Question

The escape speed from an asteroid whose radius is10 Kmis only 10 m/s . If you throw a rock away from the asteroid at a speed of 20 m/s, what will be its final speed?

Short answer

The final speed is 17.32 m/s.

Step-by-step solution

Identification of given data

• The Radius of an asteroid is 10 km.
• The escape speed$v_e=10m/s$
• The initial speed is $v_i=20m/s$

Significance of Escape speed 

The Escape speed is required to free an object from the gravitational force created by a mass object.

Determination of Final speed

The principle of the conservation of energy is the addition of final kinetic and potential energy is equal to the addition of initial kinetic and potential energy,

The expression will be,

$$\begin{gathered} K{E}_i+U_f=K{E}_j+U_j................\left(1\right) \\ Where \\ K{E}_f=FinalKinrticenergy \\ {U}_f=finalpotentialenergy \\ K{E}_j=initialKineticenergy \\ {U}_i=initialpotentialenergy \end{gathered}$$

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

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

$U_i=\frac{UMm}{R}$

Substitute this in Equation (1),

role="math" localid="1654085773539" $$\begin{gathered} 0+0=\frac{1}{2}\mathrm{m}{{\mathrm{v}}_{\mathrm{e}}}^2-\frac{\mathrm{GMm}}{\mathrm{R}} \\ \frac{\mathrm{GMm}}{\mathrm{R}}=\frac{1}{2}\mathrm{m}{{\mathrm{v}}_{\mathrm{e}}}^2 \\ \frac{\mathrm{GM}}{\mathrm{R}}=\frac{1}{2}{\left(10\right)}^2 \\ \frac{\mathrm{GM}}{\mathrm{R}}=50......\left(2\right) \end{gathered}$$

From Equation (1),

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

Substitute the values in equation (3), and we get,

role="math" localid="1654085414714" $$\begin{gathered} \frac{\mathbf{1}}{\mathbf{2}}{\left(20\right)}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{v}}_{\mathbf{f}}}^{\mathbf{2}}\mathbf{+}\mathbf{50} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}{{\mathbf{v}}_{\mathbf{f}}}^{\mathbf{2}}\mathbf{=}\mathbf{200}\mathbf{-}\mathbf{50} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{{\mathbf{v}}_{\mathbf{f}}}^{\mathbf{2}}\mathbf{=}\mathbf{300} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{v}}_{\mathbf{f}}\mathbf{=}\mathbf{17}\mathbf{.}\mathbf{32}\mathbf{}\mathit{m}\mathbf{/}\mathit{s} \end{gathered}$$

Hence, the final speed is $17.32m/s$.