IB Physics HL · 鼎睿学苑

Unit D.1: Gravitational fields单元 D.1：引力场

The opening unit of Theme D "Fields". Newton's law of universal gravitation, gravitational field strength and field lines, and orbital motion in which gravity supplies the centripetal force, leading to Kepler's third law. The HL extension introduces gravitational potential and potential energy, the field–potential relationship, and escape speed with the energy analysis of bound and unbound orbits. The same field-and-potential machinery reappears almost unchanged for electric fields in D.2, so mastering it here pays off twice.主题 D"场"的开篇。万有引力定律、引力场强度与场线，以及引力提供向心力的轨道运动，由此推出开普勒第三定律。HL 扩展引入引力势与引力势能、场与势的关系，以及逃逸速度，并对束缚轨道与非束缚轨道作能量分析。同一套"场与势"框架在 D.2 电场中几乎原样重现，因此在此掌握将受益两次。

IB Physics · Theme D.1 · First Assessment 2025 Papers 1 · 2 6 Topics · SL + HL mix6 个核心专题 · SL + HL 混合
Read me first先读这里

How to use this guide本指南使用说明

D.1 has one idea behind almost every mark: a point (or spherically symmetric) mass produces an inverse-square field $\propto 1/r^2$, while its potential falls off as $1/r$. Get those two scalings straight and the unit unlocks. The SL core is the force law, field strength, and orbits; the HL extension adds the negative-valued potential and the energetics of escape and orbit. Watch the difference between "distance from the centre" $r$ and "height above the surface".D.1 几乎每一分背后都是同一个观念：点质量（或球对称质量）产生反平方场 $\propto 1/r^2$，而其势按 $1/r$ 衰减。把这两个标度关系理清，整单元就通了。SL 核心是力定律、场强度与轨道；HL 扩展加入取负值的势以及逃逸与轨道的能量学。注意区分"到中心的距离"$r$ 与"距表面的高度"。

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If you are cramming如果你在临阵磨枪

Memorise $F = \dfrac{GMm}{r^2}$ and $g = \dfrac{GM}{r^2}$ (force per unit mass). For a circular orbit, set gravity equal to the centripetal force to get $v = \sqrt{GM/r}$ and $T^2 = \dfrac{4\pi^2}{GM} r^3$. HL: potential $V_g = -\dfrac{GM}{r}$ is negative and energy $E_p = m V_g$; escape speed $v_{esc} = \sqrt{2GM/r}$.

背熟 $F = \dfrac{GMm}{r^2}$ 与 $g = \dfrac{GM}{r^2}$（单位质量受力）。圆轨道令引力等于向心力，得 $v = \sqrt{GM/r}$ 与 $T^2 = \dfrac{4\pi^2}{GM} r^3$。HL：势 $V_g = -\dfrac{GM}{r}$ 取负，能量 $E_p = m V_g$；逃逸速度 $v_{esc} = \sqrt{2GM/r}$。

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If you are going for a 7如果你目标是 7 分

Be able to derive $T^2 \propto r^3$ from the force–centripetal balance, and the field–potential link $g = -\dfrac{\Delta V_g}{\Delta r}$. Know why potential is negative (zero is chosen at infinity), why work done against gravity equals $m\,\Delta V_g$, and why total orbit energy $E = -\dfrac{GMm}{2r}$ is negative for any bound orbit. State the assumption "spherically symmetric, treat as point mass at the centre" whenever you invoke these formulas.

能由"引力=向心力"推出 $T^2 \propto r^3$，并掌握场与势的关系 $g = -\dfrac{\Delta V_g}{\Delta r}$。理解势为何取负（零势取在无穷远）、克服引力做功为何等于 $m\,\Delta V_g$，以及任意束缚轨道总能量 $E = -\dfrac{GMm}{2r}$ 为何为负。每次用这些公式前都要写明"球对称、视为中心处点质量"的假设。

