Question

(a) What is the escape speed on a spherical asteroid whose radius is500 kmand whose gravitational acceleration at the surface is $\mathbf{3}\mathbf{.}\mathbf{0}\raisebox{1ex}{$\mathbf{m}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{s}}^{\mathbf{2}}$}\right.$ ? (b) How far from the surface will a particle go ifit leaves the asteroid’s surface with a radial speed of $\mathbf{1000}\mathbf{}\mathbf{\text{m/s}}$? (c)With what speed will an object hit the asteroid if it is dropped from$\mathbf{1000}\mathbf{}\mathbf{\text{km}}$above the surface?

Short answer

1. The escape speed on a spherical asteroid whose radius is $\text{500}\text{km}$ and whose gravitational acceleration at the surface is$\text{3.0}{\text{m/s}}^{\text{2}}$ will be $\text{1.7}\times {\text{10}}^{\text{3}}\text{m/s}$.
2. The distance from the surface that the particle will go if it leaves the asteroid’s surface with a radial speed of$100\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$is$\text{2.5}\times {\text{10}}^{\text{5}}\text{m}$.
3. The speed with which an object will hit the asteroid if it is dropped from above the surface is$\text{1.4}\times {\text{10}}^{\text{3}}\text{m/s}$ .

Step-by-step solution

Step 1: Given

The radius of the spherical asteroid$500$

The gravitational acceleration$\text{3.0}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$

Determining the concept

Using the principle of conservation of energy, find theescape speed on a spherical asteroid whose radius is$500$and has gravitational acceleration at the surface equal to$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\raisebox{1ex}{$\mathbf{m}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{s}}^{\mathbf{2}}$}\right.$, and thespeed with which an object will hit the asteroid if it is dropped fromabove the surface.

Similarly,using the principle of conservation of energy, find the distance from the surface that the particle will go if it leaves the asteroid’s surface with a radial speed of.According to the law ofconservation of energy, energy can neither be created nor be destroyed.

Formulae are as follows:

$$\begin{gathered} {\mathit{U}}_{\mathbf{i}}\mathbf{+}{\mathit{K}}_{\mathbf{i}}\mathbf{=}{\mathit{U}}_{\mathbf{f}}\mathbf{+}{\mathit{K}}_{\mathbf{f}} \\ \mathit{U}\mathbf{=}\mathbf{-}\frac{\mathbf{G}\mathbf{M}\mathbf{m}}{\mathbf{R}} \\ \mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}} \end{gathered}$$

where, , m are masses, R is the radius, v is velocity, G is gravitational constant, K is kinetic energy and U is potential energy.

(a) Determining the escape speed on a spherical asteroid whose radius is   and whose gravitational acceleration at the surface is 3.0 ms2

Now,

$U_i+K_i=U_f+K_f$

As

$$\begin{gathered} U_i=-\frac{GMm}{R} \\ {K}_i=\frac{1}{2}m{v}^2 \\ {\text{U}}_{\text{f}}=0 \\ {\text{K}}_{\text{f}}=0 \\ -\frac{\text{GMm}}{\text{R}}+\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}=0 \end{gathered}$$

As

$$\begin{gathered} \frac{\text{GM}}{\text{R}}={\text{a}}_{\text{g}}\text{R} \\ -{\text{a}}_{\text{g}}\text{Rm}+\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}=0 \\ -a_{\text{g}}\text{R}+\frac{\text{1}}{\text{2}}{\text{v}}^{\text{2}}=0 \\ \text{v}=\sqrt{{\text{2a}}_{\text{g}}\text{R}} \end{gathered}$$

As

$a_g=3.0\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2,andR=500\times {10}^3\text{m}$}\right.$

$\begin{array}{rcl}v& =& \sqrt{2\left(3.0\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.\right)}\left(500\times {10}^3\right)\\ & =& 1.7\times {10}^3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$

Therefore, the escape speed on a spherical asteroid whose radius is and whose gravitational acceleration at the surface it $3.0\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.$will be $1.7\times {10}^3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

(b) Determining the distance from the surface that the particle will go if it leaves the asteroid’s surface with a radial speed of  100ms

Now,

$U_i+K_i=U_f+K_f$

As

$$\begin{gathered} U_i=-\frac{GMm}{R} \\ {K}_i=\frac{1}{2}m{v}^2 \\ {\text{U}}_{\text{f}}=0 \\ {U}_f=-\frac{GMm}{\left(R+h\right)} \end{gathered}$$

h = distance above the surface

$-\frac{GMm}{R}+\frac{1}{2}m{v}^2=-\frac{GMm}{\left(R+h\right)}$

As

$$\begin{gathered} GM=a_g{R}^2 \\ -a_{g}Rm+\frac{1}{2}m{v}^2=-\frac{a_g{R}^{2}m}{\left(R+h\right)} \\ -a_{g}R+\frac{1}{2}{v}^2=-\frac{a_g{R}^2}{\left(R+h\right)} \\ h=\frac{2a_g{R}^2}{2a_{g}R-v^2}-R \\ h=\frac{\text{2}\left(\text{3.0}\frac{\text{m}}{{\text{s}}^{\text{2}}}\right){\left(500\times {10}^{3}m\right)}^{\text{2}}}{\text{2}\left(\text{3.0}\frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\left(500\times {10}^{3}m\right)\text{-}{\left(\text{1000}\frac{\text{m}}{\text{s}}\right)}^{\text{2}}}\left(500\times {10}^{3}m\right) \end{gathered}$$

$h=2.5\times {10}^{\text{5}}\text{m}$

Therefore, the distance from the surface that the particle will go if it leaves the asteroid’s surface with a radial speed of is $2.5\times {10}^{\text{5}}\text{m}$.

(c) Determining the speed with which an object willhit the asteroid if it is dropped from   above the surface

Now,

$U_i+K_i=U_f+K_f$

As

$$\begin{gathered} U_i=-\frac{GMm}{\left(R+h\right)} \\ {\text{K}}_{\text{i}}=0 \\ {U}_f=-\frac{GMm}{R} \\ {K}_f=\frac{1}{2}m{v}^2 \\ -\frac{GMm}{\left(R+h\right)}=-\frac{GMm}{R}+\frac{1}{2}m{v}^2 \end{gathered}$$

As

$$\begin{gathered} GM=a_g{R}^2 \\ -\frac{a_g{R}^{2}m}{\left(R+h\right)}=\frac{a_g{R}^{2}m}{R}+\frac{1}{2}m{v}^2 \\ -\frac{a_g{R}^2}{\left(R+h\right)}=\frac{a_g{R}^2}{R}+\frac{1}{2}{v}^2 \\ v=\sqrt{2a_{g}R-\frac{2a_g{R}^2}{\left(R+h\right)}} \\ v=\sqrt{2\left(\text{3.0}\frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\left(500\times {10}^3\right)}-\frac{2\left(\text{3.0}\frac{\text{m}}{{\text{s}}^{\text{2}}}\right){\left(500\times {10}^3\right)}^2}{\left[\left(500\times {10}^3\right)+\left(500\times {10}^3\right)\right]} \end{gathered}$$

$\text{v}=1.4\times {10}^3\text{m/s}$

Hence, the speed with which an object will hit the asteroid if it is dropped from1000km above the surface is $1.4\times {10}^3\text{m/s}$.

Therefore, using the law of conservation of energy and the formula for gravitational potential energy and kinetic energy, the speed and height can be found.