This value of [latex]\theta[/latex] is the smallest positive value that makes the derivative equal to zero. Therefore, in the absence of air resistance, the best angle to fire a projectile (to maximize the range) is at a [latex]45^\circ[/latex] angle. The distance it travels is given by

[latex]\large{{\bf{s}}\bigg(\frac{2v_{0}\sin{45}}{g}\bigg)=\frac{v^2_0\sin{90}}{g}\,{\bf{i}}=\frac{v^2_0}{g}\,{\bf{j}}}.[/latex]

Therefore, the range for an angle of [latex]45^\circ[/latex] is [latex]v^2_0/g[/latex].

Kepler’s Laws

During the early 1600s, Johannes Kepler was able to use the amazingly accurate data from his mentor Tycho Braheto formulate his three laws of planetary motion, now known as Kepler’s laws of planetary motion. These laws also apply to other objects in the solar system in orbit around the Sun, such as comets (e.g., Halley’s comet) and asteroids. Variations of these laws apply to satellites in orbit around Earth.

Figure 4. Kepler’s first and second laws are pictured here. The Sun is located at a focus of the elliptical orbit of any planet. Furthermore, the shaded areas are all equal, assuming that the amount of time measured as the planet moves is the same for each region.

Kepler’s third law is especially useful when using appropriate units. In particular, [latex]1\text{ }\textit{astronomical unit}[/latex] is defined to be the average distance from Earth to the Sun, and is now recognized to be [latex]149,597,870,700\text{ }m[/latex] or, approximately [latex]93,000,000[/latex]mi. We therefore write [latex]1\text{ }A.U. = 93,000,000[/latex]mi. Since the time it takes for Earth to orbit the Sun is [latex]1[/latex] year, we use Earth years for units of time. Then, substituting [latex]1[/latex] year for the period of Earth and [latex]1[/latex]A.U. for the average distance to the Sun, Kepler’s third law can be written as

[latex]T^2_p=D^3_p[/latex]

for any planet in the solar system, where [latex]T_p[/latex] is the period of that planet measured in Earth years and [latex]D_p[/latex] is the average distance from that planet to the Sun measured in astronomical units. Therefore, if we know the average distance from a planet to the Sun (in astronomical units), we can then calculate the length of its year (in Earth years), and vice versa.

Kepler’s laws were formulated based on observations from Brahe; however, they were not proved formally until Sir Isaac Newton was able to apply calculus. Furthermore, Newton was able to generalize Kepler’s third law to other orbital systems, such as a moon orbiting around a planet. Kepler’s original third law only applies to objects orbiting the Sun.

Proof

Let’s now prove Kepler’s first law using the calculus of vector-valued functions. First we need a coordinate system. Let’s place the Sun at the origin of the coordinate system and let the vector-valued function [latex]{\bf{r}}\,(t)[/latex] represent the location of a planet as a function of time. Newton proved Kepler’s law using his second law of motion and his law of universal gravitation. Newton’s second law of motion can be written as [latex]{\bf{F}}=m{\bf{a}}[/latex], where [latex]{\bf{F}}[/latex] represents the net force acting on the planet. His law of universal gravitation can be written in the form [latex]{\bf{F}}=-\frac{GmM}{\left\Vert{\bf{r}}\right\Vert^{2}}\cdot\frac{{\bf{r}}}{\left\Vert{\bf{r}}\right\Vert}[/latex], which indicates that the force resulting from the gravitational attraction of the Sun points back toward the Sun, and has magnitude [latex]\frac{GmM}{\left\Vert{\bf{r}}\right\Vert^{2}}[/latex] (Figure 9).

Figure 5. The gravitational force between Earth and the Sun is equal to the mass of the earth times its acceleration.

