10. Projectile and satellite motion

Alyssa J. Pasquale, Ph.D.; David R. Fazzini, Ph.D.; Carley Bennett, Ph.D.

10. Projectile and satellite motion

Summary

Projectile motion describes the motion of an object in free fall, and allows us to analyze the motion of baseballs, rockets, missiles, satellites, planets, stars, and other celestial bodies. This chapter builds on the topics of motion and Newton’s laws that have been introduced in previous chapters, and brings the section on kinematics to a close.

Projectile motion

Projectile motion describes the motion of an object that moves in free fall. That is, the only force acting on the object is gravity. Remember that free fall is an ideal scenario that ignores dissipative forces such as friction and air drag. The study of objects that travel like this is known as ballistics.

Ballistics (the physics of projectile motion) allows us to understand and quantify the motion of things such as baseballs after they leave the bat, rockets that are only powered internally very briefly before coming back to the Earth’s surface again, as well as the motion of planets, other celestial bodies, and artificial satellites.

In the simplest scenario, an object can be thrown straight up so that it moves in a completely vertical path. That’s the scenario we’ve already discussed in chapter three of this textbook. Alternatively, an object can be thrown upward, but at an angle to the horizon. This gives the object both vertical and horizontal motion.

In the absence of air drag and friction (which will be assumed unless otherwise stated), the only force that acts on an object during projectile motion is gravity. Gravity only works in the vertical direction. To understand how the horizontal motion of the projectile works, we must remember Newton’s first law: inertia. The projectile will continue to move at the same speed in the horizontal direction, even while the vertical speed changes due to gravity. Fortunately, we can do an analysis of the horizontal motion separately from the analysis of the vertical motion.

In the video below, two identical steel balls are launched from a platform at exactly the same time. One is dropped in a completely vertical manner, and the other is pushed, giving it horizontal motion in addition to the vertical drop.

Both of the balls hit the surface of the lab table at exactly the same time, indicating that gravity acted on the objects in an identical manner. The ball that had horizontal motion did not exhibit any effects in its vertical motion, which means that the horizontal motion for this ball did not affect its vertical motion.

Fortunately, we are able to independently calculate the horizontal and vertical motion of a projectile in free fall. This will enable us to determine what equations we can use to analyze the motion in each direction. Each of the forces and related kinematic constants and equations (acceleration, velocity, and distance) are listed in the table below.

direction	force	acceleration	velocity	displacement
horizontal	none	0 m/s²	$\vec{v}_{final} = \vec{v}_{initial}$	$\vec{d} = \vec{v}t$
vertical	gravity	-9.8 m/s²	$\vec{v}_{final} = \vec{v}_{initial} + \vec{a}t$	$\vec{d} = \frac{1}{2}\vec{a}t^2 + \vec{v}_{initial}t$

In the horizontal direction, there are no forces at work on the object. Because of this, the law of inertia tells us that the object will continue moving with the same speed in the horizontal direction. As an equation, this is expressed as: $\vec{v}_{final} = \vec{v}_{initial}$ , the velocity of the projectile at any moment in time is equal to the initial velocity. Because there is no accelerating force, the distance the object moves horizontally can be analyzed using the equation $\vec{d} = \vec{v}t$ : displacement equals velocity times time.

In the vertical direction, the force acting on the object is gravity. The acceleration due to gravity is -9.8 m/s². (Remember that the negative sign indicates that the force and acceleration point down.) Because there is acceleration at work, the equations of motion in the vertical direction will be different from the equations in the horizontal direction.

The vertical velocity of the projectile at any given moment in time is equal to $\vec{v}_{final} = \vec{v}_{initial} + \vec{a}t$ , which is the initial velocity plus acceleration (due to gravity) times time. The displacement that the object moves vertically can be analyzed using the equation $\vec{d} = \frac{1}{2}\vec{a}t^2 + \vec{v}_{initial}t$ . It is important to note that the equations of non-accelerated motion cannot be used when there is a force acting on an object!

In the video below, a ballistic cart launches a ping-pong ball directly upward with no horizontal motion. The ball moves upward as it slows down (vertically) under the influence of gravity. It stops for a brief instant of time as it turns around and moves back down to the cart, speeding up (vertically) as it does so. This is a special case of projectile motion: where there is no horizontal motion at all. This type of motion was introduced and discussed in chapter three.

