High School Physics

Circular Motion and Gravitation圆周运动与万有引力

When an object moves in a circle at constant speed, it is nonetheless accelerating — the velocity vector changes direction even as its magnitude stays fixed. That centripetal acceleration demands a net force pointed toward the center, and every rotating system from a car rounding a curve to a moon orbiting a planet obeys this logic. This guide builds from uniform circular motion ($a_c = v^2/r$) through vertical circles and banked curves, then ties the inward-force concept to Newton's law of universal gravitation ($F = Gm_1 m_2 / r^2$) and gravitational fields, and finally to satellite orbits and Kepler's three laws. Worked examples use real numbers throughout.当物体以恒定速率做圆周运动（匀速圆周运动）时，它仍在加速——速度矢量方向不断改变，即使大小保持不变。这一向心加速度（向心加速度）要求存在指向圆心的合力（向心力），从汽车转弯到月球绕行星运行，每个旋转系统都遵循这一逻辑。本指南从匀速圆周运动（$a_c = v^2/r$）出发，经竖直圆与倾斜弯道，再把向心力与万有引力定律（万有引力，$F = Gm_1 m_2 / r^2$）及引力场（引力场）相连，最终延伸到卫星轨道（轨道）与开普勒三定律（开普勒定律）。全部例题均用真实数字演算。

7 sections7 节内容 US NGSS · ON · BC · ABUS NGSS · ON · BC · AB Honors block on Kepler's laws and orbits开普勒定律与轨道为荣誉级
Read me first先读这里

How to use this guide如何使用本指南

Circular motion and gravitation appears in every curriculum we map to, but the placement differs. US NGSS treats both the centripetal-force concept (HS-PS2-1, via Newton's second law applied to circular motion) and universal gravitation (HS-PS2-4) as assessed core. Ontario places the circular-dynamics content in SPH4U (Grade 12) Strands B and D, not in the Grade 11 SPH3U course. BC Physics 12 carries both uniform circular motion and gravitational fields as core Content. Alberta Physics 20 splits the topic: circular motion lives in Unit C (GO1) and gravitation in Unit B (GO2 and C GO1). The table below tells you which sections are core for you; each row cites the curriculum source.圆周运动与万有引力出现在我们对照的每套大纲中，但位置各不相同。US NGSS 将向心力概念（HS-PS2-1，通过牛顿第二定律应用于圆周运动）和万有引力（HS-PS2-4）均列为被评估的核心内容。安大略将圆周动力学内容安排在 SPH4U（12 年级）B 单元和 D 单元，而非 11 年级的 SPH3U 课程。BC Physics 12 将匀速圆周运动与引力场均作为核心内容。阿尔伯塔 Physics 20 将该主题拆分：圆周运动在 C 单元（GO1），万有引力在 B 单元（GO2 及 C GO1）。下表告诉你哪些节属于你的核心；每行均注明课纲来源。

If you are in…如果你在… Focus on these sections重点学习 Defer / lighter可推迟 / 减负 Source依据
🇺🇸 US NGSS HS Physical Science美国 NGSS 物理科学 §1–§3 (circular motion, centripetal force) under HS-PS2-1; §4–§5 (universal gravitation and fields) under HS-PS2-4. Both PEs are assessed core.§1–§3（圆周运动、向心力）属 HS-PS2-1；§4–§5（万有引力与引力场）属 HS-PS2-4。两个 PE 均为被评估的核心。 §6–§7 (Kepler's laws, satellite orbits): valuable and closely connected, but NGSS HS-PS2-4 is limited to "systems with two objects" and does not explicitly assess orbital mechanics derivations.§6–§7（开普勒定律、卫星轨道）：有价值且关联密切，但 NGSS HS-PS2-4 限于"两物体系统"，未明确评估轨道力学推导。 ngss_hs_ps_extract.md — HS-PS2-1 and HS-PS2-4 PEs— HS-PS2-1 与 HS-PS2-4 表现期望
🇨🇦 ON Grade 12 — SPH4U安大略 12 年级 — SPH4U All seven sections. SPH4U Strand B covers uniform circular motion; Strand D covers gravitational fields, universal gravitation, and satellite dynamics. The full unit is Grade-12 core.全部 7 节。SPH4U B 单元涵盖匀速圆周运动；D 单元涵盖引力场、万有引力与卫星动力学。全单元为 12 年级核心。 Nothing — this topic is the heart of SPH4U Strands B and D无 — 本主题是 SPH4U B 单元与 D 单元的核心 science_11-12_physics_extract.md — SPH4U Strand B (B2, B3) and Strand D (D2, D3)— SPH4U B 单元（B2、B3）与 D 单元（D2、D3）
🇨🇦 BC Grade 12 — Physics 12BC 12 年级 — Physics 12 All seven sections. Physics 12 Content explicitly lists "uniform circular motion: centripetal force and acceleration; changes to apparent weight" (horizontal and vertical circles) and "gravitational field and Newton's law of universal gravitation" with satellite motion elaboration.全部 7 节。Physics 12 内容明确列出"匀速圆周运动：向心力与向心加速度；表观重力变化"（水平圆与竖直圆）以及"引力场与牛顿万有引力定律"及卫星运动细化。 Nothing — BC Physics 12 pairs circular motion and gravitational fields as a unified Content block无 — BC Physics 12 将圆周运动与引力场配对为统一内容块 physics_11-12_extract.md — Physics 12 Content and Elaborations— Physics 12 内容与细化说明
🇨🇦 AB Grade 11 — Physics 20阿尔伯塔 11 年级 — Physics 20 All seven sections. Physics 20 Unit B GO2 covers universal gravitation; Unit C GO1 covers circular motion, satellite orbits, and Kepler's laws. Both are Grade-11 core.全部 7 节。Physics 20 B 单元 GO2 涵盖万有引力；C 单元 GO1 涵盖圆周运动、卫星轨道与开普勒定律。两者均为 11 年级核心。 Nothing — Alberta treats circular motion and gravitation as a single combined unit in Physics 20无 — 阿尔伯塔在 Physics 20 中将圆周运动与万有引力作为一个合并单元处理 physics_20-30_extract.md — Physics 20 Unit B GO2 (20–B2.1k–6k) and Unit C GO1 (20–C1.1k–7k)— Physics 20 B 单元 GO2（20–B2.1k–6k）与 C 单元 GO1（20–C1.1k–7k）
🇺🇸 AP / IB feeder trackAP / IB 衔接轨道 All seven sections plus every going-deeper derivation. AP Physics 1 / C and IB Physics HL all assume centripetal-force fluency and universal-gravitation derivations from early in the course.全部 7 节，并完成每个"深入"推导。AP Physics 1 / C 与 IB Physics HL 在课程早期就默认你熟练向心力与万有引力推导。 Nothing — this unit is the foundation that AP Physics Unit 5 (rotation) and the gravitation / orbital-mechanics units in AP and IB build on无 — 本单元是 AP Physics Unit 5（转动）以及 AP、IB 引力与轨道力学各单元所依赖的基础 ngss_hs_ps_extract.md — see the feeder links in "What This Feeds Into"— 见"本单元的去向"中的衔接链接

Once you have located your row, use the two cards below for the speed at which you should work through the recommended sections.找到所在行后，用下面两张卡片决定推进速度。

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If you are cramming the night before如果你在临阵磨枪

Memorise four things: $a_c = v^2/r$ (centripetal acceleration, always points inward); $F_c = mv^2/r$ (net force in circular motion, not a separate "centripetal" force but whichever real force points inward); $F = Gm_1 m_2/r^2$ (universal gravitation); and that orbital speed comes from setting $F_g = F_c$. Read every cram-cheat box. Skip the Kepler derivations if you are short on time.背熟四件事：$a_c = v^2/r$（向心加速度，始终指向圆心）；$F_c = mv^2/r$（圆周运动中的合力，并非独立的"向心力"，而是指向圆心的那个真实力）；$F = Gm_1 m_2/r^2$（万有引力）；轨道速度由令 $F_g = F_c$ 求得。读每个速记框，时间不足时跳过开普勒推导。

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If you are going for the top mark如果你目标顶分

Always draw a free-body diagram and identify which real force (tension, gravity, normal force, friction) provides the net centripetal force. In vertical circles the net centripetal force changes with position — set it up freshly at the top and bottom separately. For gravitation, always write $r$ as the orbital radius (centre-to-centre), not altitude above surface. AB Physics 20 expects you to derive orbital speed and period from Newton's law of universal gravitation, and to use Kepler's third law quantitatively.始终画受力图并确定哪个真实力（张力、重力、法向力、摩擦力）提供净向心力。在竖直圆中，净向心力随位置变化——在最高点与最低点分别重新建立方程。处理引力时，始终把 $r$ 写作轨道半径（圆心到圆心），而非距地面高度。AB Physics 20 要求你从万有引力定律推导轨道速度与周期，并定量使用开普勒第三定律。

