{"id":565,"date":"2025-10-09T15:22:16","date_gmt":"2025-10-09T15:22:16","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=565"},"modified":"2025-10-24T13:33:30","modified_gmt":"2025-10-24T13:33:30","slug":"motion-in-space-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/motion-in-space-fresh-take\/","title":{"raw":"Motion in Space: Fresh Take","rendered":"Motion in Space: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Describe how particles move in space using velocity and acceleration vectors<\/li>\r\n \t<li>Explain the tangential and normal components of acceleration<\/li>\r\n \t<li>Use vector-valued functions to describe and analyze planetary motion<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Motion Vectors in the Plane and in Space<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Vector-valued functions describe how objects move through space over time. The position function [latex]{\\bf{r}}(t)[\/latex] and its derivatives give you everything you need to understand motion: where the object is, how fast it's moving, and how its motion is changing.<\/p>\r\n<p class=\"whitespace-normal break-words\">For position function [latex]{\\bf{r}}(t)[\/latex]:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Velocity vector<\/strong>: [latex]{\\bf{v}}(t)={\\bf{r}}'(t)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Acceleration vector<\/strong>: [latex]{\\bf{a}}(t)={\\bf{v}}'(t)={\\bf{r}}''(t)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Speed<\/strong>: [latex]v(t)=\\left\\Vert{\\bf{v}}(t)\\right\\Vert=\\left\\Vert{\\bf{r}}'(t)\\right\\Vert[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Two Dimensions vs. Three Dimensions<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The formulas work the same way in both cases. You just add or remove the [latex]{\\bf{k}[\/latex] component.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>In 2D<\/strong>: [latex]{\\bf{r}}(t)=x(t){\\bf{i}}+y(t){\\bf{j}}[\/latex]<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Velocity: [latex]{\\bf{v}}(t)=x'(t){\\bf{i}}+y'(t){\\bf{j}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Acceleration: [latex]{\\bf{a}}(t)=x''(t){\\bf{i}}+y''(t){\\bf{j}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Speed: [latex]v(t)=\\sqrt{(x'(t))^{2}+(y'(t))^{2}}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>In 3D<\/strong>: [latex]{\\bf{r}}(t)=x(t){\\bf{i}}+y(t){\\bf{j}}+z(t){\\bf{k}}[\/latex]<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Velocity: [latex]{\\bf{v}}(t)=x'(t){\\bf{i}}+y'(t){\\bf{j}}+z'(t){\\bf{k}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Acceleration: [latex]{\\bf{a}}(t)=x''(t){\\bf{i}}+y''(t){\\bf{j}}+z''(t){\\bf{k}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Speed: [latex]v(t)=\\sqrt{(x'(t))^{2}+(y'(t))^{2}+(z'(t))^{2}}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Understanding the Geometry<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">The <strong>velocity vector<\/strong> is always tangent to the path. It points in the direction of motion.<\/li>\r\n \t<li class=\"whitespace-normal break-words\">The <strong>acceleration vector<\/strong> points toward where the object is turning. Think of a car going around a curve\u2014you feel pushed toward the outside, but acceleration points inward.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Speed<\/strong> is the magnitude of velocity. It tells you how fast without indicating direction.<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">Speed can stay constant while velocity changes. Why? Because velocity includes both speed and direction. Even at constant speed, turning changes your velocity vector because your direction changes.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Start with the position function [latex]{\\bf{r}}(t)[\/latex]. Take the first derivative to get velocity. Take the second derivative to get acceleration. Find the magnitude of velocity to get speed.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">A particle moves in a path defined by the vector-valued function [latex]{\\bf{r}}\\,(t)=(t^{2}-3t)\\,{\\bf{i}}+(2t-4)\\,{\\bf{j}}+(t+2)\\,{\\bf{k}}[\/latex], where [latex]t[\/latex] measures time in seconds and where distance is measured in feet. Find the velocity, acceleration, and speed as functions of time.[reveal-answer q=\"fs-id1167794933164\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794933164\"][latex]\\begin{align*}\\hspace{5cm} {\\bf{v}}(t)&amp;={\\bf{r}}'(t)=(2t-3)\\,{\\bf{i}}+2\\,{\\bf{j}}+{\\bf{k}} \\\\{\\bf{a}}(t)&amp;={\\bf{v}}'(t)=2\\,{\\bf{I}} \\\\v(t)&amp;=\\left\\Vert{\\bf{r}}'(t)\\right\\Vert=\\sqrt{(2t-3)^{2}+2^{2}+1^{2}}=\\sqrt{4t^{2}-12t+14} \\end{align*} [\/latex]\r\n\r\n&nbsp;\r\n\r\nThe units for velocity and speed are feet per second, and the units for acceleration are feet per second squared.