HL flagHL 标记说明 Sections D1.4 (gravitational potential and potential energy), D1.5 (field–potential relationship and work done) and D1.6 (escape speed and orbit energy) are HL extension content. SL students may safely skip the HL-flagged blocks; sections D1.1–D1.3 are SL + HL core.D1.4（引力势与引力势能）、D1.5（场与势的关系及做功）与 D1.6（逃逸速度与轨道能量）为 HL 扩展内容。SL 学生可跳过带 HL 标记的段落；D1.1–D1.3 为 SL + HL 共同核心。
Topic D1.1 · Newton's Law of Universal GravitationTopic D1.1 · 万有引力定律

Newton's Law of Universal Gravitation万有引力定律 D.1 SL+HL

The force law. Every point mass attracts every other along the line joining them. From the data booklet, F = GMm/r²: $$ F = \frac{G M m}{r^{2}}. $$
  • $G = 6.67 \times 10^{-11}\ \mathrm{N\,m^{2}\,kg^{-2}}$ is the universal gravitational constant (data booklet).
  • Inverse-square: double $r$ and the force falls to a quarter; triple $r$ and it falls to a ninth.
  • $r$ is the centre-to-centre separation, not the surface gap. For a sphere, all the mass acts as if concentrated at the centre.
  • The force is always attractive and Newton's third law makes it equal and opposite on the two bodies.
力定律。任意两点质量沿连线相互吸引。由数据手册，F = GMm/r²： $$ F = \frac{G M m}{r^{2}}. $$
  • $G = 6.67 \times 10^{-11}\ \mathrm{N\,m^{2}\,kg^{-2}}$ 为万有引力常量（数据手册）。
  • 反平方：$r$ 加倍则力降为四分之一；$r$ 增至三倍则力降为九分之一。
  • $r$ 是中心到中心的距离，不是表面间隙。对于球体，全部质量等效集中于中心。
  • 力恒为吸引；牛顿第三定律使两物体受力等大反向。
Worked Example D1.1 (force between two masses)D1.1 例题（两质量间的引力）

Two $5.0 \times 10^{3}\ \mathrm{kg}$ trucks park with their centres $8.0\ \mathrm{m}$ apart. Find the gravitational force between them. Then find the new force if the separation is halved.两辆 $5.0 \times 10^{3}\ \mathrm{kg}$ 的卡车并排停放，中心相距 $8.0\ \mathrm{m}$。求二者间的引力。再求距离减半后的新引力。

Identify. Both masses equal, $M = m = 5.0 \times 10^{3}\ \mathrm{kg}$, $r = 8.0\ \mathrm{m}$. Use $F = GMm/r^{2}$.

识别。两质量相等，$M = m = 5.0 \times 10^{3}\ \mathrm{kg}$，$r = 8.0\ \mathrm{m}$。用 $F = GMm/r^{2}$。

Set up. Substitute $G = 6.67 \times 10^{-11}\ \mathrm{N\,m^{2}\,kg^{-2}}$.

列式。代入 $G = 6.67 \times 10^{-11}\ \mathrm{N\,m^{2}\,kg^{-2}}$。

Execute.

计算。

$$ F = \frac{(6.67 \times 10^{-11})(5.0 \times 10^{3})^{2}}{(8.0)^{2}} \approx 2.6 \times 10^{-5}\ \mathrm{N}. $$

Halved separation. Inverse-square: halving $r$ multiplies $F$ by $4$, so $F' \approx 1.0 \times 10^{-4}\ \mathrm{N}$.

距离减半。反平方：$r$ 减半使 $F$ 变为 $4$ 倍，故 $F' \approx 1.0 \times 10^{-4}\ \mathrm{N}$。

Evaluate. Tiny because $G$ is minuscule; everyday objects barely attract. Gravity only dominates when at least one mass is astronomical.

评估。极小，因为 $G$ 极小；日常物体几乎不相互吸引。只有至少一方为天文级质量时引力才显著。

Going deeper: why a sphere acts as a point mass深入：球体为何等效为点质量

Newton's shell theorem states that a spherically symmetric shell of matter attracts an external point mass exactly as if all the shell's mass were at its centre. A solid sphere is a stack of such shells, so the whole planet acts as a point mass at its centre for any external object.

