1. Chapter 11 - Gravity

2. Example 3 The International Space Station Travels in a roughly circular orbit around the earth. If it’s altitude is 385 km above the earth’s surface, how long do you have to wait between sightings? The International Space Station Travels in a roughly circular orbit around the earth. If it’s altitude is 385 km above the earth’s surface, how long do you have to wait between sightings? Sightings only occur at night, and only if the space station is above the horizon. Therefore, the minimum time is about equal to the orbital period. To find the orbital period, we use Kepler’s third law with M s in Equation 11-6 replaced by the mass of the earth M E. The numerical calculation is simplified using GM E = R(2/E)g (Equation 8) Sightings only occur at night, and only if the space station is above the horizon. Therefore, the minimum time is about equal to the orbital period. To find the orbital period, we use Kepler’s third law with M s in Equation 11-6 replaced by the mass of the earth M E. The numerical calculation is simplified using GM E = R(2/E)g (Equation 8)

3. At an altitude h = 385 km, r = R E + h = 6760 km. substitute r = R E + h and solve for the period: Use GM E = R E 2 g to write T in terms of g.

4. Example 4 A projectile is fired straight up from the surface of the earth with an initial speed of v i = 8 km/s. Find the maximum height it reaches. The maximum height is found using energy conservation. We take the surface of the earth as the initial point, with U i = - GM E m/R E and K i = ½ mv 2. At the greatest height, K f = 0.

6. Example 5 A projectile is fired straight up from the surface of the earth with an initial speed v i = 15 km/s. Find the speed of the projectile when it si very far from the earth, neglecting air resistance. Very far from the earth means r >> R E. The initial speed is greater than the escape speed of 11 km/s, so the total energy of the projectile is positive and the projectile will escape the earth with some final kinetic energy.

8. Example 6 Show that the total energy of a satellite in a circular orbit is half its potential energy.

9. Example 7 Two point particles, each of mass M, are fixed on the y axis at y = +a and y = -a. Find the gravitational field at all points on the x axis as a function of x. The distance between P and either particle is r = √( x 2 + a2). The resultant field g is the vector sum of the fields g1 and g2 due to each particle.

11. Example 8 A uniform rod of mass M and length L (here we go again…) is centered on the origin and lies along the x axis. Find the gravitational field due to the rod at all points on the x axis for x > L/2

13. Example 9 A planet with a hollow core consists of a uniform, thick spherical shell with a mass M, outer radius R, and inner radius R/2 a)What amount of mass is closer than ¾ R to the center of the planet? b)What is the gravitational field a distance ¾ R from the center?

14. A) The mass of M’ (mass of the shell with outer radius ¾R and inner radius ½ R) is the desity ρ times the volume, V B) The gravitational field at r = ¾ R is due to only the mass of M’:

15. Example 10 A solid sphere with a radius R and mass M is spherically symmetrical but not uniform. Its density, ρ, defined as its mass per unit volume is proportional to the distance from the center r for r < R. That is, ρ = Cr for r < R, where is a constant. a)Find C b)Find g r for r > R c)Find g r at r = R/2

16. A) B) C)