The Space Shuttle, including Astronaut Sandra Magnus and all its inhabitants (animate and inanimate) follows a nearly circular path as it orbits Earth. The shuttle is constantly changing direction and therefore she is always accelerating. This acceleration is proportional to the square of the velocity and inversely proportional to the distance from the center of the earth.

Newton’s 2nd Law (N2L) tells Sandra and her crew that for an object to be accelerating, there must be a Net Force exerted on it. What could be this force?? Well, it is our good friend the force of gravity (Fg) and this force vector points directly toward the center of Earth. This certainly means that everything aboard the shuttle is constantly accelerating toward the center of the Earth because that is the direction of the net force!!! Everything, therefore, is in Free Fall!!

Now, since the famous astronaut ice cream is falling at the same rate as the astronauts, they stay in the same position relative to each other. It seems as if they are floating, or that they are in a state of weightlessness. Weightlessness is NOT a lack of gravity!! The Fg due to an object is proportional to the inverse square of the distance to the object’s centre. Because the shuttle and other “satellites” are considered to be “Low-Earth Orbiting” , the acceleration due to gravity is only about 10% less than that on the Earth’s surface. And normally, we will consider this distance negligible in the vastness of space.

🎤 Important🎤

If the shuttle is constantly feeling a tug toward the Earth, why doesn’t it come crashing down to the surface? Well, the shuttle is always falling but, because of its velocity tangential to the orbital path, it’s always missing the Earth!!

This was first postulated by one of Isaac Newton’s greatest thought experiments!!

He thought a cannon could fire a ball with enough velocity that it’s trajectory would match the curvature of Earth. It would therefore remain the same height above the Earth but fall around, not toward, Earth! We would essentially cross the boundaries of projectile motion and into circular motion!

Science fiction has become reality! We launched the Voyager Probe that is still telling us about how weird interstellar space is 43 years later!! We can trace the origins of space flight to Isaac Newton’s Law of Universal Gravitation; the idea that all objects in the universe attract each other through the force of gravity. This is the gravitational force exerted by an object with mass on any other object with mass, and so it goes!! These gravitational components are even responsible for holding Earth together!!

Now we all know, from skiers to show dogs, that in order to make a high speed, sharp turn we must have a larger acceleration than when making a high speed gentle turn. This is specifically evident it the “racing line” that expert drivers try to find when maximizing the radius of turns on a race track. This is because acceleration also represents the change in direction of the velocity vector. So, what is this relationship between an objects acceleration, it’s speed and the radius of the circular path?

🏎 Let’s Delve!! 🛵

An object traveling in a circular path at constant speed is experiencing Uniform Circular Motion (UCM). Because the object’s speed is constant, it’s acceleration has no component tangent to its trajectory!! This tells us that the entire acceleration vector must point directly to the center of curvature!! We call this centripetal acceleration taken from the Latin for “center-seeking”

Now, it is of immense importance in the AP world and beyond to understand that in order to move with a constant velocity in a circular path, we must accelerate!! Why? It’s because our velocity vector is always changing direction. This is in direct contrast to linear motion with constant velocity where acceleration is zero (0).

Let’s consider the motion of the great hornbill in flight throughout one-quarter of a revolution. If she is initially moving North such that after turning 90 degrees, it’s moving at the same speed, only now to the East. It’s velocity vector originally had a zero (0) x-component but ends with an entire eastward velocity. The opposite occurs in the vertical direction. The loss of speed by one component is perfectly offset by a gain in speed in the other component so that the overall speed remains constant!! Finally, we have a formula to calculate the centripetal acceleration of an object following a circular path.

a(c) = v^2/r

This is read as, the magnitude of the centripetal acceleration is proportional to the square of the speed and inversely proportional to its radius. Therefore, there is greater a(c) the greater the hornbill’s speed and smaller the radius (the tighter the turn). This equation also applies to an object traveling at a constant speed around a portion of a circle. An object moving in non-uniform circular motion has a component along the direction of motion as well as perpendicular. No matter what, the hornbill and other objects moving in a circle have an acceleration directed toward the center of the circular path!!