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7949627&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=frfgju2zMDk&amp;video_target=tpm-plugin-oqnq7v8s-frfgju2zMDk\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.14_transcript.html\">transcript for \u201cCP 3.14\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2>Components of the Acceleration Vector<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The acceleration vector doesn't just tell you how motion is changing\u2014it reveals <em>how<\/em> that change happens. It splits into two perpendicular pieces: one affecting speed, the other affecting direction.<\/p>\r\n<p class=\"whitespace-normal break-words\">The acceleration vector lies in the plane formed by [latex]{\\bf{T}}(t)[\/latex] and [latex]{\\bf{N}}(t)[\/latex]. It can be written as:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{a}}(t)=a_{\\bf{T}}{\\bf{T}}(t)+a_{\\bf{N}}{\\bf{N}}(t)[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">where:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Tangential component<\/strong> [latex]a_{\\bf{T}}=\\frac{{\\bf{v}}\\cdot{\\bf{a}}}{\\left\\Vert{\\bf{v}}\\right\\Vert}[\/latex] measures how quickly speed is changing<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Normal component<\/strong> [latex]a_{\\bf{N}}=\\frac{\\left\\Vert{\\bf{v}}\\times{\\bf{a}}\\right\\Vert}{\\left\\Vert{\\bf{v}}\\right\\Vert}[\/latex] measures how quickly direction is changing<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>What Each Component Means<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]a_{\\bf{T}}[\/latex] tells you if you're speeding up or slowing down. Positive means speeding up. Negative means slowing down.<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]a_{\\bf{N}}[\/latex] tells you how sharply you're turning. Larger values mean tighter curves.<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Alternative Formula for Normal Component<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">If calculating the cross product feels tedious, use:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]a_{\\bf{N}}=\\sqrt{{\\left\\Vert{\\bf{a}}\\right\\Vert}^{2}-a^{2}_{\\bf{T}}}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Understanding Centripetal Acceleration<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The normal component is also called the <strong>centripetal component<\/strong>. Picture driving on a circular track at constant speed. Your speed isn't changing, so [latex]a_{\\bf{T}}=0[\/latex]. But you're constantly turning, so [latex]a_{\\bf{N}} \\neq 0[\/latex]. The acceleration points toward the center of the track. You feel pushed outward\u2014that's your body resisting the inward acceleration.<\/p>\r\n<p class=\"whitespace-normal break-words\">Imagine driving into a curve while braking. Your acceleration vector points partly backward (slowing down) and partly toward the inside of the curve (turning). The tangential component captures the braking. The normal component captures the turning.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Find [latex]{\\bf{v}}(t)[\/latex] and [latex]{\\bf{a}}(t)[\/latex] first. Compute [latex]a_{\\bf{T}}[\/latex] using the dot product formula. Then find [latex]a_{\\bf{N}}[\/latex] using either the cross product or the simpler square root formula.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">An object moves in a path defined by the vector-valued function [latex]{\\bf{r}}\\,(t)=4t\\,{\\bf{i}}+t^{2}\\,{\\bf{j}}[\/latex], where [latex]t[\/latex] measures time in seconds.\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>Find [latex]a_{\\bf{T}}[\/latex] and [latex]a_{\\bf{N}}[\/latex] as functions of [latex]t[\/latex].<\/li>\r\n \t<li>Find [latex]a_{\\bf{T}}[\/latex] and [latex]a_{\\bf{N}}[\/latex] at time [latex]t=-3[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1167794933165\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794933165\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]a_{\\bf{T}}=\\frac{2t}{\\sqrt{t^{2}+4}},\\ a_{\\bf{N}}=\\frac{2}{\\sqrt{t^{2}+4}}[\/latex]<\/li>\r\n \t<li>[latex]a_{\\bf{T}}\\,(-3)=-\\frac{6\\sqrt{13}}{13},\\ a_{\\bf{N}}\\,(-3)=\\frac{2\\sqrt{13}}{13}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7949629&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=lvhavI9RGdo&amp;video_target=tpm-plugin-1p09hgv8-lvhavI9RGdo\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.15_transcript.html\">transcript for \u201cCP 3.15\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2>Projectile Motion<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\">Projectile motion describes objects moving under gravity's influence alone\u2014think baseballs, arrows, or cannonballs. We ignore air resistance. The math becomes straightforward when you recognize that gravity only affects vertical motion, leaving horizontal motion unchanged.