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PhET Projectile Motion Simulation

Change the angle of a cannon and the starting velocity of a projectile to see how it changes the path of a projectile that it launches. You can also change the height of the cannon to see how launch angle affects the horizontal distance of the projectile.

PhET Gravity and Orbits Simulation

You can see how the orbit of a planet about a star changes as you play around with the masses of the star and the planet.

Video Transcript

Hello there! Welcome to lecture 10: projectile and satellite motion!

Projectile motion describes the motion of an object in free-fall, and allows us to analyze the motion of rockets, missiles, satellites, planets, stars, and other celestial bodies. This discussion builds on our previous discussions of motion and Newton’s laws, and brings our discussions of kinematics to a close.

Each of the following concepts will be discussed in this video: projectile motion, solving projectile motion equations, and satellites and orbits.

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 from previous lectures 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 rockets that are only powered internally very briefly before coming back to the Earth’s surface again, as well as the motion of planets and other celestial bodies as well as artificial satellites.

In the simplest scenario, an object can be thrown in a completely vertical path. That’s the scenario we’ve already discussed in lecture three. Alternatively, an object can be thrown upward, but at an angle to the horizon. This gives the object both vertical and horizontal motion.

As mentioned, 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.

Let’s look at a demo to convince ourselves that vertical and horizontal motion act independently of each other. In this video, two otherwise 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 affect the vertical motion.

Let’s consider the physics that act in each dimension: horizontally and vertically. This will enable us to determine what equations we can use to analyze the motion.

In the horizontal direction, there are no forces at work on the object. This means that the law of inertia holds: the object will continue moving at the same speed in the same direction. As an equation, we can say that 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: distance equals velocity times time. 

In the vertical direction, the force acting on the object is gravity. The acceleration due to gravity is negative 9.8 meters per second-squared. (Remember that the negative sign indicates that the force points down.) Because there is acceleration at work, the equations of motion in the vertical direction will be different. The velocity of the projectile at any given moment in time is equal to the initial velocity plus acceleration times time. The distance the object moves vertically can be analyzed using the equation distance equals one-half times acceleration times time-squared plus initial velocity times time.

Let’s look at another video showing projectile motion. This uses the ballistic cart we’ve seen in other lectures discussing inertia and linear motion. In one case, the ballistic cart launches a ping-pong ball directly upward. The ball moves upward as it slows down 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 as it does so. This is a special case of projectile motion: where there is no horizontal motion at all. We have already discussed this type of motion in lecture three.

On the other hand, the ballistic cart can launch the ball while it moves with a horizontal speed. 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.

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. Understanding this motion enables scientists and engineers to design rockets, satellites, and other projectiles that have parabolic motion.

Speaking of ballistic design… 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, we can consider that 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 these videos, I launched a projectile at fifteen degrees, thirty degrees, forty-five degrees, sixty degrees, and seventy-five degrees with respect to the horizon. 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.

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

Solving projectile motion equations

Let’s use the information we’ve just gained, as well as some past information about kinematics we’ve learned in previous lectures, to analyze the motion of projectiles. We can do projectile motion analysis using kinematics equations or using conservation of energy equations. Generally speaking, kinematics equations are best used when we need to calculate timing; for example, how long something stays in the air. Conservation of energy is best used when we need to calculate heights; how high a projectile will be at any given speed.

For example, say a projectile is launched from ground level with an initial horizontal velocity of 10 meters per second and an initial vertical velocity of 30 meters per second. We want to solve for three different things: (1) how high does the projectile go? (2) How long will the projectile stay in the air? (3) How far will the projectile go in the horizontal direction?

The first question asks about height. Let’s use kinematics first and then use conservation of energy to solve this question. We must use equations concerning vertical motion to answer this question.

Our two kinematics equations are: d equals one-half a times t-squared plus v-initial times t, and v-final equals v-initial plus a times t. While we eventually want to use the distance equation to solve for height, at this moment we don’t know the time 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.

We know that at the top of its motion, the projectile will instantaneously have zero vertical velocity. Now we can plug in some numbers. Zero equals 30 meters per second plus negative 9.8 meters per second-squared times time. Subtract 30 meters per second from both sides. Then divide both sides of the equation by negative 9.8 meters per second-squared. We see that time is equal to 3.06 seconds. It takes 3.06 seconds to reach the peak of the motion.

Now we can plug that back into the distance equation. Plug in the acceleration, time, and initial velocity. Distance equals one-half times negative 9.8 meters per second-squared times 3.06 seconds, squared, plus 30 meters per second times 3.06 seconds. After doing the math, we see that the distance, in other words, the maximum vertical height of the projectile, is 45.92 meters.