On the other hand, the ballistic cart can launch the ball while it moves with a horizontal velocity. We learned in our discussion of inertia that the ping-pong ball will continue moving at the same horizontal speed, even while the vertical motion of the ball is influenced by gravity. This is why the ball is caught by the cart again as it moves back down to its initial height, shown in the video below.

The shape of the motion of a projectile moving with horizontal and vertical motion is known as a parabola. We can therefore say that a projectile has parabolic motion. A graph of the position of a projectile (qualitatively similar to the ball launched by the ballistic cart) with respect to time is shown in Figure 10.1. Understanding this motion enables scientists and engineers to design rockets, satellites, and other projectiles that have parabolic motion.

A graph of position (0-80 m) plotted on the y-axis and time (0-8 s) plotted on the x-axis. There are 16 data points plotted. A fit line shows a parabolic trend. The parabola's vertex is located at (4,80). — Figure 10.1 – A graph of parabolic motion with position plotted as a function of time. This figure, created by Alyssa J. Pasquale, Ph.D., is licensed under CC BY-NC-SA 4.0.

An important takeaway from the video of the steel balls dropping, and the videos of the ballistic cart launches, is that the horizontal and vertical components of motion can be calculated and analyzed independently. This greatly simplifies the study of projectile motion!

When air drag cannot be ignored

It should be noted that when air drag cannot be ignored, the equations describing the trajectory of the projectile are not as simple. This is because the drag force always points in a direction tangent to the trajectory and opposite to the instantaneous velocity at every instant. This means that the vertical and horizontal forces are continuously changing. Therefore, the vertical and horizontal analyses are not independent and must be analyzed simultaneously. Such an analysis is beyond the scope of this textbook.

Maximizing horizontal range

Because a projectile can be launched at so many different angles with respect to the horizon, what angle would we choose if we wanted a projectile to have the maximum horizontal range?

If we conceptualize the motion of a projectile, an object launched directly upward will spend the most time in the air, but will have the least horizontal motion as it travels in a completely vertical line. On the other extreme, if something were to be launched at just a few degrees with respect to the horizon it would not spend as much time in the air, minimizing its horizontal range. To maximize the horizontal range, we must choose an angle somewhere in between zero and ninety degrees.

In the video below, Dr. Pasquale launched a projectile at fifteen degrees, thirty degrees, forty-five degrees, sixty degrees, and seventy-five degrees with respect to the horizon. Each of the projectiles has an identical initial velocity. In each case, we can look at the parabolic motion of the projectile. Because the initial height of the projectile changes when the launcher is tilted at different angles, we will only consider the motion back to the same starting height, rather than considering the motion back to the height of the table.

As expected, very steep angles maximize the height of the ball, but don’t move much horizontally. Very shallow angles spend less time in the air, causing them to lose out on horizontal motion. At forty-five degrees, we maximize the horizontal range by creating the best tradeoff between time in the air and horizontal motion.

Figure 10.2 shows a graph of the horizontal and vertical positions for projectiles launched at the same angles as in the video. These positions were calculated using the equations of motion. The graph demonstrates the vertical and horizontal ranges of each launch angle. (This data is qualitatively similar to the data shown in the video above, but is not identical due to the small amount of air drag that is present.)

A graph of vertical position (0-160 cm) plotted on the y-axis and horizontal position (0-320 cm) plotted on the x-axis. There are datasets for 5 different launch angles. Each data set contains 17 data points and a parabolic fit line. The data for the 75 degree launch angle is the tallest parabola but has the shortest horizontal range. 60 degree launch has a shorter parabola with a larger horizontal range. The 45 degree launch has a medium height parabola with the largest horizontal range. The 30 degree launch angle has a short vertical range and a medium horizontal range. The 15 degree launch angle has the shortest vertical range and short horizontal range. — Figure 10.2 – A graph of the parabolic motion of projectiles launched at different starting angles. The projectile launched at 45 degrees has a maximum horizontal range. This figure, created by Alyssa J. Pasquale, Ph.D., is licensed under CC BY-NC-SA 4.0.

In the absence of air drag, a launch angle of forty-five degrees will maximize the horizontal range of a projectile.

