91Ó°ÊÓ

The radius of the spherical asteroid$500$

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

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.

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×{10}^3\text{m}$}\right.$

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

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

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

$h=2.5×{10}^{\text{5}}\text{m}$

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

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

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

$\text{v}=1.4×{10}^3\text{m/s}$

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

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.