Now let’s use conservation of energy. Because we are ignoring air drag, we know that the gravitational potential energy plus the kinetic energy is going to be equal to a constant. Ignoring horizonal motion, when the projectile starts moving, it has all kinetic energy and no potential energy. At the top of its peak, it has all potential energy and no kinetic energy. This means that the KE will be completely converted to GPE as the ball moves to its maximum vertical position.

In other words, GPE at the peak equals KE at the start. Because energy is a scalar, we can now ignore the sign of the gravitational acceleration. Let’s plug in the equations for GPE and KE and solve. M times g times h equals one-half times m times v-squared. We can divide both sides by mass to cancel out the m terms. Divide both sides by g and we find that h equals one-half times v-squared divided by g. Plug in the initial vertical velocity and 9.8 meters per second-squared, and we see that height equals one-half times 30 meters per second, squared, divided by 9.8 meters per second-squared. Height equals 45.92 meters. As expected, this agrees with our kinematic calculation.

See how conservation of energy saves a step? However… using kinematics is beneficial in that we gain two important pieces of information: time and height. Either way, let’s solve for the last two questions we asked.

Now let’s determine how long the projectile will remain in the air. Based on our kinematic calculation, we know that the projectile takes 3.06 seconds to get to the top of the motion. Because the motion is symmetric, if it takes 3.06 seconds to get up, it will take another 3.06 seconds to get back down again. So we know that the projectile will spend 6.12 seconds in the air.

If we had only used conservation of energy to solve for height, we could still determine the time in the air by using the velocity equation for vertical motion. It would require us to have an understanding of the symmetry of projectile motion. In other words, when the projectile reaches the ground again, right before it stops moving, it will have a final velocity of 30 meters per second, but this time pointing down.

Use v-final equals v-initial plus a times t. V-final equals negative 30 meters per second. V initial is positive 30 meters per second. A is negative 9.8 meters per second-squared. Plug and chug and we find that the time is 6.12 seconds.

Finally, we can answer the question of how far the projectile moves horizontally. Because the projectile moves for 6.12 seconds, and it has a horizontal velocity of 10 meters per second, we can use v equals d over to t to calculate the distance. Distance equals 10 meters per second times 6.12 seconds, giving a horizontal range of 61.2 meters.

Satellites and orbits

A satellite is a projectile that falls around the Earth, rather than falling back down to its surface. Let’s consider a thought experiment. We build an extremely tall ladder, possibly several kilometers high, and stand at the top with several baseballs. Assuming we have enough oxygen to survive this experiment, we can throw the baseballs and see what happens.

To start, we drop a baseball. It falls straight down to the Earth’s surface in a vertical line. If we now give the ball a gentle throw in the horizontal direction, it will still fall to the Earth’s surface, but the parabolic motion gives it a gentle curve. As we throw our remaining baseballs faster and faster, we will see a point in which the curve of the parabola equals to the curve of the planet. In that case, we have turned the baseball into a satellite!

At what speed must we throw the baseball to obtain this satellite motion? We turn to rotational motion to help. The centripetal force of the satellite must be supplied by gravity. Set the centripetal force equation equal to Newton’s law of universal gravitation. M times v-squared over r equals G times m times m-earth divided by r-squared. The mass of the satellite cancels out on both sides. We can also cancel out one of the r terms on each side. Then we take the square root of both sides and solve for v. Orbital velocity equals the square root of G times mass of the Earth divided by r. R is the radius of motion, which is equal to the radius of the Earth plus the orbital height. 

Let’s calculate the orbital velocity of the international space station, which orbits approximately 400 kilometers above the Earth’s surface. The mass of the Earth is 6 times 10 to the 24 kilograms, and the radius of the Earth is approximately 6400 kilometers. The orbital velocity therefore equals the square root of: 6.67 times 10 to the negative 11 times 6 times 10 to the 24, divided by 6800 times ten to the three. After doing the math, we find that the orbital velocity is approximately 7,672 meters per second. That’s fast!

Now that we know how to calculate the orbital velocity of a satellite, we can start to get an idea of the shape of a satellite’s orbit. The orbit of a satellite describes its path around a planet or star. All orbits are elliptical. A special case of an elliptical orbit is the circular orbit. Our baseball example, and calculations for the international space station, described circular orbits.

In a circular orbit, 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.

Now let’s consider an elliptical orbit. The 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. However, 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.

Thanks for taking the time to learn about projectile and satellite motion! Until next time, stay well.