Honors flag.荣誉级标记。 Sections 6 and 7 (Kepler's Laws and Satellites) carry the Honors chip for the US NGSS track, where HS-PS2-4 Assessment Boundary is "limited to systems with two objects" and does not explicitly require orbital-mechanics derivations. Kepler's laws and satellite orbit calculations are core, not honors, in Ontario SPH4U (Strand D), BC Physics 12 (gravitational dynamics and energy relationships elaboration), and AB Physics 20 (outcomes 20–C1.5k–20–C1.7k). If your row above sends you to §6 and §7, treat them as required content.§6 和 §7（开普勒定律与卫星）在 US NGSS 轨道上标 Honors，因为 HS-PS2-4 的评估边界"限于两物体系统"，并未明确要求轨道力学推导。开普勒定律与卫星轨道计算在安大略 SPH4U（D 单元）、BC Physics 12（引力动力学与能量关系细化）、AB Physics 20（结果 20–C1.5k–20–C1.7k）中均为核心而非荣誉内容。如果你的行指向 §6 和 §7，就把它们视为必学内容。
Section 1 · Uniform circular motion第 1 节 · 匀速圆周运动

Uniform Circular Motion匀速圆周运动

Constant speed does not mean zero acceleration — direction keeps changing.速率不变不等于加速度为零 — 方向一直在变。
  • Uniform circular motion (UCM)匀速圆周运动 — motion in a circle at constant speed $v$. Speed is constant; velocity is not (direction changes).— 以恒定速率 $v$ 做圆周运动。速率不变；速度不恒定（方向改变）。
  • Period $T$ and frequency $f$周期 $T$ 与频率 $f$ $T = 1/f$ — $T$ is the time for one complete revolution; $f$ is the number of revolutions per second (Hz).— $T$ 是完成一圈所需时间；$f$ 是每秒转数（Hz）。
  • Speed from radius and period由半径与周期求速率 — the circumference $2\pi r$ is covered in one period $T$:— 周长 $2\pi r$ 在一个周期 $T$ 内走完：
$$ v = \frac{2\pi r}{T} = 2\pi r f $$ Alberta 20–C1.1k describes UCM as "a special case of two-dimensional motion"; 20–C1.3k requires quantitative relationships among speed, frequency, period and radius.阿尔伯塔 20–C1.1k 把匀速圆周运动描述为"二维运动的一种特例"；20–C1.3k 要求速率、频率、周期与半径之间的定量关系。
Worked Example 1 · Speed of a revolving object例题 1 · 旋转物体的速率

A ball on a string moves in a horizontal circle of radius $0.80$ m, completing $5.0$ revolutions per second. Find (a) the period and (b) the speed of the ball.一只绳上的球在半径 $0.80$ m 的水平圆上运动，每秒完成 $5.0$ 圈。求 (a) 周期与 (b) 球的速率。

(a) Period is the reciprocal of frequency.(a) 周期是频率的倒数。

$$ T = \frac{1}{f} = \frac{1}{5.0} = 0.20 \text{ s}. $$

(b) Speed from circumference divided by period.(b) 速率由周长除以周期得到。

$$ v = \frac{2\pi r}{T} = \frac{2\pi (0.80)}{0.20} = \frac{5.03}{0.20} \approx 25 \text{ m/s}. $$

The ball travels $2\pi(0.80) \approx 5.0$ m per revolution, completing $5.0$ per second, so $v \approx 25\ \mathrm{m/s}$. That is a fast whirl — the speed is real even though the ball keeps returning to the same spot.球每圈走 $2\pi(0.80) \approx 5.0$ m，每秒 $5.0$ 圈，故 $v \approx 25\ \mathrm{m/s}$。转得很快——速率是真实的，即便球不断回到同一位置。

An object moves in a circle at constant speed. Which statement is correct?一物体以恒定速率做圆周运动。哪句正确？
§1 · Q1
Its velocity is constant because its speed is constant速度恒定，因为速率恒定
Its acceleration is zero because its speed is constant加速度为零，因为速率恒定
Its velocity changes direction even though its speed is constant速度方向不断改变，尽管速率恒定
There is no net force because the speed is not changing无合力，因为速率不变
Velocity is a vector: both magnitude (speed) and direction matter. In UCM the speed is constant but the direction changes continuously, so the velocity changes and there is a non-zero acceleration.速度是矢量：大小（速率）与方向都重要。在匀速圆周运动中速率恒定，但方向不断改变，故速度变化，加速度不为零。
Constant speed does not mean constant velocity — direction must also be constant. In a circle the direction changes every instant, so velocity changes and acceleration is non-zero.速率不变不等于速度不变——方向也必须不变。在圆周运动中方向每时每刻都在变，故速度变化，加速度不为零。
A car completes one lap of a circular track of radius $50$ m in $31.4$ s. What is its speed? (Use $\pi \approx 3.14$.)一辆车在半径 $50$ m 的圆形跑道上用 $31.4$ s 完成一圈。速率是多少？（取 $\pi \approx 3.14$。）
§1 · Q2
$50\ \mathrm{m/s}$
$10\ \mathrm{m/s}$
$1.6\ \mathrm{m/s}$
$314\ \mathrm{m/s}$
$v = 2\pi r / T = 2(3.14)(50) / 31.4 = 314 / 31.4 = 10\ \mathrm{m/s}$.$v = 2\pi r / T = 2(3.14)(50) / 31.4 = 314 / 31.4 = 10\ \mathrm{m/s}$。
Circumference $= 2\pi r = 2(3.14)(50) = 314$ m. Speed $= 314\ \mathrm{m} / 31.4\ \mathrm{s} = 10\ \mathrm{m/s}$.周长 $= 2\pi r = 2(3.14)(50) = 314$ m。速率 $= 314\ \mathrm{m} / 31.4\ \mathrm{s} = 10\ \mathrm{m/s}$。
Section 2 · Centripetal acceleration & force第 2 节 · 向心加速度与向心力

Centripetal Acceleration and Centripetal Force向心加速度与向心力

The inward acceleration and the real force that provides it.指向圆心的加速度，以及提供它的真实力。 $$ a_c = \frac{v^2}{r} = \frac{4\pi^2 r}{T^2} \qquad \text{(centripetal acceleration, directed inward)}$$ $$ F_c = ma_c = \frac{mv^2}{r} \qquad \text{(net centripetal force, directed inward)}$$
  • Direction.方向。 Both $a_c$ and the net force always point toward the center — perpendicular to the velocity, never tangent to the circle.$a_c$ 与合力始终指向圆心——垂直于速度方向，绝非沿切线。
  • "Centripetal force" is not a new force."向心力"不是新力。 It is whatever real force (tension, gravity, normal force, friction) happens to point inward. Identify that real force on the free-body diagram, then set it equal to $mv^2/r$.它就是恰好指向圆心的那个真实力（张力、重力、法向力、摩擦力）。在受力图上找到该真实力，再令其等于 $mv^2/r$。
  • Doubling $v$ at fixed $r$.固定 $r$ 时速度加倍。 Since $a_c \propto v^2$, doubling the speed requires four times the centripetal force. Doubling $r$ at fixed $v$ halves $a_c$.因为 $a_c \propto v^2$，速度加倍需要四倍向心力。固定 $v$ 时 $r$ 加倍使 $a_c$ 减半。
AB 20–C1.2k: "explain, qualitatively and quantitatively, that the acceleration in uniform circular motion is directed toward the centre of a circle." NGSS HS-PS2-1 ties the centripetal case to Newton's second law (net force = $ma$).AB 20–C1.2k："定性与定量地解释匀速圆周运动中的加速度指向圆心。"NGSS HS-PS2-1 将向心力情形与牛顿第二定律（合力 $= ma$）相联系。
Worked Example 2 · Ball on a string例题 2 · 绳上的球

A $0.50$ kg ball moves in a horizontal circle of radius $1.2$ m at $6.0$ m/s. Find (a) the centripetal acceleration and (b) the tension in the string providing it.一只 $0.50$ kg 的球以 $6.0$ m/s 在半径 $1.2$ m 的水平圆上运动。求 (a) 向心加速度与 (b) 提供它的绳中张力。

(a) Centripetal acceleration.(a) 向心加速度。

$$ a_c = \frac{v^2}{r} = \frac{(6.0)^2}{1.2} = \frac{36}{1.2} = 30 \text{ m/s}^2. $$

(b) The tension is the only inward force (ignore gravity for a horizontal circle on a horizontal surface).(b) 张力是唯一的向内力（水平面上的水平圆，忽略重力分量）。

$$ F_c = ma_c = (0.50)(30) = 15 \text{ N}. $$

The string must exert $15$ N inward to keep the ball on the circle. If the string breaks (tension falls to zero) the ball flies off in a straight line tangent to the circle — Newton's first law taking over.绳必须施加 $15$ N 向内的力使球保持在圆上。若绳断裂（张力降为零），球将沿圆的切线方向飞出——牛顿第一定律接管。

