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, ris the distance between them (center to center), F_gis the attractive force, and Gis the constant of proportionality. Consider a projectile of

constant mass mbeing fired vertically from Earth (see Figure 3.12). Let trepresent 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 ris the distance between the projectile and the center of Earth, Ris the radius of Earth, Mis 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 velocityof Earth.
5. If g= 9.81 m/sec ²and 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?

Short 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-by-step solution

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 thatrole="math" localid="1663971026881" $\frac{\mathbf{dv}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\frac{{\mathbf{gR}}^{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}$

Step 2(b): obtain vdvdt=-gR2r2

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  v2=2gR2r+vo2-2gR

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

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 role="math" localid="1663971538088" ${\mathbf{v}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{2}{\mathbf{gR}}^{\mathbf{2}}}{\mathbf{r}}\mathbf{+}{{\mathbf{v}}_{\mathbf{o}}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{gR}$

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 thatthe velocity of the projectile remains positive if and only ifrole="math" localid="1663971976334" ${{\mathbf{v}}_{\mathbf{o}}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{gR}\mathbf{>}\mathbf{0}$.

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.