Now is a good time to introduce the concepts of frequency, f, and period, T, of revolution. Frequency is a rate of any per second. It could be beats or flaps per second. The Period is the time is takes for one revolution. These two values are inversely related!! The unit for frequency is simply a placeholder and doesn’t appear in the Period.

Since velocity, as always, is distance divided by time, and the distance around a circle is its circumference. We have a formula for tangential velocity!!

v = 2(pi)r/T

Now, reconnecting with Sandra and the astronauts, the feeling of weightlessness means that any surface Sandra and the other astronauts touch has this same acceleration as Sandra so that it is impossible to exert a Normal Force on her.

This helps dispel the common misconception that an object in orbit is beyond the pull of the Earth’s gravity. But if gravity weren’t exerted by Earth on an orbiting satellite, it would have zero (0) acceleration and couldn’t stay in circular orbit. It would maintain constant velocity and fly off into space in a straight line. F(g) provides a(c)!! Sandra seems to be weightless because she is really a satellite of her own following the same orbit as the Shuttle and there is nothing pushing her toward any of the walls!

🏇🏿 Centripetal Force ⛷

Think of Jill on her Vespa making a sharp turn. She follows a curved path, so the direction of the velocity vector is always changing. Because the velocity is always changing, the Vespa is always accelerating, with a component of the acceleration vector directed toward the center of the circle or curve. We know that this acceleration depends on the radius of the curve and the Vespa’s velocity. And our good friend Isaac Newton tells us that an object accelerates because of a net external force exerted on it and this F(net) must point in the direction of the acceleration!!

What forces provide the acceleration needed to make an object follow a curved path?

1. The Force of Static Friction!! This could be between a track athlete’s shoes and the track or the tyres of a race car and the track. An example would be Jill on the Vespa.

2. The Force of Tension!! This could be the tension in a string swinging keys around a circle on a lanyard.

3. The Force of Gravity!!! This is the case of our orbiting satellites, moons and everything else in the universe!!

4. The Normal Force (or the Lift Force in case of an airplane)!! This could be a loop on a rollercoaster!!

Since the acceleration is always toward the center of the circle, N2L says that at any instant the F(net) exerted on the Vespa must also be directed toward the center of the circle. Furthermore, the magnitude of F(net) must equal mass times the centripetal acceleration!!

F(c) = ma(c)

This equation says that in UCM, the magnitude of the F(net) exerted on the Vespa is equal to the Vespa’s mass times centripetal acceleration. This leads to our equation for centripetal force.

F(c) = mv^2/r

Banking and Lift and Conical Motion!!

The Fn that we feel when we stand in the ground is analogous to the F(l), or the lift force that a plane feels whilst in flight! So let’s look at how an airplane makes a turn and in the process, discuss the concepts of banking and multiple components contributing to the Fc.

An airplane banks (dips its wings) to one side to turn in that direction. By banking the plane, the lift force, a force exerted perpendicular to the direction of flight due to the motion of air over the airplane’s wings, ends up with both a vertical and horizontal component. The horizontal component of this force provides the centripetal acceleration needed to make the plane move around in a circle.

(a) If an airplane of mass m is traveling at speed v and is banked by an angle (theta), derive an expression for the radius of the turn.

(b) Perry the Pilot has a mass of m(p) and a weight Fg = m(p)*g. If the pilot is sitting on a scale as t he plane makes a banked turn, what does the scale read?

After we draw our FBDs similar to our conical motion lab with the pigs or a car rounding a banked turn, we see that, besides Fg, a second force has a vertical component balancing Fg so the airplane doesn’t accelerate up and down and a horizontal component that provides the a(c). In our problem, want to find the radius of the turn and the Fn that the scale exerts on Perry. This is referred to as Perry’s apparent weight or how much she feels she weighs throughout the turn.

Obviously, we write N2L I component form. The (x) component involves the radius of the turn but we don’t know F(l). We find F(l) from the (y) component equation and then we substitute the value of F(l) into the (x) equation and solve for r. Obviously the airplane’s acceleration is directed toward the center of the turn and is therefore a(c).

(x): F(l) sin(theta) = mv^2/r

(y): F(l) cos(theta) = mg

Ultimately (see derivation), we end up with

r = v^2/g tan(theta)

The component form of N2L for Perry is very similar to that for the plane with the F(l) replaced by the normal force, F(n).