A $2.0$ kg object moves in a circle of radius $0.5$ m at $4.0$ m/s. What net force is needed?一个 $2.0$ kg 的物体以 $4.0$ m/s 在半径 $0.5$ m 的圆上运动。需要多大的合力？
§2 · Q1
$8$ N
$16$ N
$4$ N
$64$ N
$F_c = mv^2/r = (2.0)(4.0)^2 / 0.5 = (2.0)(16) / 0.5 = 32/0.5 = 64$ N.$F_c = mv^2/r = (2.0)(4.0)^2 / 0.5 = (2.0)(16) / 0.5 = 32/0.5 = 64$ N。
Use $F_c = mv^2/r$. Square the speed first: $v^2 = 16\ \mathrm{m^2/s^2}$.用 $F_c = mv^2/r$。先把速度平方：$v^2 = 16\ \mathrm{m^2/s^2}$。
A car rounds a flat circular curve. What provides the centripetal force?一辆车在平坦圆弯道上行驶。什么提供向心力？
§2 · Q2
Static friction between the tires and road轮胎与路面之间的静摩擦力
A new "centripetal" force created by the circular motion圆周运动产生的新的"向心力"
The normal force from the road路面对车的法向力
The engine thrust pointing inward发动机推力指向圆心
On a flat curve the normal force is vertical (balancing gravity), and engine thrust is forward (tangential). The only force with a component pointing toward the center of the curve is static friction between the tires and the road surface.在平坦弯道上，法向力竖直（平衡重力），发动机推力向前（沿切线）。唯一有分量指向弯道圆心的力是轮胎与路面之间的静摩擦力。
"Centripetal force" is not a new separate force; it is always an already-existing force redirected inward. On a flat curve, friction is the inward force."向心力"不是新的独立力；它始终是某个已存在的力被引向圆心。在平坦弯道上，摩擦力是向内的力。
Section 3 · Vertical circles & banked curves第 3 节 · 竖直圆与倾斜弯道

Vertical Circles and Banked Curves竖直圆与倾斜弯道

The net inward force changes with position in a vertical circle.在竖直圆中，净向内力随位置变化。
  • At the top of a vertical circle竖直圆最高点 (object on the inside of the loop): gravity and tension both point inward (toward center). $T + mg = mv^2/r$. Minimum speed occurs when $T = 0$: $v_{\min} = \sqrt{gr}$.（物体在圆圈内侧）：重力与张力都指向圆心。$T + mg = mv^2/r$。最小速度在 $T = 0$ 时出现：$v_{\min} = \sqrt{gr}$。
  • At the bottom of a vertical circle竖直圆最低点 : normal force or tension points up (inward); gravity points down (outward). $N - mg = mv^2/r$, so $N = mg + mv^2/r > mg$ — you feel heavier.：法向力或张力向上（向内）；重力向下（向外）。$N - mg = mv^2/r$，故 $N = mg + mv^2/r > mg$——感觉更重。
  • Banked curve (ideal banking, no friction)倾斜弯道（理想坡度，无摩擦）: the horizontal component of the normal force provides centripetal force. $\tan\theta = v^2/(rg)$.：法向力的水平分量提供向心力。$\tan\theta = v^2/(rg)$。
BC Physics 12 Content elaboration: "uniform circular motion: both horizontal and vertical circles; changes to apparent weight." AB 20–C1.4k: "explain, qualitatively, uniform circular motion in terms of Newton's laws of motion."BC Physics 12 内容细化："匀速圆周运动：水平圆与竖直圆；表观重力的变化。"AB 20–C1.4k："用牛顿运动定律定性解释匀速圆周运动。"
Worked Example 3 · Roller-coaster loop例题 3 · 过山车竖直圆

A roller-coaster car of mass $600$ kg moves through a vertical loop of radius $12$ m. Using $g = 9.8\ \mathrm{m/s^2}$, find (a) the minimum speed at the top of the loop and (b) the normal force on the riders at the bottom if the speed there is $18$ m/s.一辆质量 $600$ kg 的过山车通过半径 $12$ m 的竖直圆圈。取 $g = 9.8\ \mathrm{m/s^2}$，求 (a) 圆圈顶部的最小速度与 (b) 若底部速度为 $18$ m/s，乘客在底部所受法向力。

(a) Minimum speed at the top: set $N = 0$.(a) 顶部最小速度：令 $N = 0$。 At the top, gravity alone provides centripetal force when $N = 0$:在顶部，当 $N = 0$ 时重力单独提供向心力：

$$ mg = \frac{mv_{\min}^2}{r} \;\Longrightarrow\; v_{\min} = \sqrt{gr} = \sqrt{9.8 \times 12} = \sqrt{117.6} \approx 10.8 \text{ m/s}. $$

(b) Normal force at the bottom: $N$ and $mg$ oppose each other.(b) 底部法向力：$N$ 与 $mg$ 方向相反。 The net upward force provides centripetal acceleration toward the center (upward at the bottom):净向上力提供指向圆心（底部时向上）的向心加速度：

$$ N - mg = \frac{mv^2}{r} \;\Longrightarrow\; N = mg + \frac{mv^2}{r} = 600(9.8) + \frac{600(18)^2}{12} $$ $$ N = 5880 + \frac{600 \times 324}{12} = 5880 + 16200 = 22080 \approx 22 \text{ kN}. $$

This is $22080 / 5880 \approx 3.8$ times the riders' weight — a classic "apparent weight" increase at the bottom of a loop.这是乘客体重的 $22080 / 5880 \approx 3.8$ 倍——圆圈底部经典的"表观重力"增大现象。

At the top of a vertical loop, if the speed is exactly the minimum ($N = 0$), what provides the centripetal force?在竖直圆最高点，若速度恰好为最小值（$N = 0$），什么提供向心力？
§3 · Q1
The normal force alone仅法向力
Gravity alone仅重力
Tension alone仅张力
No force is needed at the top最高点不需要力
When $N = 0$ (the minimum-speed condition), the track exerts no normal force. Only gravity acts on the car, directed downward toward the center. So gravity alone is the centripetal force: $mg = mv^2/r$.当 $N = 0$（最小速度条件）时，轨道不施加法向力。只有重力作用在车上，向下指向圆心。故重力单独作为向心力：$mg = mv^2/r$。
At minimum speed $N = 0$ by definition. Gravity (always downward) points toward the center at the top of the loop, so gravity alone is the centripetal force.最小速度时 $N = 0$（定义如此）。重力（始终向下）在圆圈顶部指向圆心，故重力单独作为向心力。
A car rounds the bottom of a dip in the road at speed $v$. Compared with a stationary reading, the normal force from the seat on the driver is:一辆车以速度 $v$ 过道路最低点。与静止时相比，座椅对驾驶员的法向力：
§3 · Q2
Less than $mg$小于 $mg$
Equal to $mg$等于 $mg$
Greater than $mg$大于 $mg$
Zero为零
At the bottom, $N - mg = mv^2/r > 0$, so $N = mg + mv^2/r > mg$. The seat pushes the driver up more than gravity pulls them down — the driver feels heavier than normal.在最低点，$N - mg = mv^2/r > 0$，故 $N = mg + mv^2/r > mg$。座椅向上推驾驶员的力大于重力向下拉的力——驾驶员感觉比平时更重。
At the bottom of a curve the net upward force must supply centripetal acceleration (inward = upward here). So $N > mg$, giving a sensation of increased weight.在弯道最低点，净向上力必须提供向心加速度（此处向内即向上）。故 $N > mg$，产生体重增加的感觉。
Section 4 · Newton's law of universal gravitation第 4 节 · 万有引力定律

Newton's Law of Universal Gravitation牛顿万有引力定律

Every mass attracts every other mass across any distance.任何质量在任何距离都吸引其他质量。 $$ F = G\frac{m_1 m_2}{r^2} $$
  • $G = 6.674 \times 10^{-11}\ \mathrm{N\cdot m^2\,kg^{-2}}$$G = 6.674 \times 10^{-11}\ \mathrm{N\cdot m^2\,kg^{-2}}$ — the universal gravitational constant, measured by Cavendish (1798).— 万有引力常量，由卡文迪什（1798）测定。
  • $r$ is the centre-to-centre distance$r$ 是质心到质心的距离 — NOT the altitude above the surface. When an object orbits Earth, $r = R_E + h$ where $h$ is the altitude.— 而非距地表高度。物体绕地球轨道时，$r = R_E + h$，$h$ 为轨道高度。
  • Inverse-square law.平方反比律。 Double the distance and the gravitational force drops to $\tfrac{1}{4}$. The force on each body is equal in magnitude and opposite in direction (Newton's third law).距离加倍，引力降至 $\tfrac{1}{4}$。两个天体上的力大小相等、方向相反（牛顿第三定律）。
  • Connecting to $g$.与 $g$ 相联系。 Near Earth's surface: $g = GM_E / R_E^2$. At altitude $h$: $g_h = GM_E/(R_E + h)^2$.近地表：$g = GM_E / R_E^2$。高度 $h$ 处：$g_h = GM_E/(R_E + h)^2$。
NGSS HS-PS2-4: "Use mathematical representations of Newton's Law of Gravitation … to describe and predict the gravitational … forces between objects." AB 20–B2.2k: "describe, qualitatively and quantitatively, Newton's law of universal gravitation."NGSS HS-PS2-4："用万有引力定律的数学表达式描述并预测物体间的引力。"AB 20–B2.2k："定性与定量地描述牛顿万有引力定律。"
Worked Example 4 · Force between Earth and the Moon例题 4 · 地球与月球间的引力

The Earth ($M_E = 5.97 \times 10^{24}$ kg) and the Moon ($M_M = 7.34 \times 10^{22}$ kg) are separated by $r = 3.84 \times 10^8$ m (centre-to-centre). Use $G = 6.674 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$. Find the gravitational force between them.地球（$M_E = 5.97 \times 10^{24}$ kg）与月球（$M_M = 7.34 \times 10^{22}$ kg）质心间距 $r = 3.84 \times 10^8$ m。取 $G = 6.674 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$。求两者间的引力。

$$ F = G\frac{M_E M_M}{r^2} = \frac{(6.674 \times 10^{-11})(5.97 \times 10^{24})(7.34 \times 10^{22})}{(3.84 \times 10^8)^2} $$

Numerator.分子。 $6.674 \times 5.97 \times 7.34 \approx 292.3$, and the powers of ten give十次幂为 $10^{-11+24+22} = 10^{35}$. So numerator故分子 $\approx 2.923 \times 10^{37}\ \mathrm{N\,m^2}$.