Symmetry of motion

In the absence of forces such as air drag, the motion of a projectile is symmetric both in time and horizontally in space. This means that

it takes a projectile the same amount of time to go from ground level to the top of its path as it does to go from the top of its path back down to ground level again,
the vertical velocity of a projectile at the end of its motion when it reaches the ground again is equal and opposite to the initial vertical velocity of the projectile at launch, and
a projectile moves the same horizontal distance between ground level and the top of its path as it does from the top of its path back down to ground level again.

From the symmetry of motion arises three key moments in the motion of a projectile, which are

the initial time a projectile is launched,
the moment the projectile reaches the top of its path, and
the time the projectile reaches ground level again.

These three key moments act as important times to calculate for different kinematic properties (height or vertical velocity) because there will always be a known variable, simplifying the equations that are used. This information is outlined in the table below.

key moment	time	vertical position (y)	horizontal position (x)	vertical velocity ( $\vec{v}_{y,final}$ )
initial launch	0	0	0	$\vec{v}_{y,initial}$
top of the path	½ of total time	maximum height	½ maximum range	0
return to ground level	total time	0	maximum range	$-\vec{v}_{y,initial}$

Solving projectile motion equations

The motion of projectiles can be analyzed using the information presented in this and previous chapters of this textbook.

For example, say a projectile is launched from ground level with an initial horizontal velocity of 10 m/s and an initial vertical velocity of 30 m/s. We want to solve for three different quantities:

How high will the projectile go?
How long will the projectile stay in the air?
How far will the projectile travel in the horizontal direction?

The first question asks about height. Because we are interested in the maximum height of the projectile, the key moment is the top of the path of the projectile. We can see from the table above (and recall from many examples in this textbook) that the vertical velocity at that key moment is 0 m/s. The two vertical kinematics equations are

$\begin{align*} \vec{d} &= \frac{1}{2}\vec{a}t^2 + \vec{v}_{initial}t\\ \vec{v}_{final} &= \vec{v}_{initial} + at \end{align*}$

While we eventually want to use the distance equation to solve for height, at this moment we don’t know the amount of time that the projectile travels, so we can’t use the equation yet. However, we can use the velocity equation to solve for the amount of time it will take the projectile to reach the top of its motion (which is, incidentally, equal to one-half of the total time of motion). Because the vertical velocity at this key moment is 0 m/s, we can solve the velocity equation for time.

$\begin{align*} 0~\textrm{m/s} &= 30~\textrm{m/s} + (-9.8~\textrm{m/s}^2)t\\ -30~\textrm{m/s} &= (-9.8~\textrm{m/s}^2)t\\ 3.06~\textrm{s} &= t \end{align*}$

It takes 3.06 s for the projectile to reach the peak of its motion. Now that we know the time, we can plug that value into the vertical distance equation to determine the height, which is

$\begin{align*} d &= \frac{1}{2}(-9.8~\textrm{m/s}^2)(3.06~\textrm{s})^2 + (30~\textrm{m/s})(3.06~\textrm{s})\\ &= -45.88~\textrm{m} + 91.8~\textrm{m}\\ &= 45.92~\textrm{m}. \end{align*}$

Now we can determine how long the projectile will remain in the air. Based on the calculation of time at the key moment of being at the top of the path, we know that the projectile takes 3.06 s to complete one half of its total motion. Because the motion is symmetric, if it takes 3.06 s to get up, it will take another 3.06 s to get back down again. So we know that the projectile will spend 6.12 s in the air.

If you are asked to solve for how long the projectile will remain in the air without needing to first calculate the initial height, it’s possible to determine this value by using the key moment of the return to ground level. At that time, the vertical position is zero, and the final vertical velocity is equal to the opposite value of the initial vertical velocity. Use the velocity equation to solve for time.

$\begin{align*} -30~\textrm{m/s} &= 30~\textrm{m/s} + (-9.8~\textrm{m/s}^2)t\\ -60~\textrm{m/s} &= (-9.8~\textrm{m/s}^2)t\\ 6.12~\textrm{s} &= t \end{align*}$

Finally, we can answer the question of how far the projectile moves horizontally. The position of maximum horizontal range occurs at the key moment of the return to ground level. We know from previous calculations that it takes the projectile 6.12 s to reach that key moment. Now we can use the equations of horizontal motion (no acceleration) to calculate the horizontal range.