Setting these two forces equal to each other, and using the fact that [latex]{\bf{a}}\,(t)={\bf{v}}'\,(t)[/latex], we obtain

[latex]\large{m{\bf{v}}'\,(t)=-\frac{GmM}{\left\Vert{\bf{r}}\right\Vert^{2}}\cdot\frac{{\bf{r}}}{\left\Vert{\bf{r}}\right\Vert}}[/latex],

which can be rewritten as

[latex]\large{\frac{d{\bf{v}}}{dt}=-\frac{GM}{\left\Vert{\bf{r}}\right\Vert^{3}}{\bf{r}}}.[/latex]

This equation shows that the vectors [latex]d{\bf{v}}/dt[/latex] and [latex]{\bf{r}}[/latex] are parallel to each other, so [latex]d{\bf{v}}/dt\,\times\,{\bf{r}}=0[/latex]. Next, let’s differentiate [latex]{\bf{r}}\,\times\,{\bf{v}}[/latex] with respect to time:

[latex]\large{\frac{d}{dt}({\bf{r}}\,\times\,{\bf{v}})=\frac{d{\bf{r}}}{dt}\,\times\,{\bf{v}}+{\bf{r}}\,\times\frac{d{\bf{v}}}{dt}={\bf{v}}\,\times\,{\bf{v}}+{\bf{0}}={\bf{0}}}.[/latex]

This proves that [latex]{\bf{r}}\,\times\,{\bf{v}}[/latex] is a constant vector, which we call [latex]{\bf{C}}[/latex]. Since [latex]{\bf{r}}[/latex] and [latex]{\bf{v}}[/latex] are both perpendicular to [latex]{\bf{C}}[/latex] for all values of [latex]t[/latex], they must lie in a plane perpendicular to [latex]{\bf{C}}[/latex]. Therefore, the motion of the planet lies in a plane.

Next we calculate the expression [latex]d{\bf{v}}/dt\,\times\,{\bf{C}}[/latex]:

[latex]\frac{d{\bf{v}}}{dt}\,\times\,{\bf{C}}=-\frac{GM}{\left\Vert{\bf{r}}\right\Vert^{3}}{\bf{r}}\,\times\,({\bf{r}}\,\times\,{\bf{v}})=-\frac{GM}{\left\Vert{\bf{r}}\right\Vert^{3}}[({\bf{r}}\cdot{\bf{v}})\,{\bf{r}}-({\bf{r}}\cdot{\bf{r}})\,{\bf{v}}].[/latex]

The last equality in the above equation is from the triple cross product formula (Introduction to Vectors in Space). We need an expression for [latex]{\bf{r}}\cdot{\bf{v}}[/latex]. To calculate this, we differentiate [latex]{\bf{r}}\cdot{\bf{r}}[/latex] with respect to time:

[latex]\frac{d}{dt}({\bf{r}}\,\cdot\,{\bf{r}})=\frac{d{\bf{r}}}{dt}\cdot{\bf{r}}+{\bf{r}}\cdot\frac{d{\bf{r}}}{dt}=2{\bf{r}}\cdot\frac{d{\bf{r}}}{dt}=2{\bf{r}}\cdot{\bf{v}}.[/latex]

Since [latex]{\bf{r}}\,\cdot\,{\bf{r}}=\left\Vert{\bf{r}}\right\Vert^{2}[/latex], we also have

[latex]\frac{d}{dt}({\bf{r}}\cdot{\bf{r}})=\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert^{2}=2\left\Vert{\bf{r}}\right\Vert\,\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert[/latex].

Combining the last two equations, we get

[latex]\begin{array}{ccc}\hfill {2{\bf{r}}\cdot{\bf{v}}} & =\hfill & {2\left\Vert{\bf{r}}\right\Vert\,\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert} \hfill \\ \hfill {{\bf{r}}\cdot{\bf{v}}}& =\hfill & {\left\Vert{\bf{r}}\right\Vert\,\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert{.}}\hfill \\ \hfill \end{array}[/latex]

Substituting into our expression for [latex]\frac{d{\bf{v}}}{dt}\,\times\,{\bf{C}}[/latex] gives us