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\"><strong>The Setup<\/strong><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\">Place the origin at the launch point. The positive [latex]y[\/latex]-axis points up. The positive [latex]x[\/latex]-axis points forward. Gravity pulls downward with acceleration [latex]g \\approx 32 \\text{ ft\/s}^2[\/latex] (English) or [latex]g \\approx 9.8 \\text{ m\/s}^2[\/latex] (metric).<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\"><strong>The Position Equation<\/strong><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\">Starting with acceleration [latex]{\\bf{a}}(t) = -g{\\bf{j}}[\/latex], integrate twice to get:<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{s}}(t)=-\\frac{1}{2}gt^{2}{\\bf{j}}+{\\bf{v}}{0}t+{\\bf{s}}{0}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\">where [latex]{\\bf{v}}{0}[\/latex] is initial velocity and [latex]{\\bf{s}}{0}[\/latex] is initial position.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\"><strong>Projectile Launched at an Angle<\/strong><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\">When fired at angle [latex]\\theta[\/latex] with initial speed [latex]v_0[\/latex] from the origin:<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{s}}(t)=v_{0}t\\cos\\theta{\\bf{i}}+(v_{0}t\\sin\\theta-\\frac{1}{2}gt^{2}){\\bf{j}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\">The [latex]{\\bf{i}[\/latex] component gives horizontal distance. The [latex]{\\bf{j}[\/latex] component gives altitude.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\"><strong>Key Quantities<\/strong><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Range<\/strong> [latex]R[\/latex]: maximum horizontal distance (at same altitude)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Height<\/strong> [latex]H[\/latex]: maximum altitude<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\"><strong>Finding Maximum Height<\/strong><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\">The projectile reaches maximum height when vertical velocity equals zero. Set [latex]v_y(t) = v_0\\sin\\theta - gt = 0[\/latex] and solve for [latex]t[\/latex]. Substitute this [latex]t[\/latex] into the position equation.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\"><strong>Finding Range<\/strong><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\">For launch and landing at the same altitude:<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]R=\\frac{v^2_0\\sin 2\\theta}{g}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\">Maximum range occurs at [latex]\\theta = 45\u00b0[\/latex], giving [latex]R = \\frac{v^2_0}{g}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section class=\"textbox interact\" aria-label=\"Interact\">Visit this <a href=\"http:\/\/www.openstax.org\/l\/20_projectile\" target=\"_blank\" rel=\"noopener nofollow\">website<\/a> for a video showing projectile motion.<\/section><section class=\"textbox example\" aria-label=\"Example\">An archer fires an arrow at an angle of [latex]40^\\circ[\/latex] above the horizontal with an initial speed of [latex]98\\text{ }m\/sec[\/latex]. The height of the archer is [latex]171.5\\text{ }cm[\/latex]. Find the horizontal distance the arrow travels before it hits the ground.[reveal-answer q=\"fs-id1168794933165\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1168794933165\"]\r\n<div style=\"text-align: center;\">[latex]967.15\\ m[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7949633&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=oFRnQ2QtMeY&amp;video_target=tpm-plugin-05gnio59-oFRnQ2QtMeY\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.16_transcript.html\">transcript for \u201cCP 3.16\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2>Kepler's Laws<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Johannes Kepler used astronomical observations to describe how planets move around the Sun. His three laws apply to any orbiting object\u2014planets, comets, asteroids, even satellites around Earth.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Three Laws<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Law of Ellipses<\/strong>: Every planet follows an elliptical orbit. The Sun sits at one focus of the ellipse.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Law of Equal Areas<\/strong>: A line from the Sun to a planet sweeps out equal areas in equal time intervals. This means planets move faster when closer to the Sun and slower when farther away.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Law of Harmonies<\/strong>: The ratio of orbital periods squared equals the ratio of orbit sizes cubed. For any two planets: [latex]\\frac{T_1^2}{T_2^2} = \\frac{D_1^3}{D_2^3}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Using Kepler's Third Law<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">When measuring distance in astronomical units (A.U.) and time in Earth years, the third law simplifies to:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]T^2_p=D^3_p[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">where [latex]T_p[\/latex] is the planet's orbital period and [latex]D_p[\/latex] is its average distance from the Sun.