Chp. 10 - Projectile and Satellite Motion - HW

Why is kinetic energy a constant for a satellite in circular orbit but not for a satellite in elliptical orbit?

KE is constant because no work is done by gravity on the satellite.

What did Kepler discover about the time periods of planets and their distances from the sun?

Kepler discovered that the period squared was proportional to the radial distance cubed. T^2/R^3

A certain satellite has a kinetic energy of 8 billion joules at perigee (the point closest to Earth) and 5 billion joules at apogee (the point farthest from Earth). As the satellite travels from apogee to perigee, how much work does the gravitational force do on it? Does its potential energy increase or decrease during this time, and by how much?

3,000,000,000 J of work was done by the gravitational force. When traveling from the apogee to perigee, the gravitational force does 3,000,000,000 J of work. When traveling from the apogee to perigee, the Potential energy decreases by 3 billion joules (on this path, the gravitational force is converting 3,000,000,000 J of energy from potential energy to kinetic energy).

How can a projectile "fall around the Earth'?

A projectile can fall around the Earth if it has sufficient tangential speed so that its curve downward is no sharper than that of Earth's curvature.

What exactly is a projectile?

A projectile is any object that is projected by some means and continues in motion by its own inertia.

At what part of an elliptical orbit does an Earth satellite have the greatest speed? The lowest speed?

A satellite has the greatest speed when nearest Earth (perigee), and the lowest speed when farthest away (apogee).

Why is it important that the projectile in the preceding question be above Earth's atmosphere?

A satellite must remain above the atmosphere because air resistance would not only slow it down, but increase at a high speed. A satellite must not have to contend with either of these.

A projectile is launched vertically at an angle of 75 degrees from the horizontal and strikes the ground a certain distance downrange. For what other angle of launch at the same speed would this projectile land just as far away?

An angle of 15 degrees would produce the same range, in accord with figure 10.11.

A baseball projected with an initial velocity of 141 m/s at an angle of 45 degrees follows a parabolic path and hits a balloon at the top of its trajectory. Ignoring air resistance, show the ball hits a balloon at a speed of 100 m/s.

At the top of its trajectory, the vertical component of velocity is zero, leaving only the horizontal component. The horizontal component at the top or anywhere along the path is the same as the initial horizontal component, 100 m/s

Students in the lab (see figure 10.5) measure the speed of a steel ball to be 8.0 m/s when launched horizontally from a 1.0 m high tabletop. Their objective is to place a 20 cm tall coffee can on the floor to catch the ball. Show that they score a bull's-eye when the can is placed 3. m from the base of the table.

Df=1/2*a*t^2+Vo*t+Do , which becomes 0.2=-4.9*t^2+0+1t=sqrt[(1.0-0.2)/4.9]=sqrt[0.80/4.9]=.404sDf=1/2*a*t^2+Vo*t+Do, which becomes Df=0+8*0.404+0=3.23m

A projectile falls beneath the straight-line path it would follow if there were no gravity. How many meters does it fall below this line if it has been traveling for 1 second? 2 seconds?

In 1 s, the projectile falls 5 m beneath the line; in 2 s, 20 m beneath.

Calculate the hang time of a person who moves 3 m horizontally during a 1.25 m high jump. What is the hang time when the person moves 6 m horizontally during this jump?

The hang time depends on only the vertical component of the initial velocity and the corresponding vertical distance attained. From d = 5t^2, a vertical 1.25 m drop corresponds to 0.5 s (t = the square root 2d/g = square root 2(1.25)/10 = 0.5 s. Double the time up and time down for hang time of 1 s. Hang time is the same whatever the horizontal distance traveled.

A ball is thrown horizontally from a cliff at a speed of 10 m/s. You predict that its speed 1 s later will be slightly greater than 14 m/s. Your friend says it will be 10 m/s. Show who is correct

The horizontal component of speed remains constant at 10 m/s. After 1 s, a vertical component of speed = gt = 9.8(1) = 9.8 m/s is added to the horizontal speed. The size of the total speed of the ball after 1 s = √(10²+9.8²) = √196.04 ≈ 14.0 m/s

For orbits of greater altitude, is the period longer or shorter?

The period of satellite at higher altitudes is longer.

Why doesn't the force of gravity change the speed of a satellite in circular orbit?

The speed of a satellite doesn't change when there's no component of gravitational force in the direction of it's motion.

At a particular point in its orbit, a satellite in an elliptical orbit has a gravitational potential energy of 5000 MJ with respect to Earth's surface and a kinetic energy of 4500 MJ. Later in its orbit, the satellite's potential energy is 6000 MJ. What is its kinetic energy at that point?

Total energy = 5000 MJ + 4500 MJ = 9500 MJ. Subtract 6000 MJ and KE = 3500 MJ

A rock is thrown upward at an angle. What happens to the horizontal component of its velocity as it rises and as it falls?

With no air resistance the horizontal component of velocity remains constant, both in rising and falling.

Escape speed from the Earth is 11.2 km/s. Is it possible to escape Earth at half this speed? At one quarter this speed? If so, how?

Yes, the escape speed can be at speeds lower than 11.2 km/s if that speed is sustained.

You're in an airplane that flies horizontally with speed 1000 km/h (280 m/s) when an engine falls off. Ignore air resistance and assume it takes 30 s for the engine to hit the ground. a) Show that the airplane is 4.5 km high b) Show that the horizontal distance that the aircraft engine moves during its fall is 8400 m. c) If the airplane somehow continues to fly as if nothing had happened, where is the engine relative to the airplane at the moment the engine hits the ground?

a) from y = 5t^2 = 5(30)^2 = 4500 m or 4.5 km high. b) In 30 seconds; d =vt =280 m/s x 30 s = 8400 m. c) The engine is directly below the airplane

Bull's-eye Bob at a hunting range fires is rifle at a target 200 meters downrange. The bullet moves horizontally from the rifle barrel with a speed of 400 m/s. a) How far does the bullet drop from a straight-line horizontal path by the time it reaches the target? b) To hit a bull's-eye why does Bob adjust his gunsight so the barrel points a bit upward when he aims at the target?

d = 0.5g*Tf^2 = 4.9*0.5^2 = 1.23 m.