which is the object's weight. Free falling is the term for an item that is solely moving due to the force of gravity, and Newton's second rule of motion describes this motion. We can use the concept of free fall acceleration and the escape speed from the planet.

Formula

$$\begin{gathered} \rho =\frac{M}{\frac{4}{3}\pi r^3} \\ {a}_g=\frac{Gm}{r^2} \\ v=\sqrt{\frac{2Gm}{r}} \end{gathered}$$

Where,

$\rho$ is density of the planet

G is the gravitational constant ( $6.67\times {10}^{-11} {\text{m}}^{\text{3}}\text{/kg} ·{\text{s}}^{\text{2}}$)

$a_g$is the gravitational acceleration

m is the mass of object

v is the escape speed

Consider radius of the planet

r = 10

The density of the planet is equal to earth. Hence, we can find the mass of the planet as

$$\begin{gathered} \frac{m}{\frac{4}{3}\pi r^3}=\frac{M_E}{\frac{4}{3}\pi R_E^3} \\ m=\frac{M_E}{\frac{4}{3}\pi R_E^3}\frac{4}{3}\pi r^3 \\ m=\frac{M_E}{R_E^3}{r}^3 \\ =\frac{6\times {10}^{24}\text{kg}}{{\left(6.4\times {10}^6\text{m}\right)}^3}\times {\left(10\text{m}\right)}^3 \\ =2.3\times {10}^7\text{kg} \end{gathered}$$

The free fall acceleration on the planet’s surface:

The expression for the acceleration due to gravity on the planet’s surface is

$$\begin{gathered} a_g=\frac{Gm}{r^2} \\ =\frac{\left(6.67\times {10}^{-11}\frac{{\text{Nm}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)\times 2.3\times {10}^7\text{kg}}{{\left(10\text{m}\right)}^2} \\ =1.5\times {10}^{-5}{\text{m/s}}^{\text{2}} \end{gathered}$$

The free fall acceleration on the planet’s surface is $a_g=2\times {10}^{-5}{\text{m/s}}^{\text{2}}$

The expression for the escape speed from the planet is

$$\begin{gathered} v=\sqrt{\frac{2Gm}{r}} \\ =\sqrt{\frac{2\left(6.67\times {10}^{-11}\frac{{\text{Nm}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)\times 2.3\times {10}^7\text{kg}}{10\text{m}}} \\ =0.0175 \text{m/s} \end{gathered}$$

The escape speed from the planet is$0.0175 \text{m/s}$