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Beyond Planets<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Newton later proved these laws using calculus and gravity. He showed the third law works for any orbital system. For a moon orbiting a planet, the formula becomes:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]P^{2}=\\frac{4\\pi^{2}a^{3}}{G(m+M)}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">where [latex]m[\/latex] and [latex]M[\/latex] are the masses, [latex]a[\/latex] is the orbit's semi-major axis, and [latex]G[\/latex] is the gravitational constant.<\/p>\r\n<p class=\"whitespace-normal break-words\">Newton proved planets follow elliptical paths by combining his laws of motion with universal gravitation. Gravity pulls toward the Sun. This force combined with the planet's velocity creates an ellipse. Hyperbolic paths exist too\u2014comets moving too fast to be captured orbit once and escape.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Titan is the largest moon of Saturn. The mass of Titan is approximately [latex]1.35\\,\\times\\,10^{23}\\text{ kg}[\/latex]. The mass of Saturn is approximately [latex]5.68\\,\\times\\,10^{26}\\text{ kg}[\/latex]. Titan takes approximately [latex]16[\/latex] days to orbit Saturn. \u00a0Use this information, along with the universal gravitation constant [latex]G=6.67\\,\\times\\,10^{-11}\\text{ m}^{3}\/\\text{kg}\\cdot{\\text{sec}}^{2}[\/latex] to estimate the distance from Titan to Saturn.[reveal-answer q=\"fs-id1168894933165\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1168894933165\"]\r\n<div style=\"text-align: center;\">[latex]\\large{a=1.224\\,\\times\\,10^{9}\\approx1,224,000\\text{ km}}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7949632&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=Mk28qcuDB6A&amp;video_target=tpm-plugin-k8n0m4oy-Mk28qcuDB6A\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.17_transcript.html\">transcript for \u201cCP 3.17\u201d here (opens in new window).<\/a><\/center><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Describe how particles move in space using velocity and acceleration vectors<\/li>\n<li>Explain the tangential and normal components of acceleration<\/li>\n<li>Use vector-valued functions to describe and analyze planetary motion<\/li>\n<\/ul>\n<\/section>\n<h2>Motion Vectors in the Plane and in Space<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Vector-valued functions describe how objects move through space over time. The position function [latex]{\\bf{r}}(t)[\/latex] and its derivatives give you everything you need to understand motion: where the object is, how fast it&#8217;s moving, and how its motion is changing.<\/p>\n<p class=\"whitespace-normal break-words\">For position function [latex]{\\bf{r}}(t)[\/latex]:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Velocity vector<\/strong>: [latex]{\\bf{v}}(t)={\\bf{r}}'(t)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Acceleration vector<\/strong>: [latex]{\\bf{a}}(t)={\\bf{v}}'(t)={\\bf{r}}''(t)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Speed<\/strong>: [latex]v(t)=\\left\\Vert{\\bf{v}}(t)\\right\\Vert=\\left\\Vert{\\bf{r}}'(t)\\right\\Vert[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Two Dimensions vs. Three Dimensions<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The formulas work the same way in both cases. You just add or remove the [latex]{\\bf{k}[\/latex] component.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>In 2D<\/strong>: [latex]{\\bf{r}}(t)=x(t){\\bf{i}}+y(t){\\bf{j}}[\/latex]<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Velocity: [latex]{\\bf{v}}(t)=x'(t){\\bf{i}}+y'(t){\\bf{j}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Acceleration: [latex]{\\bf{a}}(t)=x''(t){\\bf{i}}+y''(t){\\bf{j}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Speed: [latex]v(t)=\\sqrt{(x'(t))^{2}+(y'(t))^{2}}[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>In 3D<\/strong>: [latex]{\\bf{r}}(t)=x(t){\\bf{i}}+y(t){\\bf{j}}+z(t){\\bf{k}}[\/latex]<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Velocity: [latex]{\\bf{v}}(t)=x'(t){\\bf{i}}+y'(t){\\bf{j}}+z'(t){\\bf{k}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Acceleration: [latex]{\\bf{a}}(t)=x''(t){\\bf{i}}+y''(t){\\bf{j}}+z''(t){\\bf{k}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Speed: [latex]v(t)=\\sqrt{(x'(t))^{2}+(y'(t))^{2}+(z'(t))^{2}}[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Understanding the Geometry<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">The <strong>velocity vector<\/strong> is always tangent to the path. It points in the direction of motion.<\/li>\n<li class=\"whitespace-normal break-words\">The <strong>acceleration vector<\/strong> points toward where the object is turning. Think of a car going around a curve\u2014you feel pushed toward the outside, but acceleration points inward.