Unit 10: Projectile and Satellite Motion

83 min

To follow an orbit close to Earth, a satellite must have a tangential speed of 8 km/s. Since the circumference of Earth is about 40,000 km, the satellite would make a complete revolution in a period of \begin{align} T &= \frac{40{,}000\,\text{km}}{8\,\text{km/s}} \nonumber\\[6pt] &= 5000\,\text{s} \quad (\text{or}\; 83\,\text{min}) \end{align}

Since the baseball is projected with an initial velocity of 141 m/s at an angle of 45°, the horizontal and the vertical components each have an initial value equal to 100 m/s. Note that \begin{equation} \sqrt{(100\,\text{m/s})^2 + (100\,\text{m/s})^2} = 141\,\text{m/s} \nonumber\\[6pt] \end{equation}

In the absence of air resistance, the baseball does not experience any horizontal acceleration and its speed in this direction remains constant. The ball’s upward speed, however, decreases (due to gravity) at a rate of 10 m/s every second. During the moment when it is at maximum height, the baseball loses all its vertical speed but continues to move horizontally at 100 m/s (see Video Q10.31).

The time the ball is airborne does not depend on its initial horizontal speed.

Here, the ball rolls off the tabletop in the horizontal direction and performs free fall with zero initial vertical velocity. Note that the vertical portion of the motion is what brings the ball down to the floor and, therefore, determines the time of flight (assuming air resistance is negligible). You can conclude that objects rolling off the table at different horizontal speeds will take the same time to hit the floor.

No. While the ball is in the air, its horizontal speed remains constant, while the train’s speed continues to change. For example, while speeding up, the train moves ahead of the ball, which then falls behind you. The opposite happens if the train is slowing down.

1. An observer on the ground will see the package follow a curved path as it falls, as shown in Video Q10.44.
2. You (on the airplane) will see the package fall straight down relative to the airplane.
3. The package will strike the ground at a point directly under the airplane at that moment.
4. In a real situation, air drag will cause the package to strike the ground at a point behind the airplane at that moment.

From the perspective of an observer on the ground, the package falls freely while maintaining a horizontal speed equal to that of the airplane. So, while falling to the ground, the package remains directly below the airplane. In this case, the package behaves like a projectile, as shown in Video Q10.44.

From your perspective in the airplane, the package falls straight down. This is because in your reference frame, the package has zero horizontal speed, while the ground appears to move horizontally backward at a speed equal to that of the airplane. Therefore, when the package strikes the ground, the airplane is directly over the point of impact.

In a real situation, however, air resistance reduces both the horizontal and the vertical speeds of a projectile. Therefore, the package strikes the ground behind the airplane.

Note that the ball loses all its vertical velocity at the top of its trajectory, while its horizontal velocity is unaffected.

When you throw a baseball upward, you give it an initial velocity with horizontal and vertical components. In the absence of air resistance, the horizontal component remains unchanged because no force acts on the ball in this direction. The vertical component, however, decreases as the ball rises above the ground, until the vertical velocity becomes zero at the highest point of the ball’s trajectory (see Video Q10.47). At this point, the ball slows to its minimum speed, which is equal to its constant horizontal speed ($v_x$ in Video Q10.47).

Bullseye Bill has to aim the barrel slightly higher than the target because the bullet does not travel in a straight line. The bullet curves toward the ground during its flight.

While in the air, a bullet moves as a projectile because it is pulled down by the force of gravity. Therefore, if the rifle’s barrel is pointed exactly at the target, the bullet will hit below the projected line, as shown in Figure Q10.48. To compensate for this deviation, the barrel should be slightly elevated.

Because of its tangential speed.

If the Moon had no rotational motion around Earth, the force of gravity would pull the Moon directly toward Earth and the two bodies would collide. If the Moon had a relatively small tangential speed, it would eventually approach and hit Earth’s surface like a projectile.

In reality, the Moon revolves around Earth in a nearly circular orbit with a radius of 384,000 km at a speed of about 1 km/s. The force of gravity provides the required centripetal force.

Planets farther from the Sun than Earth.

Planets farther from the Sun move slower and in larger orbits than Earth. According to Kepler’s third law, the square of the orbital period of a planet is directly proportional to the cube of its average distance from the Sun, or \begin{equation} T^2 \propto r^3 \end{equation} Thus, the period of rotation $T$ increases with the distance $r$ from the Sun. The outer planets in the solar system therefore have periods of rotation longer than one Earth year (yr):

• Mars: 1.9 yr
• Jupiter: 12 yr
• Saturn: 29 yr
• Uranus: 84 yr
• Neptune: 165 yr

The net force in each case is proportional to the object’s mass.

A freely falling object of mass $m$ experiences a net downward force equal to the force gravity on the object. By applying Newton’s second law to this motion, you have

\begin{align} m a &= F_g \nonumber\\[6pt] \Rightarrow\quad ma &= G\, \frac{m M_\text{E}}{R_\text{E}^2} \nonumber\\[6pt] \Rightarrow\quad a &= \frac{G M_\text{E}}{R_\text{E}^2} = g \label{u10_q56_ag} \end{align}

Equation 10.7 shows that all freely falling objects experience equal acceleration ($g$) and gain speed at the same rate, regardless of their mass.

The centripetal force ($F_c$) required for rotational motion is provided by the force of gravity ($F_g$) on the satellite. Note that both forces are proportional to the satellite’s mass ($m$), so you can write

\begin{align} F_c &= F_g \nonumber\\[6pt] \Rightarrow\quad \frac{m v^2}{r} &= G\, \frac{m M_\text{E}}{r^2} \nonumber\\[6pt] \Rightarrow\quad v^2 &= \frac{G M_\text{E}}{r} \nonumber\\[6pt] \Rightarrow\quad v &= \sqrt{\frac{G M_\text{E}}{r}}\label{u10_q56_v} \end{align}

Equation 10.8 shows that the satellite’s speed ($v$) is independent of its own mass but dependent on Earth’s mass ($M_\text{E}$).

December

The total mechanical energy of the Sun–Earth system is a conserved quantity, and it is equal to the sum of Earth’s kinetic energy (KE) and gravitational potential energy (PE). In other words, $\text{KE} + \text{PE}$ is the same at any point on Earth’s orbit around the Sun. Since PE is smaller at locations closer to the Sun, KE (and therefore speed) increases at such locations.

Work is done on an object when at least part of the applied force is along the direction of motion. In a circular orbit, the force of gravity continuously acts perpendicular to the direction of motion and, therefore, no work is done on the satellite. However, in an elliptical orbit, there is always a component of the gravitational pull parallel to the satellite’s path (except at the apogee and perigee points). Therefore, work is done on the satellite in this case.

11.2 km/s

Escape speed is the minimum initial speed required for an object launched vertically upward to overcome the force of Earth’s gravity and not fall back to the ground. This speed is equal to 11.2 km/s.

An object launched straight up at 11.2 km/s will eventually stop at a distance far from Earth when all the object’s initial KE is converted into gravitational PE. When the object starts to fall back toward Earth, the process is reversed: PE is now converted into KE. Just before it hits Earth’s surface, the object reaches its initial KE again. Therefore, the maximum speed of impact on Earth’s surface is equal to the launch speed of 11.2 km/s (the escape speed). If launched at a greater speed, the object will not return to Earth.

The first half falls straight down, while the other half leaves Earth.

Based on the law of conservation of momentum, if one half of the satellite comes to rest after the explosion, the other half (of equal mass) will move away at twice the original speed. As a result, the part that freezes in orbit simply falls straight down under the influence of its weight. The remaining section has enough speed to escape its orbit about Earth and become a satellite of the Sun.