Fn = m(p)g/cos(theta)

Now let’s analyze our results! Part (a) says that the faster the plane’s speed for a given bank angle (theta), the larger the radius of the turn. To make a right turn at high speed (large v and small r), the quantity tan(theta) has to be as large as possible. So Perry needs a steep bank to make a tight turn. The greater the bank angle (theta), the greater the value of tan(theta). Again, notice the disappearance of mass from relevance. A bird and a jet observe the same physics no matter their difference in mass!!

Our result in Part (b) shows the danger of too steep a bank angle. As (theta) increases, cos (theta) decreases, so the reading on the scale, Fn, becomes larger and larger. Perry’s apparent weight increases and she feels heavier than normal. The same is true for every part of Perry’s body, including her blood!! If the bank angle is too steep, her heart can’t pump this heavy blood to her brain. As a result, pilots will lose consciousness when their weight is about 4-5 times mg!!

Gravitation!!

Newton’s Law of Gravitation begins to explain the orbits of planets and satellites. The simplest type of orbits to analyze is a circular orbit an many 🌍 satellites, including the International Space Station (ISS) and GPS navigation satellites are in circular or nearly circular orbits.

Isaac Newton knew what was required to put a satellite in orbit three centuries before the first 🌎 satellite was set in orbit. It was his thought experiment that is a ball is thrown at just the right speed, the surface of the Earth would fall away below the ball so that the ball remains the same height above the surface. It is 🌎’s gravitation that causes the ball to accelerate toward 🌎’s center. If the speed is just right, the result is UCM as it keeps falling and missing 🌎!!

We can find the speed v required for circular orbit of radius r from a = v^2/r. For a satellite of mass m orbiting 🌎, the acceleration is provided by 🌎’s gravitational force from N2L!!

F(Earth on satellite) = ma(c)

F(g) = Gm(1)m(2)/r^2 = m(1)v^2/r

Solve for v

v = sqrt(Gm/r)

where G = 6.67 x 10^-11 Nm^2/kg^2

Many low Earth orbiting satellites (LEOS) have a height of a few hundred kilometers which is only a short distance compared to the Earth’s radius of 6370 km. So we can find the speed of LEOS by replacing r with Earth’s radius.

The result is v = 7.91 x 10^3 m/s

(17,700 mph or 28,500 kph)

Form this equation, we see that increasing orbits radius decreases the speed of the circular orbit!! The 🌚 orbits the 🌎 at 3.48 x 10^8 m, about 60 times the Earth’s radius and its orbital speed is 1020 m/s. You can now understand why speed decreases with increasing orbital radius by whirling a ball on the edge of a string and increasing and decreasing the radius. A satellite that orbits close to 🌍 experiences a substantial gravitational pull and moves at a higher speed while a planet with a larger radius of orbit experiences less gravitational force and moves at lower speed!!

Another way to describe how rapidly a satellite moves around its circular orbit is in terms of orbital period, T, which is the time required to complete one orbit. In UCM the speed v is constant so the orbital period, T, is the circumference 2(pi)r of the orbit divided by the speed.

T = 2(pi)r/v

This results in a formula for period by substituting orbital velocity.

T = 2(pi)r(sqrt(r/Gm(Earth))

Conveniently squaring both sides to remove the radical we have a proportional formula!!

T^2 = 4(pi)^2r^3/Gm(Earth)

This equation yields the period for LEOS as T = 5.06 x 10^3 s

84.4 minutes.

In words, the square of the period is directly proportional to the cube of the radius.

From v = sqrt(Gm/r) we also know that orbital speed is inversely proportional to the square root of the radius.

Consider a satellite that orbits at r = 4(r(Earth)) from 🌎’s center. Since r is “4^3” and equals 64 times greater. So, T^2 is also 4 times greater than for an object in LEOS which means that the T = sqrt(64) = 8 times larger!!

The orbital speed is proportional to 1/sqrt(r) which is 1/sqrt(4) = 1/2 as great for a satellite in LEO.

The proportionality tell us that for a satellite with r = 4r(Earth), the orbital period is 8 x 84.4 minutes = 675 minutes (11.2 hours) and the orbital speed is 7910/2 = 3950 m/s!!