The escape speed from a very small asteroid is only 24 m/s. If you throw a rock away from the asteroid at a speed of 35 m/s, what will be its final speed?

• The escape speed is 24 m/s
• The initial speed is 35 m/s

The escape speed is calculated for finding the minimum speed or minimum velocity to escape 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 addition of initial kinetic and potential energy,

The expression will be,

$$\begin{gathered} K{E}_f+U_f=K{E}_i+U_j.......\left(1\right) \\ Where \end{gathered}$$

$$\begin{gathered} K{E}_f=FinalKineticEnergy \\ {U}_f=finalpotentialenergy \\ K{E}_j=initialKineticenergy \\ {U}_j=initialpotentialenergy \end{gathered}$$

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

$$\begin{gathered} K{E}_f=U_f=0 \\ K{E}_i=\frac{1}{2}m{v_e}^2 \end{gathered}$$

$$\begin{gathered} where,m=mass,v_e=escapespeed \\ {U}_i=\frac{GMm}{R} \end{gathered}$$

Substitute this in Equation (1),

$$\begin{gathered} 0+0=\frac{1}{2}m{v_e}^2-\frac{GMm}{R} \\ \frac{GMm}{R}=\frac{1}{2}m{v_e}^2 \\ \frac{GM}{R}=\frac{1}{2}{v_e}^2 \\ \frac{GM}{R}=\frac{1}{2}{\left(24\right)}^2 \\ \frac{GM}{R}=288..........\left(2\right) \end{gathered}$$

From Equation (1),

$$\begin{gathered} \frac{1}{2}m{v_i}^2+0=\frac{1}{2}m{v_f}^2+\frac{GMm}{R} \\ =\frac{1}{2}{v_i}^2=\frac{1}{2}{v_f}^2+\frac{GM}{R}...............\left(3\right) \end{gathered}$$

From Equation (3),

$$\begin{gathered} \frac{1}{2}{\left(35\right)}^2=\frac{1}{2}{v_f}^2+288 \\ \frac{1}{2}{v_f}^2=612.5-288 \\ {v_f}^2=649 \\ {v}_f=25.47m/s \end{gathered}$$

Hence, the final speed is $25.47m/s$