$\begin{align*} d &= (10~\textrm{m/s})(6.12~\textrm{s})\\ &= 61.2~\textrm{m} \end{align*}$

Satellites and orbits

A satellite is a projectile that falls around the Earth, rather than falling back down to its surface. A thought experiment may help to conceptualize what this means. Consider that Jason Q. Physics builds an extremely tall tower, possibly several kilometers high, and stands at the top with several baseballs. Assuming that Jason has enough oxygen to survive this experiment, he can throw the baseballs at various speeds and see what happens.

This thought experiment is animated with a video, below. (Note: the video is not to scale!) To start, Jason drops a baseball. It falls straight down to the Earth’s surface in a vertical line. If he now gives the ball a gentle throw in the horizontal direction, it will still fall to the Earth’s surface, but the motion gives it a gentle curve. As he throws the remaining baseballs faster and faster, the baseball will travel a greater distance along the Earth’s surface. At some point, we can no longer ignore the curvature of the Earth’s surface. Eventually, the trajectory of the motion of the baseball will match the curvature of the planet. That trajectory is known as an orbit. An orbit describes the curved path that a satellite takes as it moves around an object of mass. In that case, Jason Physics has turned the baseball into a satellite!

When we were dealing with baseballs near the surface of the Earth, the force of gravity (both magnitude and direction) was considered constant: 9.8 m/s², vertically downward. For the baseballs launched from Jason Physics’ tower, the gravitational force varies in magnitude (weaker at higher altitudes) and direction (points toward the center of the Earth). When these factors are considered, the resulting trajectory of the orbit becomes elliptical in shape.

At what speed must the baseball be thrown to obtain this satellite motion? We can turn to rotational motion to calculate an exact answer. The centripetal force of the satellite must be supplied by gravity. Set the centripetal force equation equal to Newton’s law of universal gravitation, and solve for velocity.

$\begin{align*} m_{satellite}\frac{v^2}{r} &= G \frac{m_{satellite}m_{Earth}}{r^2}\\ v^2 &= G \frac{m_{Earth}}{r}\\ v &= \sqrt{G \frac{m_{Earth}}{r}} \end{align*}$

In this equation, $v$ is the orbital velocity (the minimum velocity required to obtain orbit around the Earth). $r$ is the radius of the satellite’s orbit (this is equal to the radius of the Earth plus the orbital height of the satellite). $G$ is the universal gravitational constant, and $m_{Earth}$ is the mass of the Earth. (If you wanted to place a satellite in orbit around a different celestial body, you would simply use the mass and radius of the celestial body you wish to orbit.)

This equation can be used to calculate the orbital velocity of the international space station, which orbits approximately 400 km above the Earth’s surface. The mass of the Earth is $6 \times 10^{24}~\textrm{kg}$ , and the radius of the Earth is approximately 6400 km. The orbital velocity therefore equals

$\begin{align*} v &= \sqrt{(6.67 \times 10^{-11}~\textrm{Nm}^2\textrm{/kg}^2) \frac{6 \times 10^{24}~\textrm{kg}}{400000~\textrm{m} + 6400000~\textrm{m}}}\\ &= 7654~\textrm{m/s} \end{align*}$

The orbit of a satellite describes its path around a planet or star. All closed orbits are elliptical. A special case of an elliptical orbit is the circular orbit. The above baseball example, and calculations for the international space station, described circular orbits.

In a circular orbit (shown in Figure 10.3, left), the height above the Earth’s surface remains constant. Considering the conservation of mechanical energy, because the orbital height remains constant, the GPE is constant throughout the satellite’s motion. Because mechanical energy is conserved, the KE must also remain constant, causing the speed of the satellite to remain the same throughout its motion.

Figure 10.3, described in the text. — Figure 10.3 – Two possible orbits of a planet around a star. A circular orbit (left) has an equal distance away from the star at any point in the planet’s orbit. An elliptical orbit (right) has different distances away from the star at different points in the planet’s orbit. This figure, created by Alyssa J. Pasquale, Ph.D., is licensed under CC BY-NC-SA 4.0.

