4. Use Kepler’s Second Law and the fact that L-fxp to determine at which points in an elliptical orbit around the Sun a planet has maximum and minimum speeds. (Section 13.5 will help.) 5. At the end of example 13.10, there’s an “Evaluate” blurb about how inside the surface of the Earth the force of gravity varies proportionally to the distance from the center, and it makes reference to the next chapter. which is about oscillation. Model the motion of the “mail pouch” through the shaft to the other side of the Earth as a half-period of simple harmonic motion to determine how long the pouch would take to fall through.

EXAMPLE 13,5 “FROM THE EARTH TO THE MOON” In Jules Verne’1865 story with this title, three men went to the moon in a shell fired from a giant cannon sunk in the earth in Florida () Find the minimum murale speed needed to shoot shell straight up to a height above the earth equal to the earth’s radius (1) Find the minimum muzzle speed that would allow a shell to escape from the earth completely (the escape speed) Neglect air resistance, the earth’s rotation, and the gravitational pull of the moon. The earth’s radius and mass are 6.37 x 10 m and my – 5.97 X 10 kg OS in part (1) (Fig 13.12h). In both cases – O and K, -0. Lembe the mass of the shell (with passengers). even EXECUTE: (a) We solve the energy-conservation equation for Ki + U – Ky + .0+12RE m+Gm 1 – CME Gmg (6.67 x 10- N m²/kg)(5.97 x 10 kg) “V REV 6 37 x 10m – 7910 m/s (-28,500 km/h – 17,700 mi/h) (1) Now – so U- O (see Fig. 13.11). Since Ky – 0, the total mechanical energy K, + U, is zero in this case. Again we solve the energy conservation equation for : SOLUTION 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 radis from the earth’s center and (b) at an infi- nite distance from earth. The energy-conservation equation is Ki+U – K + Uy, with U given by Eq. (139). In Fig. 13.12 point 1 is at – R. where the shell leaves the cannon with speed (the target variable). Point 2 is where the shell reaches its maximum height; in part(a) -2(Fig. 13.12a), and {mvi+ ( mm ) – 0 + 0 26mg 13.12 Our sketches for this problem. (a) (b) 2(6.67 X 10-N m²/kg)(5.97 x 104 kg) 6,37 x 10 m – 1.12 x 10 m/s (- 40,200 km/h – 25,000 mi/h) 2. Projectile, nasson 2. Projectile mossm 1-2RE Earth, mass me 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 carth; however, before launch it’s already moving at 410 m/s to the cast because of the earth’s rotation. Launching to the east takes advantage of this free contribution toward escape speed To generalize, the initial speed, needed for a body to escape from the surface of a spherical body of mass M and radius R ignoring air resistance is -V2GMR escape speedThis equation yields 5.03 x 10 m/s for Mars, 6.02 x 10m’s for Jupiter, and 6.18 X 10 m/s for the sun Borth massol More on Gravitational Potential Energy As a final note, let’s show that when we are close to the earth’s surface, Eq. (13.9) reduces to the familiar U – my from Chapter 7. We first rewrite Eq. (13.8) as Won Commi-

The magnitude of the gravitational force on mis then 13.25 A hole through the center of the earth (assumed to be uni- form). When an object is a distance from the center, only the mass inside a sphere of radius r exerts a net gravitational force on it. Cross section through earth Spherical region of radius EVALUATE: Inside this uniform-density sphere, F, is directly proportional to the distance from the center, rather than to 1/r as it is outside the sphere. At the surface Re, we have F – G ym/R/, as we should. In the next chapter we’ll learn how to compute the time it would take for the mail pouch to emerge on the other side of the earth.