Denominator.分母。 $(3.84)^2 = 14.75$, times乘以 $10^{16}$: denominator分母 $\approx 1.475 \times 10^{17}\ \mathrm{m^2}$.

$$ F \approx \frac{2.923 \times 10^{37}}{1.475 \times 10^{17}} \approx 1.98 \times 10^{20} \text{ N}. $$

This mutual force ($\approx 2 \times 10^{20}$ N) keeps the Moon in its roughly circular orbit. By Newton's third law the Moon pulls on Earth with the same magnitude — causing the ocean tides.这一相互引力（$\approx 2 \times 10^{20}$ N）使月球保持在近似圆形的轨道上。由牛顿第三定律，月球以同等大小的力拉地球——引发海洋潮汐。

The gravitational force between two objects is $F$. If the distance between them is doubled (masses unchanged), what is the new force?两物体间的引力为 $F$。若它们之间的距离加倍（质量不变），新引力是多少？
§4 · Q1
$2F$
$F/2$
$F/4$
$4F$
$F \propto 1/r^2$. Doubling $r$ means $r^2$ becomes $4r^2$, so $F$ becomes $F/4$. The inverse-square relationship is the defining feature of gravity.$F \propto 1/r^2$。$r$ 加倍意味着 $r^2$ 变为 $4r^2$，故 $F$ 变为 $F/4$。平方反比关系是引力的决定性特征。
Gravity follows an inverse-square law: $F \propto 1/r^2$. Double the distance, divide the force by $2^2 = 4$.引力遵循平方反比律：$F \propto 1/r^2$。距离加倍，力除以 $2^2 = 4$。
An object of mass $m$ is at Earth's surface (radius $R$). It is moved to an altitude $R$ above the surface. By what factor does the gravitational force on it change?质量为 $m$ 的物体在地球表面（半径 $R$）。将其移至距地表 $R$ 的高度。它所受引力变为原来的几倍？
§4 · Q2
Decreases by factor of $2$减为原来的 $1/2$
Decreases by factor of $4$减为原来的 $1/4$
Decreases by factor of $\sqrt{2}$减为原来的 $1/\sqrt{2}$
Stays the same不变
At the surface $r = R$; at altitude $R$ the centre-to-centre distance is $R + R = 2R$. Force $\propto 1/r^2$: ratio is $R^2/(2R)^2 = 1/4$. The force becomes $F/4$.在地表 $r = R$；高度 $R$ 处质心距为 $R + R = 2R$。力 $\propto 1/r^2$：比值为 $R^2/(2R)^2 = 1/4$。力变为 $F/4$。
The centre-to-centre distance at altitude $R$ is $2R$ (not $R$). Doubling $r$ gives $F/4$ by the inverse-square law.高度 $R$ 处质心距为 $2R$（而非 $R$）。$r$ 加倍则由平方反比律得 $F/4$。
Going deeper — connecting $G$ and $g$ (AB 20–B2.5k)深入 — 将 $G$ 与 $g$ 相联系（AB 20–B2.5k）

At Earth's surface, the gravitational force on a mass $m$ equals $mg$ by definition of $g$, and also equals $GmM_E/R_E^2$ by universal gravitation. Setting them equal:在地表，质量 $m$ 所受引力按 $g$ 的定义等于 $mg$，同时按万有引力等于 $GmM_E/R_E^2$。令两者相等：

$$ mg = \frac{GmM_E}{R_E^2} \;\Longrightarrow\; g = \frac{GM_E}{R_E^2}. $$

This shows that $g$ is not a mysterious constant but rather a consequence of Newton's law and Earth's mass and radius. Plugging in $G = 6.674 \times 10^{-11}$, $M_E = 5.97 \times 10^{24}$ kg, $R_E = 6.37 \times 10^6$ m gives $g \approx 9.81\ \mathrm{m/s^2}$, recovering the familiar value. At a different planet, $g_{\mathrm{planet}} = GM_{\mathrm{planet}}/R_{\mathrm{planet}}^2$. Alberta 20–B2.5k requires you to "relate, qualitatively and quantitatively, using Newton's law of universal gravitation, the gravitational constant to the local value of the acceleration due to gravity."这表明 $g$ 并非神秘常数，而是牛顿定律与地球质量、半径的必然结果。代入 $G = 6.674 \times 10^{-11}$、$M_E = 5.97 \times 10^{24}$ kg、$R_E = 6.37 \times 10^6$ m 得 $g \approx 9.81\ \mathrm{m/s^2}$，恢复了熟悉的值。在其他行星上，$g_{\mathrm{planet}} = GM_{\mathrm{planet}}/R_{\mathrm{planet}}^2$。阿尔伯塔 20–B2.5k 要求你"用牛顿万有引力定律定性与定量地将引力常量与当地重力加速度联系起来"。

Section 5 · Gravitational fields第 5 节 · 引力场

Gravitational Fields引力场

A field replaces "action at a distance" with a property of space.场用"空间属性"取代"超距作用"。 $$ \vec{g} = \frac{\vec{F}}{m} = -\frac{GM}{r^2}\hat{r} \qquad (\mathrm{N/kg} = \mathrm{m/s^2},\ \text{directed toward the mass}) $$
  • Gravitational field strength $g$引力场强 $g$ at a point is the force per unit mass that a small test mass would experience there. Units: N/kg (same as m/s²).在某点处是单位质量的小试探质量在该点所受的力。单位：N/kg（与 m/s² 相同）。
  • Field lines场线 point inward toward the mass; more closely spaced lines mean stronger field. For a sphere the field outside is identical to a point mass at the center.指向质量；场线越密场越强。对球体，外部引力场与位于球心的点质量产生的场相同。
  • NGSS HS-PS2-4 ClarificationNGSS HS-PS2-4 澄清说明 : "Emphasis is on both quantitative and conceptual descriptions of gravitational and electric fields." AB 20–B2.4k: "define the term 'field' as a concept that replaces 'action at a distance' and apply the concept to describe gravitational effects."："强调对引力场与电场的定量与概念性描述。"AB 20–B2.4k："将'场'定义为取代'超距作用'的概念，并将其应用于描述引力效应。"
Worked Example 5 · Gravitational field at altitude例题 5 · 高空引力场强

Find the gravitational field strength at an altitude of $400$ km above Earth's surface. Use $G = 6.674 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$, $M_E = 5.97 \times 10^{24}$ kg, $R_E = 6.37 \times 10^6$ m.求距地表 $400$ km 高度处的引力场强。取 $G = 6.674 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$，$M_E = 5.97 \times 10^{24}$ kg，$R_E = 6.37 \times 10^6$ m。

Centre-to-centre distance.质心到质心的距离。 $r = R_E + h = (6.37 + 0.40) \times 10^6 = 6.77 \times 10^6$ m.$r = R_E + h = (6.37 + 0.40) \times 10^6 = 6.77 \times 10^6$ m。

$$ g_h = \frac{GM_E}{r^2} = \frac{(6.674 \times 10^{-11})(5.97 \times 10^{24})}{(6.77 \times 10^6)^2} = \frac{3.984 \times 10^{14}}{4.583 \times 10^{13}} \approx 8.69\ \mathrm{m/s^2}. $$

At $400$ km altitude (the orbit of the ISS) gravity is about $89\%$ of its surface value — far from negligible. This is why the ISS crew are in free fall, not "weightless" in the sense of having no gravity.在 $400$ km 高度（国际空间站轨道处）引力约为地表值的 $89\%$——远非可以忽略。这就是为何国际空间站宇航员处于自由落体状态，而非"无重力"。

The gravitational field at a point is $5.0\ \mathrm{N/kg}$. What force does a $3.0$ kg mass experience there?某点的引力场强为 $5.0\ \mathrm{N/kg}$。一个 $3.0$ kg 的质量在该点受到多大的力？
§5 · Q1
$5.0$ N
$15$ N
$1.7$ N
$8.0$ N
$F = mg = (3.0)(5.0) = 15$ N. The field strength is force per unit mass, so multiply by the mass.$F = mg = (3.0)(5.0) = 15$ N。场强是单位质量所受的力，故乘以质量。
Field strength $g$ = force / mass, so force = $g \times m = 5.0 \times 3.0 = 15$ N.场强 $g$ = 力 / 质量，故力 $= g \times m = 5.0 \times 3.0 = 15$ N。
As you move farther from a planet, what happens to the gravitational field strength?随着你远离行星，引力场强如何变化？
§5 · Q2
It decreases, following an inverse-square relationship按平方反比关系减小
It decreases linearly with distance与距离成正比地减小
It stays constant because gravity extends to infinity保持不变，因为引力延伸至无穷远
It increases because the field lines spread out增大，因为场线扩散
$g = GM/r^2$, so $g \propto 1/r^2$. Double the distance and the field strength drops to one-quarter. This is the inverse-square law in action.$g = GM/r^2$，故 $g \propto 1/r^2$。距离加倍，场强降至四分之一。这就是平方反比律的体现。
$g = GM/r^2$ follows an inverse-square law, not a linear decrease. Gravity extends to infinity but weakens rapidly with distance.$g = GM/r^2$ 遵循平方反比律，而非线性减小。引力延伸至无穷远，但随距离迅速减弱。
Section 6 · Orbits & Kepler's laws Honors — US NGSS第 6 节 · 轨道与开普勒定律 荣誉 — US NGSS