In an elliptical orbit, a satellite will orbit around a planet or a star, with the planet or star at one of the focal points of the ellipse. As the satellite moves, the distance away from the planet or star changes (Figure 10.3, right). Mechanical energy is conserved. This means that as the GPE decreases, when the satellite is closest to the planet, the KE will increase, and vice versa. Therefore, when the satellite is closest to the planet, it will be moving at the fastest speed. When the satellite is farthest from the planet, it will be moving at the slowest speed.

Kepler’s laws of planetary motion

The discovery of elliptical orbits came about in the early 1600s thanks to Johannes Kepler, before Newton published his theory on universal gravitation. Kepler was an apprentice of Tycho Brahe, a Danish astronomer, who made very accurate observations of the positions of stars and planets over a span of decades in the late 1500s (without the aid of a telescope). Kepler used Brahe’s data and plotted the orbit of Mars from the vantage point of someone looking down on the solar system from above. Kepler tried to use the equation of a circle to fit the points along the orbit and found that he couldn’t. He was only able to fit the orbital path if he used the equation for an ellipse rather than a circle. The discovery that planets orbit in ellipses rather than circles is the first of what became known as Kepler’s three laws of planetary motion.

Kepler’s first law

Kepler’s first law of planetary motion (the law of orbits) states that planets orbit their sun in ellipses with the sun at one focus of the ellipse (such as the orbit shown in Figure 10.3, right). This law is valid for any satellite in a closed orbit around a host body, whether it be a planet around a star or a moon around a planet. It should be noted that circular orbits are possible. A circle is just a special case of the ellipse in which the two foci are at the same point (Figure 10.3, left).

Kepler’s second law

Kepler’s second law of planetary motion (the law of areas) states that the position vector of the planet relative to its sun will sweep out equal areas in equal time intervals. As seen in Figure 10.4, the planet takes a certain time ( $\Delta t$ ) to move from point A to point B. When the planet is at point C, it will move to point D in the same time interval ( $\Delta t$ ). The areas of region ABO and region CDO are the same.

A drawing of an elliptical orbit with a star at one focal point. A planet will travel along the elliptical orbit. Two pie-shaped slices of equal area in the ellipse, with the star at the point, are highlighted. One of the areas has a long path length along the orbit (labeled A to B). One of the areas has a short path length along the orbit (labeled C to D). Both of these distances will be traveled in the same interal of time (labeled Δt). — Figure 10.4 – An elliptical orbit showing two equal areas (ABO and CDO) where the planet will have equal transit times from point to point (A to B and C to D), leading to different orbital speeds. This figure, created by Alyssa J. Pasquale, Ph.D., is licensed under CC BY-NC-SA 4.0.

This turns out to be a consequence of the conservation of angular momentum. While the mathematical details are too involved for this textbook, a qualitative description is warranted. Like the figure skaters who pull in their arms and legs to minimize their rotational inertia to make themselves rotate faster, the planet’s rotational inertia about the sun is smaller when it is close to the sun. When the planet is farther away, its rotational inertia is greater because its mass is farther from the center of rotation (the sun). This increase in rotational inertia causes the rotational speed around the sun to decrease. Thus, the planet moves faster when closer to its sun and slower when farther away from its sun.

Kepler’s third law

Kepler’s third law of planetary motion (the law of periods) states a mathematical relationship between the time it takes for the planet to complete one orbit (the orbital period) and the semi-major axis of the elliptical orbit (see Figure 10.5).

A drawing of an ellipse. The two focal points are shown and labeled. The semi-minor axis (shortest distance between the center and the edge of the ellipse) and semi-major axis (longest distance between the center and edge of the ellipse) are highlighted and labeled. — Figure 10.5 – An ellipse, showing the two focal points and the semi-minor and semi-major axes. This figure, created by Alyssa J. Pasquale, Ph.D., is licensed under CC BY-NC-SA 4.0.

Specifically, the relationship states that the square of the orbital period is directly proportional to the cube of the semi-major axis: $T^2 \propto a^3$ . For a circular orbit, the semi-major axis just becomes the radius of the circle. With just a little effort, Kepler’s third law can be easily shown for a circular orbit (see the textbox below for the derivation). We start by recognizing that the gravitational force is responsible for the centripetal force that produces the circular motion.