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Speed<\/strong> is the magnitude of velocity. It tells you how fast without indicating direction.<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Speed can stay constant while velocity changes. Why? Because velocity includes both speed and direction. Even at constant speed, turning changes your velocity vector because your direction changes.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Start with the position function [latex]{\\bf{r}}(t)[\/latex]. Take the first derivative to get velocity. Take the second derivative to get acceleration. Find the magnitude of velocity to get speed.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">A particle moves in a path defined by the vector-valued function [latex]{\\bf{r}}\\,(t)=(t^{2}-3t)\\,{\\bf{i}}+(2t-4)\\,{\\bf{j}}+(t+2)\\,{\\bf{k}}[\/latex], where [latex]t[\/latex] measures time in seconds and where distance is measured in feet. Find the velocity, acceleration, and speed as functions of time.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794933164\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167794933164\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{align*}\\hspace{5cm} {\\bf{v}}(t)&={\\bf{r}}'(t)=(2t-3)\\,{\\bf{i}}+2\\,{\\bf{j}}+{\\bf{k}} \\\\{\\bf{a}}(t)&={\\bf{v}}'(t)=2\\,{\\bf{I}} \\\\v(t)&=\\left\\Vert{\\bf{r}}'(t)\\right\\Vert=\\sqrt{(2t-3)^{2}+2^{2}+1^{2}}=\\sqrt{4t^{2}-12t+14} \\end{align*}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>The units for velocity and speed are feet per second, and the units for acceleration are feet per second squared.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7949627&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=frfgju2zMDk&amp;video_target=tpm-plugin-oqnq7v8s-frfgju2zMDk\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.14_transcript.html\">transcript for \u201cCP 3.14\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2>Components of the Acceleration Vector<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The acceleration vector doesn&#8217;t just tell you how motion is changing\u2014it reveals <em>how<\/em> that change happens. It splits into two perpendicular pieces: one affecting speed, the other affecting direction.<\/p>\n<p class=\"whitespace-normal break-words\">The acceleration vector lies in the plane formed by [latex]{\\bf{T}}(t)[\/latex] and [latex]{\\bf{N}}(t)[\/latex]. It can be written as:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{a}}(t)=a_{\\bf{T}}{\\bf{T}}(t)+a_{\\bf{N}}{\\bf{N}}(t)[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">where:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Tangential component<\/strong> [latex]a_{\\bf{T}}=\\frac{{\\bf{v}}\\cdot{\\bf{a}}}{\\left\\Vert{\\bf{v}}\\right\\Vert}[\/latex] measures how quickly speed is changing<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Normal component<\/strong> [latex]a_{\\bf{N}}=\\frac{\\left\\Vert{\\bf{v}}\\times{\\bf{a}}\\right\\Vert}{\\left\\Vert{\\bf{v}}\\right\\Vert}[\/latex] measures how quickly direction is changing<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>What Each Component Means<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]a_{\\bf{T}}[\/latex] tells you if you&#8217;re speeding up or slowing down. Positive means speeding up. Negative means slowing down.<\/li>\n<li class=\"whitespace-normal break-words\">[latex]a_{\\bf{N}}[\/latex] tells you how sharply you&#8217;re turning. Larger values mean tighter curves.<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Alternative Formula for Normal Component<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">If calculating the cross product feels tedious, use:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]a_{\\bf{N}}=\\sqrt{{\\left\\Vert{\\bf{a}}\\right\\Vert}^{2}-a^{2}_{\\bf{T}}}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Understanding Centripetal Acceleration<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The normal component is also called the <strong>centripetal component<\/strong>. Picture driving on a circular track at constant speed. Your speed isn&#8217;t changing, so [latex]a_{\\bf{T}}=0[\/latex]. But you&#8217;re constantly turning, so [latex]a_{\\bf{N}} \\neq 0[\/latex]. The acceleration points toward the center of the track. You feel pushed outward\u2014that&#8217;s your body resisting the inward acceleration.<\/p>\n<p class=\"whitespace-normal break-words\">Imagine driving into a curve while braking. Your acceleration vector points partly backward (slowing down) and partly toward the inside of the curve (turning). The tangential component captures the braking. The normal component captures the turning.