Orbits and Kepler's Laws Honors — US NGSS轨道与开普勒定律 荣誉 — US NGSS

Curriculum note.课纲提示。 Kepler's laws and orbital mechanics are core content in Ontario SPH4U (Strand D), BC Physics 12 ("gravitational dynamics and energy relationships" elaboration: "satellite motion, orbit changes"), and AB Physics 20 (outcomes 20–C1.5k through 20–C1.7k). The Honors chip applies only to the US NGSS track, where HS-PS2-4's Assessment Boundary is "limited to systems with two objects" and explicit orbital-mechanics derivations are not required.开普勒定律与轨道力学在安大略 SPH4U（D 单元）、BC Physics 12（"引力动力学与能量关系"细化："卫星运动、轨道变化"）以及 AB Physics 20（结果 20–C1.5k 至 20–C1.7k）中均为核心内容。Honors 标记仅适用于 US NGSS 轨道，其 HS-PS2-4 评估边界"限于两物体系统"，不明确要求轨道力学推导。
Kepler's three laws, briefly.开普勒三定律，简述。
  • Law 1 (Ellipses)第一定律（椭圆轨道） — planets orbit the Sun in ellipses with the Sun at one focus. For circular orbits (a special case) the "ellipse" degenerates to a circle.— 行星以太阳为一个焦点在椭圆轨道上运行。对圆形轨道（特殊情形），"椭圆"退化为圆。
  • Law 2 (Equal Areas)第二定律（等面积） — the line from the Sun to the planet sweeps equal areas in equal times. A planet moves fastest when closest (perihelion) and slowest when farthest (aphelion).— 行星到太阳的连线在相等时间内扫过相等面积。近日点最快，远日点最慢。
  • Law 3 (Periods)第三定律（周期） — $T^2 \propto r^3$, specifically $T^2 / r^3 = 4\pi^2 / (GM)$ (constant for all objects orbiting the same central mass $M$).— $T^2 \propto r^3$，具体为 $T^2 / r^3 = 4\pi^2 / (GM)$（对绕同一中心质量 $M$ 运行的所有天体为常数）。
$$ \frac{T^2}{r^3} = \frac{4\pi^2}{GM} \qquad \text{(Kepler's Third Law)} $$ AB 20–C1.7k: "explain, qualitatively, how Kepler's laws were used in the development of Newton's law of universal gravitation." BC Physics 12: "gravitational dynamics and energy relationships — satellite motion, orbit changes, launch velocity, escape velocity."AB 20–C1.7k："定性解释开普勒定律如何被用于发展牛顿万有引力定律。"BC Physics 12："引力动力学与能量关系——卫星运动、轨道变化、发射速度、逃逸速度。"
Worked Example 6 · Orbital period from Kepler's Third Law例题 6 · 由开普勒第三定律求轨道周期

Mars orbits the Sun at an average radius of $r_M = 2.28 \times 10^{11}$ m. Earth's average orbital radius is $r_E = 1.50 \times 10^{11}$ m and its period is $T_E = 365$ days. Use Kepler's Third Law to find Mars's orbital period.火星以平均轨道半径 $r_M = 2.28 \times 10^{11}$ m 绕太阳运行。地球平均轨道半径 $r_E = 1.50 \times 10^{11}$ m，周期 $T_E = 365$ 天。用开普勒第三定律求火星的轨道周期。

Both orbit the same Sun, so $T^2/r^3$ is the same constant for both.两者绕同一太阳运行，故 $T^2/r^3$ 对两者为同一常数。

$$ \frac{T_M^2}{r_M^3} = \frac{T_E^2}{r_E^3} \;\Longrightarrow\; T_M^2 = T_E^2 \left(\frac{r_M}{r_E}\right)^3 = 365^2 \times \left(\frac{2.28}{1.50}\right)^3. $$ $$ \left(\frac{2.28}{1.50}\right)^3 = (1.52)^3 \approx 3.51. \qquad T_M^2 = 133225 \times 3.51 \approx 4.68 \times 10^5. $$ $$ T_M \approx \sqrt{4.68 \times 10^5} \approx 684 \text{ days} \approx 1.87 \text{ years}. $$

The accepted value is $687$ days (1.88 years) — excellent agreement. No knowledge of $G$ or $M_\odot$ was needed; the ratio $r_M/r_E$ alone determines the ratio of periods.公认值为 $687$ 天（1.88 年）——吻合极好。无需知道 $G$ 或 $M_\odot$；仅 $r_M/r_E$ 的比值即可决定周期之比。

According to Kepler's Second Law, a planet in an elliptical orbit moves fastest when:根据开普勒第二定律，椭圆轨道上的行星在何处运动最快？
§6 · Q1
It is farthest from the Sun距太阳最远时
It is at one of the foci位于某个焦点时
Its orbital speed is always constant轨道速率始终不变
It is closest to the Sun距太阳最近时
Equal areas in equal times: when the planet is closest (perihelion) the radius is short, so the planet must move faster to sweep the same area as at aphelion. Fastest at perihelion, slowest at aphelion.等时间等面积：行星距太阳最近（近日点）时半径最短，必须运动最快才能扫过与远日点相同的面积。近日点最快，远日点最慢。
Kepler's Second Law: equal areas in equal times. The area swept per unit time is constant, so when the planet is closest (shortest radius) it must move fastest.开普勒第二定律：等时间等面积。单位时间扫过的面积恒定，故行星距太阳最近（半径最短）时必须运动最快。
Planet X has twice the orbital radius of Planet Y (same star). How does the orbital period of X compare to that of Y? (Use Kepler's Third Law.)X 行星的轨道半径是 Y 行星（同一恒星）的两倍。X 的轨道周期与 Y 相比如何？（用开普勒第三定律。）
§6 · Q2
$T_X = 2T_Y$$T_X = 2T_Y$
$T_X = 4T_Y$$T_X = 4T_Y$
$T_X = 2\sqrt{2}\,T_Y$$T_X = 2\sqrt{2}\,T_Y$
$T_X = 8T_Y$$T_X = 8T_Y$
$T^2 \propto r^3$. If $r_X = 2r_Y$ then $T_X^2/T_Y^2 = (2)^3 = 8$, so $T_X/T_Y = \sqrt{8} = 2\sqrt{2} \approx 2.83$.$T^2 \propto r^3$。若 $r_X = 2r_Y$，则 $T_X^2/T_Y^2 = (2)^3 = 8$，故 $T_X/T_Y = \sqrt{8} = 2\sqrt{2} \approx 2.83$。
Apply $T^2/r^3 = \mathrm{const}$: $(T_X/T_Y)^2 = (r_X/r_Y)^3 = 2^3 = 8$, so $T_X = \sqrt{8}\,T_Y = 2\sqrt{2}\,T_Y$.用 $T^2/r^3 = \text{常数}$：$(T_X/T_Y)^2 = (r_X/r_Y)^3 = 2^3 = 8$，故 $T_X = \sqrt{8}\,T_Y = 2\sqrt{2}\,T_Y$。
Going deeper — deriving Kepler's Third Law from Newton's law of universal gravitation (AB 20–C1.5k)深入 — 从牛顿万有引力定律推导开普勒第三定律（AB 20–C1.5k）

For a circular orbit of radius $r$ around a central mass $M$, the gravitational force provides the centripetal force:对于绕中心质量 $M$ 的半径 $r$ 圆形轨道，引力提供向心力：

$$ \frac{GmM}{r^2} = \frac{mv^2}{r} \;\Longrightarrow\; v^2 = \frac{GM}{r}. $$

The orbital speed is also $v = 2\pi r / T$. Substitute:轨道速率也等于 $v = 2\pi r / T$。代入：

$$ \left(\frac{2\pi r}{T}\right)^2 = \frac{GM}{r} \;\Longrightarrow\; \frac{4\pi^2 r^2}{T^2} = \frac{GM}{r} \;\Longrightarrow\; T^2 = \frac{4\pi^2}{GM} r^3. $$

This is Kepler's Third Law. The ratio $T^2/r^3 = 4\pi^2/(GM)$ is constant for all satellites orbiting the same central mass $M$, regardless of the satellite's own mass $m$. This result, combined with measured planetary periods and radii, gave Newton a way to calculate the mass of the Sun and later the mass of any planet with a moon. Alberta outcome 20–C1.5k requires "explain, quantitatively, planetary and natural and artificial satellite motion, using circular motion to approximate elliptical orbits" — this derivation is its core.这就是开普勒第三定律。比值 $T^2/r^3 = 4\pi^2/(GM)$ 对所有绕同一中心质量 $M$ 运行的卫星为常数，与卫星自身质量 $m$ 无关。这一结果结合实测行星周期与轨道半径，为牛顿提供了计算太阳质量，乃至任何有卫星的行星质量的方法。阿尔伯塔结果 20–C1.5k 要求"用圆形轨道近似椭圆轨道，定量解释行星与天然及人造卫星运动"——这一推导是其核心。

