Question 13.5: "FROM THE EARTH TO THE MOON" In Jules Verne’s 1865 story wit...

IDENTIFY and SET UP:

Once the shell leaves the cannon muzzle, only the (conservative) gravitational force does work. Hence we can use conservation of mechanical energy to find the speed at which the shell must leave the muzzle so as to come to a halt (a) at two earth radii from the earth’s center and (b) at an infinite distance from earth. The energy-conservation equation is K_{1}+U_{1}=K_{2}+U_{2}, with U given by Eq. (13.9).

U=-\frac{G m_{\mathrm{E}} m}{r}             (13.9)

In Fig. 13.12 point 1 is at r_1 = R_E, where the shell leaves the cannon with speed v_1 (the target variable). Point 2 is where the shell reaches its maximum height; in part (a) r_2 = 2R_E (Fig. 13.12a), and in part (b) r_2 = ∞ (Fig 13.12b). In both cases v_2 = 0 and K_2 = 0.

Let m be the mass of the shell (with passengers).

EXECUTE:

(a) We solve the energy-conservation equation for v_2:
\begin{gathered}K_{1}+U_{1}=K_{2}+U_{2} \\\frac{1}{2} m v_{1}^{2}+\left(-\frac{G m_{\mathrm{E}}m}{R_{\mathrm{E}}}\right)=0+\left(-\frac{G m_{\mathrm{E}} m}{2 R_{\mathrm{E}}}\right) \\v_{1}=\sqrt{\frac{G m_{\mathrm{E}}}{R_{\mathrm{E}}}}=\sqrt{\frac{\left(6.67 \times 10^{-11}\mathrm{~N}\cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\right)\left(5.97 \times 10^{24} \mathrm{~kg}\right)}{6.37\times 10^{6} \mathrm{~m}}} \\=7910 \mathrm{~m} / \mathrm{s}(=28,500 \mathrm{~km} /\mathrm{h}=17,700 \mathrm{mi} / \mathrm{h})\end{gathered}

(b) Now r_2 = ∞ so U_2 = 0 (see Fig. 13.11). Since K_2= 0, the total mechanical energy K_2 + U_2 is zero in this case. Again we solve the energy-conservation equation for v_1:

EVALUATE: Our results don’t depend on the mass of the shell or the direction of launch. A modern spacecraft launched from Florida must attain essentially the speed found in part (b) to escape the earth; however, before launch it’s already moving at 410 m/s to the east because of the earth’s rotation. Launching to the east takes advantage of this “free” contribution toward escape speed.

To generalize, the initial speed v_1 needed for a body to escape from the surface of a spherical body of mass M and radius R (ignoring air resistance) is v_{1}=\sqrt{2 G M / R} (escape speed). This equation yields 5.03 \times 10^{3} \mathrm{~m} / \mathrm{s} \text { for Mars, } 6.02 \times 10^{4} \mathrm{~m} / \mathrm{s} Jupiter, and 6.18 \times 10^{5} \mathrm{~m} / \mathrm{s} for the sun.