Escape Velocity. According to Newton’s law of gravitation,

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, \(F_{g}=G M_{1} M_{2} / r^{2}\), where \(M_{1}\) and \(M_{2}\) 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 y the velocity of the projectile.

(a) Show that the motion of the projectile, under Earth’s gravitational force, is governed by the

equation

\(\frac{d v}{d t}=-\frac{g R^{2}}{r^{2}}\),

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 \(g=G M / R^{2}\).

(b) Use the fact that \(dr/dt=v\) to obtain

\(v \frac{d v}{d r}=-\frac{g R^{2}}{r^{2}}\).

(c) If the projectile leaves Earth’s surface with velocity \(v_{0}\), show that

\(v^{2}=\frac{2 g R^{2}}{r}+v_{0}^{2}-2 g R\).

(d) Use the result of part (c) to show that the velocity of the projectile remains positive if and only if \(v_{0}^{2}-2 g R>0\). The velocity \(v_{e}=\sqrt{2 g R}\) is called the escape velocity of Earth.

(e) If \(g=9.81 \mathrm{~m} / \mathrm{sec}^{2}\) and \(R=6370 \mathrm{~km}\) for Earth, what is Earth’s escape velocity?

(f) 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?

Equation Transcription:

Text Transcription:

F_{g}=GM_{1}M_{2  r^{2}

M_{1}

M_{2}

F_{g}

\frac{d v}{d t}=-\frac{g R^{2}}{r^{2}}

g=G M / R^{2}

dr/dt=v

v {dv}over{dr}=-{gR^{2}}{r^{2}}

v_{0}

v^{2}={2gR^{2}}over{r}+v_{0}^{2}-2gR

v_{0}^{2}-2gR>0

v_{e}=\sqrt{2 g R}

g=9.81 m/sec^{2}

R=6370 \mathrm{~km}

g_{m}=g/6

R_{m}=1738