Section 7 · Satellites & problem-solving Honors — US NGSS第 7 节 · 卫星与解题策略 荣誉 — US NGSS

Satellites and Problem-Solving Strategy Honors — US NGSS卫星与解题策略 荣誉 — US NGSS

Orbital speed, period, and the geostationary condition.轨道速度、周期与地球同步条件。

Set gravitational force equal to centripetal force to get orbital speed and period:令引力等于向心力，得到轨道速度与周期：

$$ \frac{GmM}{r^2} = \frac{mv^2}{r} \;\Longrightarrow\; v_{\mathrm{orb}} = \sqrt{\frac{GM}{r}} \qquad T = \frac{2\pi r}{v} = 2\pi\sqrt{\frac{r^3}{GM}} $$
  • Geostationary orbit地球同步轨道 — $T = 24$ h exactly; the satellite hovers over the same point on Earth's equator. Solve $T = 24$ h to find $r \approx 42\,200$ km from Earth's center ($\approx 35\,800$ km altitude).— $T$ 恰好等于 $24$ h；卫星悬停在地球赤道上方同一点。令 $T = 24$ h 解出 $r \approx 42\,200$ km（距地球中心，$\approx 35\,800$ km 轨道高度）。
  • Satellite mass cancels out.卫星质量相消。 The orbital speed and period depend only on $r$ and $M$ (central mass), not on the satellite mass $m$. A heavier satellite in the same orbit has the same speed and period.轨道速度与周期只依赖 $r$ 与中心质量 $M$，与卫星质量 $m$ 无关。同一轨道上较重的卫星速度与周期相同。
  • Problem-solving steps.解题步骤。 (1) Draw FBD; identify which real force (tension, gravity, normal) provides the net centripetal force. (2) Write $F_{\mathrm{net,inward}} = mv^2/r$. (3) Substitute given quantities; solve for the unknown.(1) 画受力图；找出哪个真实力（张力、引力、法向力）提供净向心力。(2) 写 $F_{\mathrm{net,inward}} = mv^2/r$。(3) 代入已知量；求解未知量。
AB 20–C1.6k: "predict the mass of a celestial body from the orbital data of a satellite in uniform circular motion around the celestial body." BC Physics 12 elaboration: "satellite motion, orbit changes, launch velocity, escape velocity."AB 20–C1.6k："由绕天体做匀速圆周运动的卫星轨道数据预测天体质量。"BC Physics 12 细化："卫星运动、轨道变化、发射速度、逃逸速度。"
Worked Example 7 · Orbital speed of the ISS例题 7 · 国际空间站的轨道速度

The International Space Station orbits at an altitude of $h = 400$ km. Using $G = 6.674 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$, $M_E = 5.97 \times 10^{24}$ kg, $R_E = 6.37 \times 10^6$ m, find the orbital speed and the orbital period.国际空间站在 $h = 400$ km 高度轨道运行。取 $G = 6.674 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$，$M_E = 5.97 \times 10^{24}$ kg，$R_E = 6.37 \times 10^6$ m，求轨道速度与轨道周期。

Orbital radius.轨道半径。 $r = R_E + h = 6.37 \times 10^6 + 4.00 \times 10^5 = 6.77 \times 10^6\ \mathrm{m}.$

Orbital speed.轨道速度。

$$ v = \sqrt{\frac{GM_E}{r}} = \sqrt{\frac{(6.674\times10^{-11})(5.97\times10^{24})}{6.77\times10^6}} = \sqrt{\frac{3.984\times10^{14}}{6.77\times10^6}} = \sqrt{5.885\times10^7} \approx 7670\ \mathrm{m/s}. $$

Orbital period.轨道周期。

$$ T = \frac{2\pi r}{v} = \frac{2\pi(6.77\times10^6)}{7670} \approx \frac{4.252\times10^7}{7670} \approx 5545\ \mathrm{s} \approx 92.4\ \mathrm{min}. $$

The ISS completes about $15.7$ orbits per day at $\approx 7.67$ km/s. Note the satellite mass cancelled out completely; only $r$ and $M_E$ determine the orbit.国际空间站每天约完成 $15.7$ 圈，速度约 $7.67$ km/s。注意卫星质量完全消去；只有 $r$ 与 $M_E$ 决定轨道。

A satellite orbits Earth at radius $r$. If it moves to orbit at radius $4r$, by what factor does its orbital speed change?一颗卫星在半径 $r$ 的轨道上绕地球运行。若它移至半径 $4r$ 的轨道，轨道速度变为原来的几倍？
§7 · Q1
Increases by factor of $4$变为原来的 $4$ 倍
Decreases by factor of $2$变为原来的 $1/2$
Decreases by factor of $4$变为原来的 $1/4$
Stays the same不变
$v = \sqrt{GM/r}$, so $v \propto 1/\sqrt{r}$. Going from $r$ to $4r$: $v_{\mathrm{new}} = v/\sqrt{4} = v/2$. The speed halves.$v = \sqrt{GM/r}$，故 $v \propto 1/\sqrt{r}$。从 $r$ 到 $4r$：$v_{\mathrm{new}} = v/\sqrt{4} = v/2$。速度减半。
$v \propto r^{-1/2}$. Increasing $r$ by factor of $4$ decreases $v$ by factor of $\sqrt{4} = 2$.$v \propto r^{-1/2}$。$r$ 增大 $4$ 倍使 $v$ 减小 $\sqrt{4} = 2$ 倍。
A satellite of mass $500$ kg and another of mass $2000$ kg orbit Earth at the same radius. Which has the greater orbital speed?质量 $500$ kg 与 $2000$ kg 的两颗卫星在同一半径上绕地球运行。哪颗轨道速度更大？
§7 · Q2
Both have the same orbital speed两颗轨道速度相同
The $2000$ kg satellite is faster$2000$ kg 的卫星更快
The $500$ kg satellite is faster$500$ kg 的卫星更快
Cannot be determined without knowing the orbit altitude不知轨道高度无法判断
$v = \sqrt{GM_E/r}$ depends only on $r$ and $M_E$, not on the satellite mass. Both satellites at the same radius have exactly the same orbital speed.$v = \sqrt{GM_E/r}$ 只依赖 $r$ 与 $M_E$，与卫星质量无关。处于同一半径的两颗卫星轨道速度完全相同。
When you derive orbital speed from $F_g = F_c$, the satellite mass $m$ cancels out. The orbital speed depends only on the orbital radius and the central mass.由 $F_g = F_c$ 推导轨道速度时，卫星质量 $m$ 相消。轨道速度只依赖轨道半径与中心质量。
Exam Tactics考试策略

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Setup discipline解题前的纪律
  • Draw a FBD before any equation.写任何方程前先画受力图。 Identify the real forces and label which one (or which component) points toward the center. Set that net inward force equal to $mv^2/r$.找出真实力，标出哪个力（或哪个分量）指向圆心。令净向内力等于 $mv^2/r$。
  • $r$ is centre-to-centre, not altitude.$r$ 是质心到质心的距离，而非轨道高度。 The most common error in gravitation problems is using the altitude $h$ instead of $r = R_E + h$. Write $r$ explicitly before substituting into any formula.引力问题最常见的错误是用轨道高度 $h$ 而非 $r = R_E + h$。代入任何公式前先显式写出 $r$。
  • Carry units; check exponents carefully.带单位；仔细核查指数。 $G$ and planetary data involve powers of ten from $10^{-11}$ to $10^{24}$. A one-exponent slip changes an answer by a factor of ten.$G$ 与行星数据涉及从 $10^{-11}$ 到 $10^{24}$ 的十次幂。指数差一就使答案差十倍。
Circular motion pitfalls (§1–§3)圆周运动常见陷阱（§1–§3）
  • "Centripetal force" is not a new force."向心力"不是新力。 Never add a separate "centripetal force" arrow to a FBD. It is always an existing force (tension, gravity, normal, friction) pointing inward. The common error is adding a fictional outward "centrifugal force" — this does not exist in an inertial frame.绝对不要在受力图上另加一个"向心力"箭头。它始终是已有的力（张力、引力、法向力、摩擦力）指向圆心。常见错误是加上虚构的向外"离心力"——在惯性参考系中这种力不存在。
  • Vertical circles: redo the FBD at each position.竖直圆：在每个位置重新画受力图。 At the top, gravity and normal force can be in the same direction (both inward). At the bottom they oppose. The equations are different at each position — never use the same equation for top and bottom without re-deriving.在顶部，重力与法向力可同向（都向内）。在底部方向相反。每个位置的方程不同——绝对不要未重新推导就把同一个方程用于顶部和底部。
  • $a_c$ always points inward, never outward.$a_c$ 始终指向圆心，绝非向外。 The centripetal acceleration is centripetal (toward center) by definition. The velocity vector is tangent to the circle; $a$ is perpendicular to $v$, pointing inward.向心加速度顾名思义（指向圆心）。速度矢量沿圆的切线方向；$a$ 垂直于 $v$，指向圆心。
Gravitation pitfalls (§4–§7) Honors — US NGSS引力常见陷阱（§4–§7）荣誉 — US NGSS
  • Inverse-square: double $r$, divide $F$ by $4$.平方反比：$r$ 加倍，$F$ 变四分之一。 The factor of $4$ catches students who only halve the force. Write the ratio $F_2/F_1 = (r_1/r_2)^2$ explicitly.只把力减半的学生会在系数 $4$ 上出错。显式写出比值 $F_2/F_1 = (r_1/r_2)^2$。
  • Satellite mass cancels.卫星质量相消。 Orbital speed $v = \sqrt{GM/r}$ and period $T = 2\pi\sqrt{r^3/(GM)}$ do not depend on the satellite's mass. Two satellites of different masses at the same orbital radius move at the same speed.轨道速度 $v = \sqrt{GM/r}$ 与周期 $T = 2\pi\sqrt{r^3/(GM)}$ 与卫星质量无关。同一轨道半径上不同质量的卫星以相同速度运行。
  • For Kepler's Third Law, use ratios.使用开普勒第三定律时用比值。 If two objects orbit the same central mass, set $T_1^2/r_1^3 = T_2^2/r_2^3$ to avoid needing $G$ and $M$. This is by far the fastest approach on exams.若两个天体绕同一中心质量运行，令 $T_1^2/r_1^3 = T_2^2/r_2^3$，从而无需 $G$ 与 $M$。这是考试中迄今最快的方法。
Answer hygiene作答规范
  • Round at the very end.最后一步再四舍五入。 Carry extra digits through intermediate steps; round only the final number to the precision the question asks.中间步骤多留几位；仅在最终答案处按题目要求精度四舍五入。
  • Sanity-check magnitudes.用数量级核验。 Orbital speeds are km/s (not m/s or km/h); gravitational forces between everyday objects are tiny ($\ll 1$ N); the ISS takes $\approx 90$ min per orbit. If your answer is wildly different, recheck $r$ (altitude vs centre-to-centre).轨道速度单位是 km/s（而非 m/s 或 km/h）；日常物体间的引力极小（$\ll 1$ N）；国际空间站每圈约 $90$ min。若答案相差悬殊，检查 $r$（轨道高度还是质心距）。
  • State which value of $G$ you used.注明所用 $G$ 的值。 Write $G = 6.674 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$ at the start of any gravitation calculation so the marker can track your arithmetic.在任何引力计算开头写明 $G = 6.674 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$，便于评分者核查你的算术。
Active Recall主动复习