[latex]\begin{array}{ccc}\hfill {\frac{d{\bf{v}}}{dt}\,\times\,{\bf{C}}} & =\hfill & {-\frac{GM}{\left\Vert{\bf{r}}\right\Vert^{3}}[({\bf{r}}\cdot{\bf{r}})\,{\bf{r}}-({\bf{r}}\cdot{\bf{r}})\,{\bf{v}}]} \hfill \\ \hfill & =\hfill & {-\frac{GM}{\left\Vert{\bf{r}}\right\Vert^{3}}\big[\left\Vert{\bf{r}}\right\Vert{\big(}\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert\big)\,{\bf{r}}-\left\Vert{\bf{r}}\right\Vert^{2}{\bf{v}}\big]}\hfill \\ \hfill & =\hfill & {-GM\Big[\frac{1}{\left\Vert{\bf{r}}\right\Vert^{2}}\big(\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert\big)\,{\bf{r}}-\frac{1}{\left\Vert{\bf{r}}\right\Vert}{\bf{v}}\Big]}\hfill \\ \hfill & =\hfill & {GM\Big[\frac{\bf{v}}{\left\Vert{\bf{r}}\right\Vert}-\frac{{\bf{r}}}{\left\Vert{\bf{r}}\right\Vert^{2}}\big(\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert\big)\Big].}\hfill \\ \hfill \end{array}[/latex]

However,

[latex]\begin{array}{ccc}\hfill {\frac{d}{dt}\,\frac{{\bf{r}}}{\left\Vert{\bf{r}}\right\Vert}} & =\hfill & {\frac{\frac{d}{dt}({\bf{r}})\left\Vert{\bf{r}}\right\Vert-{\bf{r}}\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert}{\left\Vert{\bf{r}}\right\Vert}} \hfill \\ \hfill & =\hfill & {\frac{\frac{d{\bf{r}}}{dt}}{\left\Vert{\bf{r}}\right\Vert}-\frac{{\bf{r}}}{\left\Vert{\bf{r}}\right\Vert^{2}}\,\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert}\hfill \\ \hfill & =\hfill & {\frac{\bf{v}}{\left\Vert{\bf{r}}\right\Vert}-\frac{{\bf{r}}}{\left\Vert{\bf{v}}\right\Vert^{2}}\,\frac{d}{dt}\left\Vert{\bf{r}}\right\Vert.}\hfill \\ \hfill \end{array}[/latex]

Therefore, we get

[latex]\large{\frac{d{\bf{v}}}{dt}\,\times\,{\bf{C}}=GM\Big(\frac{d}{dt}\,\frac{\bf{r}}{\left\Vert{\bf{r}}\right\Vert}\Big)}.[/latex]

Since [latex]\bf{C}[/latex] is a constant vector, we can integrate both sides and obtain

[latex]{\bf{v}}\,\times\,{\bf{C}}=GM\frac{{\bf{r}}}{\left\Vert{\bf{r}}\right\Vert}+{\bf{D}},[/latex]

where [latex]\bf{D}[/latex] is a constant vector. Our goal is to solve for [latex]\left\Vert{\bf{r}}\right\Vert[/latex]. Let’s start by calculating [latex]{\bf{r}}\cdot({\bf{v}}\,\times\,{\bf{C}})[/latex]:

[latex]{\bf{r}}\cdot({\bf{v}}\,\times\,{\bf{C}})={\bf{r}}\cdot\Big(GM\frac{\bf{r}}{\left\Vert{\bf{r}}\right\Vert}+{\bf{D}}\Big)=GM\frac{\left\Vert{\bf{r}}\right\Vert^{2}}{\left\Vert{\bf{r}}\right\Vert}+{\bf{r}}\cdot{\bf{D}}=GM\left\Vert{\bf{r}}\right\Vert+{\bf{r}}\cdot{\bf{D}}.[/latex]

However, [latex]{\bf{r}}\cdot({\bf{v}}\,\times\,{\bf{C}})=({\bf{r}}\,\times\,{\bf{v}})\,\times\,{\bf{C}}[/latex], so

[latex]({\bf{r}}\,\times\,{\bf{v}})\,\times\,{\bf{C}}=GM\left\Vert{\bf{r}}\right\Vert+{\bf{r}}\cdot{\bf{D}}[/latex].