牛顿壳层定理表明：球对称物质壳层对外部点质量的吸引，恰如全部壳层质量集中于其中心。实心球是这种壳层的叠加，因此整颗行星对任何外部物体都等效为中心处的点质量。

A second consequence: inside a uniform shell the net gravitational field is zero. This is the gravitational analogue of the electrostatic result that the field inside a charged conducting shell vanishes, and it is why $r$ must be measured from the centre.

第二个推论：均匀壳层内部净引力场为零。这是"带电导体壳层内部电场为零"的引力类比，也是 $r$ 必须从中心起算的原因。

The distance between two masses is tripled. The gravitational force between them becomes:两质量间的距离增至三倍。它们之间的引力变为：
D1.1 · Q1
$3$ times larger原来的 $3$ 倍
$\tfrac{1}{3}$ as large原来的 $\tfrac{1}{3}$
$\tfrac{1}{9}$ as large原来的 $\tfrac{1}{9}$
$9$ times larger原来的 $9$ 倍
$F \propto 1/r^{2}$. Tripling $r$ divides $F$ by $3^{2} = 9$.$F \propto 1/r^{2}$。$r$ 变 $3$ 倍使 $F$ 除以 $3^{2} = 9$。
Gravitation is inverse-square, not inverse-linear. The factor is $1/3^{2} = 1/9$.引力是反平方，不是反一次方。系数为 $1/3^{2} = 1/9$。
A satellite of mass $m$ orbits a planet of mass $M$. By Newton's third law, the force the satellite exerts on the planet is:质量 $m$ 的卫星绕质量 $M$ 的行星运动。由牛顿第三定律，卫星对行星的作用力为：
D1.1 · Q2
Smaller, by the factor $m/M$较小，差一个因子 $m/M$
Zero, since the planet is far heavier为零，因为行星重得多
Larger, since $M > m$较大，因为 $M > m$
Equal in magnitude and opposite in direction大小相等、方向相反
Gravitation is a Newton-third-law pair: $\dfrac{GMm}{r^{2}}$ acts on each body. The forces are equal and opposite; only the resulting accelerations differ (the planet's is tiny because its mass is huge).引力是牛顿第三定律对：$\dfrac{GMm}{r^{2}}$ 作用于双方。受力等大反向；只是由此产生的加速度不同（行星加速度极小，因其质量巨大）。
The formula is symmetric in $M$ and $m$, so both bodies feel the same magnitude of force. Their accelerations differ, not their forces.公式关于 $M$ 与 $m$ 对称，故双方受力大小相同。不同的是加速度，不是力。
Topic D1.2 · Gravitational Field StrengthTopic D1.2 · 引力场强度

Gravitational Field Strength and Field Lines引力场强度与场线 D.1 SL+HL

Field strength = force per unit mass. From the data booklet, g = F/m: $$ g = \frac{F}{m} = \frac{G M}{r^{2}}. $$
  • $g$ is a vector pointing toward the source mass; units $\mathrm{N\,kg^{-1}}$, numerically equal to $\mathrm{m\,s^{-2}}$.
  • At a planet's surface, set $r = R$ (the planet radius): $g_{\text{surf}} = \dfrac{GM}{R^{2}}$.
  • Field lines point radially inward toward a point mass; closer spacing means a stronger field. They never cross.
  • Over a small region near a surface the field is essentially uniform: parallel, evenly spaced lines, $g \approx 9.81\ \mathrm{N\,kg^{-1}}$ on Earth.
场强度 = 单位质量受力。由数据手册，g = F/m： $$ g = \frac{F}{m} = \frac{G M}{r^{2}}. $$
  • $g$ 是指向源质量的矢量；单位 $\mathrm{N\,kg^{-1}}$，数值等于 $\mathrm{m\,s^{-2}}$。
  • 行星表面令 $r = R$（行星半径）：$g_{\text{surf}} = \dfrac{GM}{R^{2}}$。
  • 场线沿径向指向点质量；线越密场越强；场线永不相交。
  • 表面附近小范围内场近似均匀：平行等距的场线，地球上 $g \approx 9.81\ \mathrm{N\,kg^{-1}}$。
Worked Example D1.2 (surface gravity of a planet)D1.2 例题（行星的表面重力）

Mars has mass $M = 6.4 \times 10^{23}\ \mathrm{kg}$ and radius $R = 3.4 \times 10^{6}\ \mathrm{m}$. Find the gravitational field strength at its surface and compare it with Earth's $9.81\ \mathrm{N\,kg^{-1}}$.火星质量 $M = 6.4 \times 10^{23}\ \mathrm{kg}$，半径 $R = 3.4 \times 10^{6}\ \mathrm{m}$。求其表面引力场强度，并与地球的 $9.81\ \mathrm{N\,kg^{-1}}$ 比较。

Identify. Surface field uses $r = R$ in $g = GM/r^{2}$.