Since the motion is circular, the gravitational force is producing a centripetal force as described in chapter 8. We then equate the gravitational force between the planet (m) and its sun (M) separated by a distance r ( $GmM/r^2$ ) to the centripetal force ( $mv^2/r$ ), where v is the orbital speed of the planet. Recalling that speed is the path divided by time, the orbital speed is just the circumference of the circle divided by the orbital period. That is,

$v = \frac{2\pi r}{T}.$

After plugging the above equation for speed into the expression for the centripetal force, we can solve for $T^2$ . The result is

$T^2 = \frac{4 \pi^2}{GM}r^3.$

Since $\pi$ , $G$ , and $M$ are all constants, this proves the relationship that $T^2 \propto a^3$ . Note that the mass of the planet ( $m$ ) doesn’t affect this relationship because it cancels out in the derivation.

Derivation of Kepler’s third law

$\begin{align*} F_{gravity} &= F_{centripetal}\\ \frac{GmM}{r^2} &= \frac{mv^2}{r}\\ \frac{GM}{r} &= v^2 \end{align*}$

Orbital speed, $v$ , is equal to path divided by time ( $T$ ). For a circular orbit, the path is equal to the circumference of the circle: $2\pi r$ . This value will be plugged into the above equation for $v$ to continue the derivation.

$\begin{align*} \frac{GM}{r} &= (\frac{2\pi r}{T})^2 = \frac{4 \pi^2 r^2}{T^2}\\ T^2GM &= 4\pi^2 r^3\\ T^2 = (\frac{4\pi^2}{GM})r^3 \end{align*}$

As an example, a planet that is four times farther from its host star will have an orbital period that is eight times greater, since $8^2 = 4^3$ .

Practice questions

Numerical analysis

A ball is launched vertically upward from the ground with an initial speed of 10 m/s. Calculate…
1. …the time it takes the ball to reach its maximum height.
2. …the maximum height of the ball.
A soccer ball is kicked horizontally from a height of 2 meters above the ground with an initial horizontal speed of 15 m/s. Calculate…
1. …how far the ball travels horizontally before hitting the ground.
2. …the vertical speed of the ball immediately before impact.
3. …the horizontal speed of the ball immediately before impact.
4. …the total magnitude of the ball’s velocity immediately before impact.
A baseball is thrown with an initial horizontal speed of 20 m/s. It lands on the ground 4 seconds later. Calculate…
1. …how far the ball travels horizontally before hitting the ground.
2. …the height the ball falls in that time.
3. …the vertical speed of the ball immediately before impact.
4. …the horizontal speed of the ball immediately before impact.
5. …the total magnitude of the ball’s velocity immediately before impact.
A rock is dropped from a bridge into a river below. If it takes 3 seconds to hit the water, how high is the bridge above the water’s surface?
A rocket is launched from the ground with an initial vertical velocity of 30 m/s upward and a horizontal velocity of 40 m/s. Calculate…
1. …the time it takes for the rocket to hit the ground.
2. …the horizontal distance traveled by the rocket.
A satellite with a mass of 500 kg is in a circular orbit around Earth at an altitude of 500 km above the surface. The tangential speed of the satellite is 7615 m/s. Calculate its kinetic energy, gravitational potential energy, and total mechanical energy.
A communication satellite (1,000 kg) is in a geostationary orbit around Earth at an altitude of approximately 35,786 km. (That is, the satellite orbits at the same speed as the rotation of the Earth around its axis.) Calculate the kinetic energy, gravitational potential energy, and total mechanical energy of the satellite.
A satellite (2,500 kg) is in an elliptical orbit around the Earth. As it moves from an altitude of 200 km to an altitude of 400 km, calculate…
1. …the GPE of the satellite at both altitudes.
2. …the change in KE as the satellite moves from low to high altitude.
3. …the change in ME as the satellite moves from low to high altitude.

Hands-on experiments

Crumple a piece of paper into a ball and set up a target (e.g., a trash can or a box) on the floor. Stand a fixed distance away from the target and try to toss the paper ball into the target. Experiment by changing the angle and force of your toss to observe how it affects the distance and accuracy of your throw.

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Conceptual Physics Copyright © 2024 by Alyssa J. Pasquale, Ph.D.; David R. Fazzini, Ph.D.; and Carley Bennett, Ph.D. is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License, except where otherwise noted.