Flashcards闪卡

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Centripetal acceleration?向心加速度？
$$a_c = \frac{v^2}{r} = \frac{4\pi^2 r}{T^2}$$ directed toward center指向圆心
Net centripetal force?净向心力？
$$F_c = \frac{mv^2}{r}$$ always an existing real force (tension, gravity, friction, normal) pointing inward始终是某个已有真实力（张力、引力、摩擦力、法向力）指向圆心
Speed from radius and period?由半径与周期求速率？
$$v = \frac{2\pi r}{T} = 2\pi r f$$
Minimum speed at top of a vertical loop?竖直圆最高点最小速度？
$$v_{\min} = \sqrt{gr}$$ when normal force $= 0$, gravity alone provides centripetal force当法向力 $= 0$ 时，重力单独提供向心力
Normal force at bottom of a vertical circle?竖直圆最低点法向力？
$$N = mg + \frac{mv^2}{r}$$ greater than $mg$; you feel heavier大于 $mg$；感觉更重
Newton's law of universal gravitation?牛顿万有引力定律？
$$F = G\frac{m_1 m_2}{r^2}$$ $G = 6.674\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$; $r$ = centre-to-centre$G = 6.674\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$；$r$ = 质心距
Gravitational field strength?引力场强？
$$g = \frac{GM}{r^2}$$ force per unit mass (N/kg = m/s²); decreases as $1/r^2$单位质量所受力（N/kg = m/s²）；以 $1/r^2$ 减弱
Connecting $g$ at surface to $G$?将地表 $g$ 与 $G$ 相联系？
$$g = \frac{GM_E}{R_E^2}$$ same formula; at altitude $h$: use $r = R_E + h$同一公式；高度 $h$ 处：用 $r = R_E + h$
Orbital speed?轨道速度？
$$v_{\mathrm{orb}} = \sqrt{\frac{GM}{r}}$$ independent of satellite mass与卫星质量无关
Orbital period?轨道周期？
$$T = 2\pi\sqrt{\frac{r^3}{GM}}$$
Kepler's Third Law?开普勒第三定律？
$$\frac{T^2}{r^3} = \frac{4\pi^2}{GM} = \text{const}$$ same constant for all satellites around the same central mass对绕同一中心质量的所有卫星为同一常数
Kepler's Second Law?开普勒第二定律？
Equal areas in equal times. Fastest at perihelion (closest), slowest at aphelion (farthest).等时间等面积。近日点最快，远日点最慢。
Inverse-square law effect on $F_g$?平方反比律对 $F_g$ 的影响？
Double $r$ $\Rightarrow$ $F$ becomes $F/4$. Triple $r$ $\Rightarrow$ $F$ becomes $F/9$.$r$ 加倍 $\Rightarrow$ $F$ 变 $F/4$。$r$ 变三倍 $\Rightarrow$ $F$ 变 $F/9$。
Ideal banked curve angle?理想倾斜弯道角度？
$$\tan\theta = \frac{v^2}{rg}$$ horizontal component of $N$ provides centripetal force; no friction needed$N$ 的水平分量提供向心力；无需摩擦力
Mixed Practice综合练习