识别。表面场在 $g = GM/r^{2}$ 中取 $r = R$。

Set up.

列式。

$$ g = \frac{G M}{R^{2}} = \frac{(6.67 \times 10^{-11})(6.4 \times 10^{23})}{(3.4 \times 10^{6})^{2}}. $$

Execute. Numerator $\approx 4.27 \times 10^{13}$; denominator $\approx 1.16 \times 10^{13}$, so $g \approx 3.7\ \mathrm{N\,kg^{-1}}$.

计算。分子 $\approx 4.27 \times 10^{13}$；分母 $\approx 1.16 \times 10^{13}$，故 $g \approx 3.7\ \mathrm{N\,kg^{-1}}$。

Evaluate. About $0.38$ of Earth's $g$, matching the well-known "Mars gravity is roughly a third of Earth's". A $60\ \mathrm{kg}$ astronaut weighs $\approx 222\ \mathrm{N}$ there versus $589\ \mathrm{N}$ on Earth.

评估。约为地球 $g$ 的 $0.38$ 倍，与"火星重力约为地球三分之一"的常识相符。$60\ \mathrm{kg}$ 的宇航员在那里重 $\approx 222\ \mathrm{N}$，而在地球为 $589\ \mathrm{N}$。

Going deeper: g as acceleration vs g as field strength深入：作为加速度的 g 与作为场强度的 g

Combining $F = mg$ (weight) with Newton's second law $F = ma$ for a freely falling body gives $mg = ma$, so $a = g$. The free-fall acceleration equals the field strength precisely because gravitational mass (in $F = GMm/r^2$) and inertial mass (in $F = ma$) are equal — the equivalence principle that later underpins general relativity.

把 $F = mg$（重力）与自由落体的牛顿第二定律 $F = ma$ 联立得 $mg = ma$，故 $a = g$。自由落体加速度恰等于场强度，正因为引力质量（在 $F = GMm/r^2$ 中）与惯性质量（在 $F = ma$ 中）相等——这就是后来支撑广义相对论的等效原理。

This is why the units $\mathrm{N\,kg^{-1}}$ and $\mathrm{m\,s^{-2}}$ are interchangeable for $g$: $1\ \mathrm{N\,kg^{-1}} = 1\ \mathrm{kg\,m\,s^{-2}\,kg^{-1}} = 1\ \mathrm{m\,s^{-2}}$.

这也是 $g$ 的单位 $\mathrm{N\,kg^{-1}}$ 与 $\mathrm{m\,s^{-2}}$ 可互换的原因：$1\ \mathrm{N\,kg^{-1}} = 1\ \mathrm{kg\,m\,s^{-2}\,kg^{-1}} = 1\ \mathrm{m\,s^{-2}}$。