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Find [latex]{\\bf{v}}(t)[\/latex] and [latex]{\\bf{a}}(t)[\/latex] first. Compute [latex]a_{\\bf{T}}[\/latex] using the dot product formula. Then find [latex]a_{\\bf{N}}[\/latex] using either the cross product or the simpler square root formula.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">An object moves in a path defined by the vector-valued function [latex]{\\bf{r}}\\,(t)=4t\\,{\\bf{i}}+t^{2}\\,{\\bf{j}}[\/latex], where [latex]t[\/latex] measures time in seconds.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>Find [latex]a_{\\bf{T}}[\/latex] and [latex]a_{\\bf{N}}[\/latex] as functions of [latex]t[\/latex].<\/li>\n<li>Find [latex]a_{\\bf{T}}[\/latex] and [latex]a_{\\bf{N}}[\/latex] at time [latex]t=-3[\/latex].<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794933165\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167794933165\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]a_{\\bf{T}}=\\frac{2t}{\\sqrt{t^{2}+4}},\\ a_{\\bf{N}}=\\frac{2}{\\sqrt{t^{2}+4}}[\/latex]<\/li>\n<li>[latex]a_{\\bf{T}}\\,(-3)=-\\frac{6\\sqrt{13}}{13},\\ a_{\\bf{N}}\\,(-3)=\\frac{2\\sqrt{13}}{13}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7949629&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=lvhavI9RGdo&amp;video_target=tpm-plugin-1p09hgv8-lvhavI9RGdo\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.15_transcript.html\">transcript for \u201cCP 3.15\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2>Projectile Motion<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\">Projectile motion describes objects moving under gravity&#8217;s influence alone\u2014think baseballs, arrows, or cannonballs. We ignore air resistance. The math becomes straightforward when you recognize that gravity only affects vertical motion, leaving horizontal motion unchanged.<\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\"><strong>The Setup<\/strong><\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\">Place the origin at the launch point. The positive [latex]y[\/latex]-axis points up. The positive [latex]x[\/latex]-axis points forward. Gravity pulls downward with acceleration [latex]g \\approx 32 \\text{ ft\/s}^2[\/latex] (English) or [latex]g \\approx 9.8 \\text{ m\/s}^2[\/latex] (metric).<\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\"><strong>The Position Equation<\/strong><\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\">Starting with acceleration [latex]{\\bf{a}}(t) = -g{\\bf{j}}[\/latex], integrate twice to get:<\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{s}}(t)=-\\frac{1}{2}gt^{2}{\\bf{j}}+{\\bf{v}}{0}t+{\\bf{s}}{0}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\">where [latex]{\\bf{v}}{0}[\/latex] is initial velocity and [latex]{\\bf{s}}{0}[\/latex] is initial position.<\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\"><strong>Projectile Launched at an Angle<\/strong><\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\">When fired at angle [latex]\\theta[\/latex] with initial speed [latex]v_0[\/latex] from the origin:<\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]{\\bf{s}}(t)=v_{0}t\\cos\\theta{\\bf{i}}+(v_{0}t\\sin\\theta-\\frac{1}{2}gt^{2}){\\bf{j}}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\">The [latex]{\\bf{i}[\/latex] component gives horizontal distance. The [latex]{\\bf{j}[\/latex] component gives altitude.<\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\"><strong>Key Quantities<\/strong><\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Range<\/strong> [latex]R[\/latex]: maximum horizontal distance (at same altitude)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Height<\/strong> [latex]H[\/latex]: maximum altitude<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\"><strong>Finding Maximum Height<\/strong><\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\">The projectile reaches maximum height when vertical velocity equals zero. Set [latex]v_y(t) = v_0\\sin\\theta - gt = 0[\/latex] and solve for [latex]t[\/latex]. Substitute this [latex]t[\/latex] into the position equation.<\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\"><strong>Finding Range<\/strong><\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\">For launch and landing at the same altitude:<\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]R=\\frac{v^2_0\\sin 2\\theta}{g}[\/latex]<\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\">Maximum range occurs at [latex]\\theta = 45\u00b0[\/latex], giving [latex]R = \\frac{v^2_0}{g}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<section class=\"textbox interact\" aria-label=\"Interact\">Visit this <a href=\"http:\/\/www.openstax.org\/l\/20_projectile\" target=\"_blank\" rel=\"noopener nofollow\">website<\/a> for a video showing projectile motion.<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">An archer fires an arrow at an angle of [latex]40^\\circ[\/latex] above the horizontal with an initial speed of [latex]98\\text{ }m\/sec[\/latex]. The height of the archer is [latex]171.5\\text{ }cm[\/latex]. Find the horizontal distance the arrow travels before it hits the ground.