Since [latex]{\bf{r}}\,\times\,{\bf{v}}={\bf{C}}[/latex], we have

[latex]\left\Vert{\bf{C}}\right\Vert^{2}=GM\left\Vert{\bf{r}}\right\Vert+{\bf{r}}\cdot{\bf{D}}[/latex].

Note that [latex]{\bf{r}}\cdot{\bf{D}}=\left\Vert{\bf{r}}\right\Vert\,\left\Vert{\bf{D}}\right\Vert\cos{\theta}[/latex], where [latex]\theta[/latex] is the angle between [latex]\bf{r}[/latex] and [latex]\bf{D}[/latex]. Therefore,

[latex]\left\Vert{\bf{C}}\right\Vert^{2}=GM\left\Vert{\bf{r}}\right\Vert+\left\Vert{\bf{r}}\right\Vert\,\left\Vert{\bf{D}}\right\Vert\cos{\theta}[/latex].

Solving for [latex]\left\Vert{\bf{r}}\right\Vert[/latex],

[latex]\left\Vert{\bf{r}}\right\Vert=\frac{\left\Vert{\bf{C}}\right\Vert^{2}}{GM+\left\Vert{\bf{D}}\right\Vert\cos{\theta}}=\frac{\left\Vert{\bf{C}}\right\Vert^{2}}{GM}\bigg(\frac{1}{1+e\cos{\theta}}\bigg)[/latex],

where [latex]e=\left\Vert{\bf{D}}\right\Vert/GM[/latex]. This is the polar equation of a conic with a focus at the origin, which we set up to be the Sun. It is a hyperbola if [latex]e\,>\,1[/latex], a parabola if [latex]e=1[/latex], or an ellipse if [latex]e\,<\,1[/latex]. Since planets have closed orbits, the only possibility is an ellipse. However, at this point it should be mentioned that hyperbolic comets do exist. These are objects that are merely passing through the solar system at speeds too great to be trapped into orbit around the Sun. As they pass close enough to the Sun, the gravitational field of the Sun deflects the trajectory enough so the path becomes hyperbolic.

[latex]_\blacksquare[/latex]

Watch the following video to see the worked solution to the above Try It

Activity: Navigating a Banked Turn

How fast can a racecar travel through a circular turn without skidding and hitting the wall? The answer could depend on several factors:

• The weight of the car;
• The friction between the tires and the road;
• The radius of the circle;
• The “steepness” of the turn;

In this project we investigate this question for NASCAR racecars at the Bristol Motor Speedway in Tennessee. Before considering this track in particular, we use vector functions to develop the mathematics and physics necessary for answering questions such as this.

A car of mass [latex]m[/latex] moves with constant angular speed [latex]\omega[/latex] around a circular curve of radius [latex]R[/latex] (Figure 10). The curve is banked at an angle [latex]\theta[/latex]. If the height of the car off the ground is [latex]h[/latex], then the position of the car at time [latex]t[/latex] is given by the function [latex]r(t)=\langle{\bf{R}}\cos{(\omega{t})},\ {\bf{R}}\sin{(\omega{t})},\ h\rangle[/latex].

Figure 6. Views of a race car moving around a track.

1. Find the velocity function [latex]{\bf{v}}\,(t)[/latex] of the car. Show that [latex]{\bf{v}}[/latex] is tangent to the circular curve. This means that, without a force to keep the car on the curve, the car will shoot off of it.
2. Show that the speed of the car is [latex]\omega{R}[/latex]. Use this to show that [latex](2\pi{r})\,/\,{\bf{v}}=(2\pi)\,/\,\omega[/latex].
3. Find the acceleration [latex]{\bf{a}}[/latex]. Show that this vector points toward the center of the circle and that [latex]|{\bf{a}}|=R\omega^{2}[/latex].
4. The force required to produce this circular motion is called the centripetal force, and it is denoted [latex]{\bf{F}}_{\text{cent}}[/latex]. This force points toward the center of the circle (not toward the ground). Show that [latex]|{\bf{F}}_{\text{cent}}|=(m|{\bf{v}}|^{2})\,/\,R[/latex]. As the car moves around the curve, three forces act on it: gravity, the force exerted by the road (this force is perpendicular to the ground), and the friction force (Figure 11). Because describing the frictional force generated by the tires and the road is complex, we use a standard approximation for the frictional force. Assume that [latex]|{\bf{f}}|=\mu|{\bf{N}}|[/latex]for some positive constant [latex]\mu[/latex]. The constant [latex]\mu[/latex] is called the coefficient of friction.
   