An astronaut moves to an altitude equal to one Earth radius above the surface (so $r = 2R$). The field strength there compared with the surface value is:宇航员升至距表面一个地球半径处（即 $r = 2R$）。该处场强与表面值相比为：
D1.2 · Q1
$\tfrac{1}{2}$ of the surface value表面值的 $\tfrac{1}{2}$
$\tfrac{1}{4}$ of the surface value表面值的 $\tfrac{1}{4}$
$\tfrac{1}{8}$ of the surface value表面值的 $\tfrac{1}{8}$
The same value与表面值相同
$g \propto 1/r^{2}$. Going from $r = R$ to $r = 2R$ multiplies $g$ by $(R/2R)^{2} = 1/4$.$g \propto 1/r^{2}$。由 $r = R$ 到 $r = 2R$，$g$ 乘以 $(R/2R)^{2} = 1/4$。
Use $r$ measured from the centre. Doubling $r$ gives a factor $1/2^{2} = 1/4$, not $1/2$.$r$ 从中心起算。$r$ 加倍给出系数 $1/2^{2} = 1/4$，不是 $1/2$。
Planet X has twice Earth's mass and twice Earth's radius. Its surface gravitational field strength compared with Earth's is:行星 X 的质量是地球的 2 倍、半径是地球的 2 倍。其表面引力场强度与地球相比为：
D1.2 · Q2
Half as large一半
The same相同
Twice as large两倍
Four times as large四倍
$g = GM/R^{2}$. Scaling $M \to 2M$ and $R \to 2R$ gives a factor $2/2^{2} = 2/4 = 1/2$. The radius scaling wins because it is squared.$g = GM/R^{2}$。$M \to 2M$、$R \to 2R$ 给出系数 $2/2^{2} = 1/2$。半径因为平方而占主导。
Both scalings matter: $g \propto M/R^{2}$. The mass factor is $2$ but the radius factor is $1/2^{2} = 1/4$; net $1/2$.两者都要算：$g \propto M/R^{2}$。质量因子为 $2$，半径因子为 $1/2^{2} = 1/4$；净得 $1/2$。
Topic D1.3 · Orbital Motion and Kepler's Third LawTopic D1.3 · 轨道运动与开普勒第三定律

Orbital Motion and Kepler's Third Law轨道运动与开普勒第三定律 D.1 SL+HL

Gravity is the centripetal force. For a circular orbit, set the gravitational force equal to the required centripetal force $\dfrac{m v^{2}}{r}$: $$ \frac{G M m}{r^{2}} = \frac{m v^{2}}{r} \;\Rightarrow\; v = \sqrt{\frac{G M}{r}}. $$ Kepler's third law (orbital period). Substitute $v = \dfrac{2\pi r}{T}$: $$ T^{2} = \frac{4 \pi^{2}}{G M}\, r^{3} \;\Rightarrow\; T^{2} \propto r^{3}. $$
  • The orbiting mass $m$ cancels: orbital speed and period depend only on $M$ and $r$.
  • Larger orbits are slower ($v \propto 1/\sqrt{r}$) and take longer ($T \propto r^{3/2}$).
引力即向心力。对圆轨道，令引力等于所需向心力 $\dfrac{m v^{2}}{r}$： $$ \frac{G M m}{r^{2}} = \frac{m v^{2}}{r} \;\Rightarrow\; v = \sqrt{\frac{G M}{r}}. $$ 开普勒第三定律（轨道周期）。代入 $v = \dfrac{2\pi r}{T}$： $$ T^{2} = \frac{4 \pi^{2}}{G M}\, r^{3} \;\Rightarrow\; T^{2} \propto r^{3}. $$
  • 绕行质量 $m$ 被约去：轨道速度与周期只取决于 $M$ 与 $r$。
  • 轨道越大越慢（$v \propto 1/\sqrt{r}$），周期越长（$T \propto r^{3/2}$）。
Worked Example D1.3 (orbital speed and period)D1.3 例题（轨道速度与周期）

A satellite orbits Earth ($M = 6.0 \times 10^{24}\ \mathrm{kg}$) in a circular orbit of radius $r = 7.0 \times 10^{6}\ \mathrm{m}$ (about $630\ \mathrm{km}$ altitude). Find its orbital speed and period.一卫星沿半径 $r = 7.0 \times 10^{6}\ \mathrm{m}$（约 $630\ \mathrm{km}$ 高度）的圆轨道绕地球（$M = 6.0 \times 10^{24}\ \mathrm{kg}$）运行。求其轨道速度与周期。

Identify. Circular orbit: use $v = \sqrt{GM/r}$, then $T = 2\pi r / v$.

识别。圆轨道：用 $v = \sqrt{GM/r}$，再用 $T = 2\pi r / v$。

Speed.

速度。

$$ v = \sqrt{\frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})}{7.0 \times 10^{6}}} \approx \sqrt{5.72 \times 10^{7}} \approx 7.6 \times 10^{3}\ \mathrm{m\,s^{-1}}. $$

Period.