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1168794933165\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1168794933165\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]967.15\\ m[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7949633&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=oFRnQ2QtMeY&amp;video_target=tpm-plugin-05gnio59-oFRnQ2QtMeY\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.16_transcript.html\">transcript for \u201cCP 3.16\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2>Kepler&#8217;s Laws<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Johannes Kepler used astronomical observations to describe how planets move around the Sun. His three laws apply to any orbiting object\u2014planets, comets, asteroids, even satellites around Earth.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Three Laws<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Law of Ellipses<\/strong>: Every planet follows an elliptical orbit. The Sun sits at one focus of the ellipse.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Law of Equal Areas<\/strong>: A line from the Sun to a planet sweeps out equal areas in equal time intervals. This means planets move faster when closer to the Sun and slower when farther away.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Law of Harmonies<\/strong>: The ratio of orbital periods squared equals the ratio of orbit sizes cubed. For any two planets: [latex]\\frac{T_1^2}{T_2^2} = \\frac{D_1^3}{D_2^3}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Using Kepler&#8217;s Third Law<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When measuring distance in astronomical units (A.U.) and time in Earth years, the third law simplifies to:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]T^2_p=D^3_p[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">where [latex]T_p[\/latex] is the planet&#8217;s orbital period and [latex]D_p[\/latex] is its average distance from the Sun.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Beyond Planets<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Newton later proved these laws using calculus and gravity. He showed the third law works for any orbital system. For a moon orbiting a planet, the formula becomes:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]P^{2}=\\frac{4\\pi^{2}a^{3}}{G(m+M)}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">where [latex]m[\/latex] and [latex]M[\/latex] are the masses, [latex]a[\/latex] is the orbit&#8217;s semi-major axis, and [latex]G[\/latex] is the gravitational constant.<\/p>\n<p class=\"whitespace-normal break-words\">Newton proved planets follow elliptical paths by combining his laws of motion with universal gravitation. Gravity pulls toward the Sun. This force combined with the planet&#8217;s velocity creates an ellipse. Hyperbolic paths exist too\u2014comets moving too fast to be captured orbit once and escape.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Titan is the largest moon of Saturn. The mass of Titan is approximately [latex]1.35\\,\\times\\,10^{23}\\text{ kg}[\/latex]. The mass of Saturn is approximately [latex]5.68\\,\\times\\,10^{26}\\text{ kg}[\/latex]. Titan takes approximately [latex]16[\/latex] days to orbit Saturn. \u00a0Use this information, along with the universal gravitation constant [latex]G=6.67\\,\\times\\,10^{-11}\\text{ m}^{3}\/\\text{kg}\\cdot{\\text{sec}}^{2}[\/latex] to estimate the distance from Titan to Saturn.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1168894933165\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1168894933165\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">[latex]\\large{a=1.224\\,\\times\\,10^{9}\\approx1,224,000\\text{ km}}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7949632&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=Mk28qcuDB6A&amp;video_target=tpm-plugin-k8n0m4oy-Mk28qcuDB6A\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP3.17_transcript.html\">transcript for \u201cCP 3.17\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":21,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-565","chapter","type-chapter","status-publish","hentry"],"part":424,"module-header":"fresh_take","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/565","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/565\/revisions"}],"predecessor-version":[{"id":1756,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/565\/revisions\/1756"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/424"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/565\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=565"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=565"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=565"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=565"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}