Q3.4-25E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Escape Velocity. According to Newton’s law of gravitation, the attractive force between two objects varies inversely as the square of the distances between them. That is, ${\mathbf{F}}_{\mathbf{g}}\mathbf{=}{\mathbf{GM}}_{\mathbf{1}}{\mathbf{M}}_{\mathbf{2}}\mathbf{/}{\mathbf{r}}^{\mathbf{2}}$ where ${\mathbf{M}}_{\mathbf{1}}{\mathbf{andM}}_{\mathbf{2}}$ are the masses of the objects, r is the distance between them (center to center), F_g is the attractive force, and G is the constant of proportionality. Consider a projectile of

constant mass m being fired vertically from Earth (see Figure 3.12). Let t represent time and v the velocity of the projectile.

1. Show that the motion of the projectile, under Earth’s gravitational force, is governed by the equation $\frac{\mathbf{dv}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\frac{{\mathbf{gR}}^{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}$, where r is the distance between the projectile and the center of Earth, R is the radius of Earth, M is the mass of Earth, and $\mathbf{g}\mathbf{=}\mathbf{GM}\mathbf{/}{\mathbf{R}}^{\mathbf{2}}$ .
2. Use the fact that dr /dt = v to obtain $\mathbf{v}\frac{\mathbf{dv}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\frac{{\mathbf{gR}}^{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}$
3. If the projectile leaves Earth’s surface with velocity v_o, show that ${\mathbf{v}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{2}{\mathbf{gR}}^{\mathbf{2}}}{\mathbf{r}}\mathbf{+}{{\mathbf{v}}_{\mathbf{o}}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{gR}$
4. Use the result of part (c) to show that the velocity of the projectile remains positive if and only if ${{\mathbf{v}}_{\mathbf{o}}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{gR}\mathbf{>}\mathbf{0}$. The velocity ${\mathbf{v}}_{\mathbf{e}}\mathbf{=}\sqrt{\mathbf{2}\mathbf{gR}}$ is called the escape velocity of Earth.
5. If g = 9.81 m/sec ² and R = 6370 km for Earth, what is Earth’s escape velocity?
6. If the acceleration due to gravity for the moon is g_m = g/6 and the radius of the moon is R_m = 1738 km, what is the escape velocity of the moon?

Step-by-Step Solution

Answer

1. Proved
2. Proved
3. Proved.
4. Proved
5. The earth’s escape velocity is 11.18 km/sec.
6. The escape velocity of the moon is 2.38 km/sec.

Step 1(a): Find the motion of the projectile

Here M₁ = M and M₂ = m

By using the newton’s second law 

$$\begin{gathered} \mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}=-{\mathrm{F}}_{\mathrm{g}} \\ \mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}=-\frac{\mathrm{GMm}}{{\mathrm{r}}^2} \\ \frac{\mathrm{dv}}{\mathrm{dt}}=-\frac{\mathrm{GM}}{{\mathrm{r}}^2} \\ \frac{\mathrm{dv}}{\mathrm{dt}}=-\frac{{\mathrm{gR}}^2}{{\mathrm{r}}^2} \text{(Because }\mathrm{G}=\frac{{\mathrm{gR}}^2}{\mathrm{M}}\text{)} \end{gathered}$$

Hence it is proved that  $\frac{\mathbf{dv}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\frac{{\mathbf{gR}}^{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}$

Step 2(b): obtain v dv dt = - gR 2 r 2

Since

$$\begin{gathered} \mathrm{v}=\frac{\mathrm{dr}}{\mathrm{dt}} \\ \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{dv}}{\mathrm{dr}}.\frac{\mathrm{dr}}{\mathrm{dt}} \\ \frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{v}\frac{\mathrm{dv}}{\mathrm{dr}} \\ \mathrm{v}\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{-{\mathrm{gR}}^2}{{\mathrm{r}}^2} \end{gathered}$$

Hence it is proved that $\mathbf{v}\frac{\mathbf{dv}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\frac{{\mathbf{gR}}^{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}$

Step 3(c): show that v 2 = 2 gR 2 r + v o 2 - 2 gR

$$\begin{gathered} \int \mathrm{vdv}=\int \frac{-{\mathrm{gR}}^2}{{\mathrm{r}}^2}\mathrm{dr} \\ \frac{{\mathrm{v}}^2}{2}=\frac{{\mathrm{gR}}^2}{\mathrm{r}}+\mathrm{C} \\ {\mathrm{v}}^2=\frac{2{\mathrm{gR}}^2}{\mathrm{r}}+2\mathrm{C} \end{gathered}$$

When $\text{v}\left(\text{R}\right){\text{ = v}}_{\text{o}}$ then $\text{C = }\frac{{{\text{v}}_{\text{o}}}^{\text{2}}}{2}\text{ - gR}$

 ${\mathrm{v}}^2=\frac{2{\mathrm{gR}}^2}{\mathrm{r}}+{{\mathrm{v}}_{\mathrm{o}}}^2-2\mathrm{gR}$

Hence it is proved that ${\mathbf{v}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{2}{\mathbf{gR}}^{\mathbf{2}}}{\mathbf{r}}\mathbf{+}{{\mathbf{v}}_{\mathbf{o}}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{gR}$

Step 4(d): obtain the required result by given conditions

The velocity of the projectile remains the same if and only if  

 $$\begin{gathered} \iff \frac{\mathbf{2}{\mathbf{gR}}^{\mathbf{2}}}{\mathbf{r}}\mathbf{+}{{\mathbf{v}}_{\mathbf{o}}}^{\mathbf{2}}\mathbf{-}\mathbf{gR}>0 \\ \iff \frac{\mathbf{2}{\mathbf{gR}}^{\mathbf{2}}}{\mathbf{r}}\mathbf{+}{{\mathbf{v}}_{\mathbf{o}}}^{\mathbf{2}}>2gR \\ \iff \underset{\mathbf{r}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\frac{\mathbf{2}{\mathbf{gR}}^{\mathbf{2}}}{\mathbf{r}}\mathbf{+}{{\mathbf{v}}_{\mathbf{o}}}^{\mathbf{2}}>2gR \\ \iff {{\mathbf{v}}_{\mathbf{o}}}^{\mathbf{2}}>2gR \\ \iff {{\mathbf{v}}_{\mathbf{o}}}^{\mathbf{2}}-2gR>0 \end{gathered}$$

Hence it is proved that the velocity of the projectile remains positive if and only if  ${{\mathbf{v}}_{\mathbf{o}}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{gR}\mathbf{>}\mathbf{0}$.

Step 5(e): Find the earth’s escape velocity

The escape velocity is 

$$\begin{gathered} {\mathrm{v}}_{\mathrm{e}}=\sqrt{2\mathrm{gR}} \\ {\mathrm{v}}_{\mathrm{e}}=\sqrt{2(0.00981\left)\right(6370)} \\ {\mathrm{v}}_{\mathrm{e}}=11.18 \mathrm{km}/\sec \end{gathered}$$

Hence, the earth’s escape velocity is 11.18 km/sec.

Step 6(f): Find the value of the moon’s escape velocity

$$\begin{gathered} {\mathrm{v}}_{\mathrm{e}}=\sqrt{2\mathrm{gR}} \\ {\mathrm{v}}_{\mathrm{e}}=\sqrt{2(0.001635\left)\right(1738)} \\ {\mathrm{v}}_{\mathrm{e}}=2.38 \mathrm{km}/\sec \end{gathered}$$

Hence, the escape velocity of the moon is 2.38 km/sec.