   Figure 7. The car has three forces acting on it: gravity (denoted by [latex]m{\bf{g}}[/latex]), the friction force [latex]\bf{f}[/latex], and the force exerted by the road [latex]\bf{N}[/latex].

   Let [latex]v_{\text{max}}[/latex] denote the maximum speed the car can attain through the curve without skidding. In other words, [latex]v_{\text{max}}[/latex] is the fastest speed at which the car can navigate the turn. When the car is traveling at this speed, the magnitude of the centripetal force is

   [latex]|\large{{\bf{F}}_{\text{cent}}|=\frac{mv^{2}_{\text{max}}}{R}}[/latex].

    

   The next three questions deal with developing a formula that relates the speed [latex]v_{\text{max}}[/latex] to the banking angle [latex]\theta[/latex].

5. Show that [latex]|{\bf{N}}|\cos{\theta}=mg+|{\bf{f}}|\sin{\theta}[/latex]. Conclude that [latex]|{bf{N}}|=(mg)\,/\,(\cos{\theta}-\mu\sin{\theta})[/latex].
6. The centripetal force is the sum of the forces in the horizontal direction, since the centripetal force points toward the center of the circular curve. Show that
   [latex]|{\bf{F}}_{\text{cent}}|=|{\bf{N}}|\sin{\theta}+|{\bf{f}}|\cos{\theta}[/latex].

   Conclude that

   [latex]|\large{{\bf{F}}_{\text{cent}}|=\frac{\sin{\theta}+\mu\cos{\theta}}{\cos{\theta}-\mu\sin{\theta}}mg}[/latex].

    

7. Show that [latex]\text{v}^{2}_{\text{max}}=((\sin{\theta}+\mu\cos{\theta}) \ / \ (\cos{\theta}-\mu\sin{\theta}))gR[/latex]. Conclude that the maximum speed does not actually depend on the mass of the car.
   Now that we have a formula relating the maximum speed of the car and the banking angle, we are in a position to answer the questions like the one posed at the beginning of the project.
   The Bristol Motor Speedway is a NASCAR short track in Bristol, Tennessee. The track has the approximate shape shown in Figure 12. Each end of the track is approximately semicircular, so when cars make turns they are traveling along an approximately circular curve. If a car takes the inside track and speeds along the bottom of turn 1, the car travels along a semicircle of radius approximately [latex]211\ ft[/latex] with a banking angle of [latex]24^{\circ}[/latex]. If the car decides to take the outside track and speeds along the top of turn 1, then the car travels along a semicircle with a banking angle of [latex]28^{\circ}[/latex]. (The track has variable angle banking.)
   

   Figure 8. At the Bristol Motor Speedway, Bristol, Tennessee (a), the turns have an inner radius of about 211 ft and a width of 40 ft (b). (credit: part (a) photo by Raniel Diaz, Flickr)

   The coefficient of friction for a normal tire in dry conditions is approximately [latex]0.7[/latex]. Therefore, we assume the coefficient for a NASCAR tire in dry conditions is approximately [latex]0.98[/latex].

   Before answering the following questions, note that it is easier to do computations in terms of feet and seconds, and then convert the answers to miles per hour as a final step.

8. In dry conditions, how fast can the car travel through the bottom of the turn without skidding?
9. In dry conditions, how fast can the car travel through the top of the turn without skidding?
10. In wet conditions, the coefficient of friction can become as low as [latex]0.1[/latex]. If this is the case, how fast can the car travel through the bottom of the turn without skidding?
11. Suppose the measured speed of a car going along the outside edge of the turn is [latex]105\text{ mph}[/latex]. Estimate the coefficient of friction for the car’s tires.