Question

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?

Short answer

The final speed is 25.47 m/s

Step-by-step solution

Identification of given data

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

Significance of escape speed

The escape speed is calculated for finding the minimum speed or minimum velocity to escape the gravitational pull of a planet.

Determination of Final speed

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 $25.47m/s$