Practice Quiz综合测验

A $0.40$ kg ball on a string moves in a horizontal circle of radius $0.60$ m at $3.0$ m/s. What is the tension in the string?一只 $0.40$ kg 的绳上球以 $3.0$ m/s 在半径 $0.60$ m 的水平圆上运动。绳中张力是多少？
Q1
$1.2$ N
$6.0$ N
$0.72$ N
$12$ N
$T = mv^2/r = (0.40)(3.0)^2/0.60 = (0.40)(9.0)/0.60 = 3.6/0.60 = 6.0$ N.$T = mv^2/r = (0.40)(3.0)^2/0.60 = (0.40)(9.0)/0.60 = 3.6/0.60 = 6.0$ N。
Tension provides the centripetal force: $T = mv^2/r$. Square the speed first.张力提供向心力：$T = mv^2/r$。先把速度平方。
A $1500$ kg car rounds a flat curve of radius $80$ m at $20$ m/s. What friction force is needed? 🇺🇸 NGSS HS-PS2-1一辆 $1500$ kg 的车以 $20$ m/s 通过半径 $80$ m 的平坦弯道。需要多大的摩擦力？🇺🇸 NGSS HS-PS2-1
Q2
$7500$ N
$375$ N
$30000$ N
$18750$ N
Friction provides centripetal force: $f = mv^2/r = (1500)(20)^2/80 = (1500)(400)/80 = 600000/80 = 7500$ N.摩擦力提供向心力：$f = mv^2/r = (1500)(20)^2/80 = (1500)(400)/80 = 600000/80 = 7500$ N。
On a flat curve, friction is the centripetal force: $f = mv^2/r$.在平坦弯道上，摩擦力是向心力：$f = mv^2/r$。
Two masses of $3.0$ kg and $6.0$ kg are separated by $2.0$ m. What is the gravitational force between them? (Use $G = 6.674 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$.) 🇺🇸 NGSS HS-PS2-4质量 $3.0$ kg 与 $6.0$ kg 的两个物体相距 $2.0$ m。它们间的引力是多少？（取 $G = 6.674 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}$。）🇺🇸 NGSS HS-PS2-4
Q3
$3.0 \times 10^{-10}$ N
$6.0 \times 10^{-11}$ N
$3.0 \times 10^{-10}$ N
$1.2 \times 10^{-9}$ N
$F = G m_1 m_2 / r^2 = (6.674\times10^{-11})(3.0)(6.0)/(2.0)^2 = (6.674\times10^{-11})(18)/4 = 6.674\times10^{-11}\times4.5 = 3.0\times10^{-10}$ N.$F = G m_1 m_2 / r^2 = (6.674\times10^{-11})(3.0)(6.0)/(2.0)^2 = (6.674\times10^{-11})(18)/4 = 6.674\times10^{-11}\times4.5 = 3.0\times10^{-10}$ N。
Apply $F = Gm_1m_2/r^2$. Don't forget to square the separation.用 $F = Gm_1m_2/r^2$。不要忘记将间距平方。
A satellite orbits a planet of mass $M$ at radius $r$. What is the orbital period? 🇨🇦 AB 20–C1.5k / BC Physics 12一颗卫星以半径 $r$ 绕质量为 $M$ 的行星运行。轨道周期是什么？🇨🇦 AB 20–C1.5k / BC Physics 12
Q4
$T = \sqrt{4\pi^2 r^2/(GM)}$
$T = 2\pi r / \sqrt{GM}$
$T = 2\pi\sqrt{GM/r^3}$
$T = 2\pi\sqrt{r^3/(GM)}$
From $v = \sqrt{GM/r}$ and $T = 2\pi r / v$: $T = 2\pi r / \sqrt{GM/r} = 2\pi r \cdot \sqrt{r/(GM)} = 2\pi\sqrt{r^3/(GM)}$.由 $v = \sqrt{GM/r}$ 与 $T = 2\pi r / v$：$T = 2\pi r / \sqrt{GM/r} = 2\pi r \cdot \sqrt{r/(GM)} = 2\pi\sqrt{r^3/(GM)}$。
Combine $v = \sqrt{GM/r}$ with $T = 2\pi r/v$ to get $T = 2\pi\sqrt{r^3/(GM)}$. This is Kepler's Third Law in disguise.将 $v = \sqrt{GM/r}$ 与 $T = 2\pi r/v$ 结合得 $T = 2\pi\sqrt{r^3/(GM)}$。这是开普勒第三定律的另一形式。
A roller-coaster loop has radius $15$ m. What is the minimum speed at the top to maintain contact? (Use $g = 9.8\ \mathrm{m/s^2}$.) 🇨🇦 BC Physics 12 (vertical circles)过山车圆圈半径 $15$ m。保持接触所需的顶部最小速度是多少？（取 $g = 9.8\ \mathrm{m/s^2}$。）🇨🇦 BC Physics 12（竖直圆）
Q5
$7.7\ \mathrm{m/s}$
$12.1\ \mathrm{m/s}$
$147\ \mathrm{m/s}$
$3.8\ \mathrm{m/s}$
At the top, minimum speed when $N = 0$: $v_{\min} = \sqrt{gr} = \sqrt{9.8 \times 15} = \sqrt{147} \approx 12.1\ \mathrm{m/s}$.在顶部，$N = 0$ 时最小速度：$v_{\min} = \sqrt{gr} = \sqrt{9.8 \times 15} = \sqrt{147} \approx 12.1\ \mathrm{m/s}$。
At the top with $N = 0$, gravity alone provides centripetal force: $mg = mv^2/r$, so $v_{\min} = \sqrt{gr}$.顶部 $N = 0$ 时，重力单独提供向心力：$mg = mv^2/r$，故 $v_{\min} = \sqrt{gr}$。
Earth's orbital radius is $1.50 \times 10^{11}$ m and period is $365$ days. Jupiter's orbital radius is $7.78 \times 10^{11}$ m. Use Kepler's Third Law to find Jupiter's period in years. 🇨🇦 ON SPH4U D / AB 20–C1.5k地球轨道半径 $1.50 \times 10^{11}$ m，周期 $365$ 天。木星轨道半径 $7.78 \times 10^{11}$ m。用开普勒第三定律求木星的周期（年）。🇨🇦 ON SPH4U D / AB 20–C1.5k
Q6
$\approx 11.9$ years
$\approx 5.2$ years
$\approx 27$ years
$\approx 3.5$ years
$T_J^2/T_E^2 = (r_J/r_E)^3 = (7.78/1.50)^3 = (5.187)^3 \approx 139.5$. $T_J = 1 \times \sqrt{139.5} \approx 11.8$ years (accepted value: $11.86$ years). ✓$T_J^2/T_E^2 = (r_J/r_E)^3 = (7.78/1.50)^3 = (5.187)^3 \approx 139.5$。$T_J = 1 \times \sqrt{139.5} \approx 11.8$ 年（公认值：$11.86$ 年）。✓
Use ratios: $(T_J/T_E)^2 = (r_J/r_E)^3$. Solve for $T_J = T_E\sqrt{(r_J/r_E)^3}$.用比值：$(T_J/T_E)^2 = (r_J/r_E)^3$。解出 $T_J = T_E\sqrt{(r_J/r_E)^3}$。
An object weighs $W$ newtons on Earth's surface. At height $R_E$ above the surface (so the distance from Earth's centre is $2R_E$), its weight is:某物体在地球表面重 $W$ 牛顿。在距地表 $R_E$ 高度（距地球中心 $2R_E$）处，其重力为：
Q7
$W/2$
$W/8$
$W/4$
$W\sqrt{2}$
$F \propto 1/r^2$. Doubling $r$ (from $R_E$ to $2R_E$) gives $F = W / 2^2 = W/4$.$F \propto 1/r^2$。$r$ 加倍（从 $R_E$ 到 $2R_E$）得 $F = W / 2^2 = W/4$。
Centre-to-centre distance goes from $R_E$ to $2R_E$ (doubles). By the inverse-square law, $F$ becomes $W/4$.质心距从 $R_E$ 变为 $2R_E$（加倍）。由平方反比律，$F$ 变为 $W/4$。
A satellite orbits Earth at $r = 8.0 \times 10^6$ m. Use $GM_E = 3.986 \times 10^{14}\ \mathrm{m^3\,s^{-2}}$ to find the orbital speed. 🇨🇦 AB 20–C1.5k / BC Physics 12一颗卫星在 $r = 8.0 \times 10^6$ m 处绕地球运行。用 $GM_E = 3.986 \times 10^{14}\ \mathrm{m^3\,s^{-2}}$ 求轨道速度。🇨🇦 AB 20–C1.5k / BC Physics 12
Q8
$4980\ \mathrm{m/s}$
$3160\ \mathrm{m/s}$
$9950\ \mathrm{m/s}$
$7060\ \mathrm{m/s}$
$v = \sqrt{GM/r} = \sqrt{3.986\times10^{14} / 8.0\times10^6} = \sqrt{4.983\times10^7} \approx 7060\ \mathrm{m/s}$.$v = \sqrt{GM/r} = \sqrt{3.986\times10^{14} / 8.0\times10^6} = \sqrt{4.983\times10^7} \approx 7060\ \mathrm{m/s}$。
Use $v = \sqrt{GM/r}$. Divide $GM$ by $r$ first, then take the square root.用 $v = \sqrt{GM/r}$。先用 $GM$ 除以 $r$，再开平方根。
Self-check自查

Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成，再勾选每一项。

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Next Steps后续单元

What This Feeds Into本单元的去向

Circular motion and gravitation is the bridge from Newtonian mechanics into astrophysics, orbital engineering, and electrostatics. The centripetal-force framework ($F = mv^2/r$) reappears in any situation involving rotation — from spinning tops and banked curves to charged particles in magnetic fields. Universal gravitation and the inverse-square law directly parallel Coulomb's law (Unit 8), making this unit conceptually foundational for the entire fields-and-forces arc of the course.圆周运动与万有引力是从牛顿力学通往天体物理学、轨道工程与静电学的桥梁。向心力框架（$F = mv^2/r$）在任何涉及旋转的情境中再次出现——从旋转陀螺、倾斜弯道到磁场中的带电粒子。万有引力与平方反比律直接类比库仑定律（第 8 单元），使本单元成为课程中"场与力"这条主线的概念基础。

Within High School Physics.在 HS Physics 内部。

The Work, Energy and Power unit introduces gravitational potential energy $U = -GMm/r$ as an extension of the surface-level $mgh$. Electrostatics (Unit 8) uses the same inverse-square structure as $F = Gm_1m_2/r^2$ but with charges instead of masses; HS-PS2-4 is the NGSS PE that pairs both. Magnetism and EM Induction (Unit 10) applies $F = qv \times B$, which produces circular paths for charged particles — the same centripetal reasoning as §2.《功、能与功率》单元将引力势能 $U = -GMm/r$ 作为地表近似 $mgh$ 的延伸引入。静电学（第 8 单元）使用与 $F = Gm_1m_2/r^2$ 相同的平方反比结构，但以电荷取代质量；HS-PS2-4 正是将两者配对的 NGSS PE。电磁感应（第 10 单元）应用 $F = qv \times B$，使带电粒子做圆周运动——与 §2 相同的向心力分析。

Across the AP, IB, and HS Math feeders in this repo.本仓库中的 AP、IB 与 HS Math 衔接单元。

AP Physics Unit 5 · Torque and Rotational Dynamics (centripetal acceleration and circular kinematics are prerequisite; this unit extends rotation to torque, moment of inertia, and angular momentum)AP Physics Unit 5 · 力矩与转动动力学（向心加速度与圆周运动学是前提；本单元将旋转延伸至力矩、转动惯量与角动量） HS Math · Right-Triangle Trigonometry (the trig needed for banked-curve angle $\tan\theta = v^2/rg$ and force-component decomposition)HS Math · 直角三角形三角学（倾斜弯道角 $\tan\theta = v^2/rg$ 与力分量分解所需的三角学）

If you are aiming for AP Physics 1 or C, fluency with $a_c = v^2/r$, $F = Gm_1m_2/r^2$, and the orbital-speed derivation is assumed from the start. AP Physics C (Mechanics) extends circular motion to non-uniform cases and introduces gravitational potential energy rigorously as $U = -GMm/r$. For IB Physics HL, Topic A.2 (Forces and Momentum) and Topic D (Fields) cover this exact scope; the field-concept treatment in §5 is precisely what IB's gravitational field maps to.备考 AP Physics 1 或 C：从第一天起就默认你熟练 $a_c = v^2/r$、$F = Gm_1m_2/r^2$ 及轨道速度推导。AP Physics C（力学）将圆周运动扩展至非匀速情形，并严格引入引力势能 $U = -GMm/r$。备考 IB Physics HL：主题 A.2（力与动量）与主题 D（场）覆盖这一确切范围；§5 的场概念处理正是 IB 引力场所对应的内容。

DINGRUI SCHOLARS

High School Physics · Circular Motion and Gravitation · v1高中物理 · 圆周运动与万有引力 · v1