周期。

$$ T = \frac{2 \pi r}{v} = \frac{2\pi (7.0 \times 10^{6})}{7.6 \times 10^{3}} \approx 5.8 \times 10^{3}\ \mathrm{s} \approx 97\ \mathrm{min}. $$

Evaluate. About $7.6\ \mathrm{km\,s^{-1}}$ and a $97$-minute period are typical of low Earth orbit (the ISS is similar). The satellite's own mass never entered the calculation.

评估。约 $7.6\ \mathrm{km\,s^{-1}}$、周期约 $97$ 分钟，是低地轨道的典型值（国际空间站类似）。卫星自身质量自始至终未参与计算。

Going deeper: deriving Kepler's third law cleanly深入：干净地推导开普勒第三定律

Start from the force–centripetal balance and write the speed as circumference over period, $v = 2\pi r / T$:

从"引力=向心力"出发，把速度写成周长除以周期 $v = 2\pi r / T$：

$$ \frac{G M m}{r^{2}} = \frac{m v^{2}}{r} = \frac{m}{r}\left(\frac{2\pi r}{T}\right)^{2} = \frac{4 \pi^{2} m r}{T^{2}}. $$

Cancel $m$ and $r$, then rearrange:

约去 $m$ 与 $r$，再整理：

$$ \frac{G M}{r^{2}} = \frac{4 \pi^{2} r}{T^{2}} \;\Rightarrow\; T^{2} = \frac{4 \pi^{2}}{G M}\, r^{3}. $$

The constant $4\pi^2/(GM)$ depends only on the central mass, so all satellites of the same planet share one $T^2/r^3$ ratio. Kepler found this empirically for the planets in 1619; Newton derived it from $1/r^2$ gravity sixty years later. For elliptical orbits, $r$ is replaced by the semi-major axis $a$.

常数 $4\pi^2/(GM)$ 只依赖中心质量，故同一行星的所有卫星共享一个 $T^2/r^3$ 比值。开普勒于 1619 年凭经验对行星发现此律；牛顿在六十年后由 $1/r^2$ 引力推导出来。对椭圆轨道，$r$ 换为半长轴 $a$。

A satellite is moved to a new circular orbit with four times the radius. Its orbital speed becomes:卫星被移到半径为原来 4 倍的新圆轨道。其轨道速度变为：
D1.3 · Q1
$4$ times the original原来的 $4$ 倍
$2$ times the original原来的 $2$ 倍
Half the original原来的一半
Unchanged不变
$v = \sqrt{GM/r} \propto 1/\sqrt{r}$. Quadrupling $r$ multiplies $v$ by $1/\sqrt{4} = 1/2$.$v = \sqrt{GM/r} \propto 1/\sqrt{r}$。$r$ 变 $4$ 倍使 $v$ 乘以 $1/\sqrt{4} = 1/2$。
Orbital speed falls with radius: $v \propto r^{-1/2}$. A factor-$4$ radius gives a factor-$1/2$ speed.轨道速度随半径减小：$v \propto r^{-1/2}$。半径变 $4$ 倍，速度变 $1/2$。
Planet B orbits its star with a semi-major axis $4$ times that of planet A. The ratio of their orbital periods $T_{B}/T_{A}$ is:行星 B 绕其恒星的半长轴是行星 A 的 $4$ 倍。它们轨道周期之比 $T_{B}/T_{A}$ 为：
D1.3 · Q2
$4$$4$
$8$$8$
$16$$16$
$2$$2$
$T^{2} \propto r^{3}$ so $T \propto r^{3/2}$. Then $T_{B}/T_{A} = 4^{3/2} = (\sqrt{4})^{3} = 2^{3} = 8$.$T^{2} \propto r^{3}$ 故 $T \propto r^{3/2}$。于是 $T_{B}/T_{A} = 4^{3/2} = 2^{3} = 8$。
Use Kepler's third law $T^{2} \propto r^{3}$, so $T \propto r^{3/2}$. With $r$ ratio $4$: $4^{3/2} = 8$.用开普勒第三定律 $T^{2} \propto r^{3}$，故 $T \propto r^{3/2}$。半径比为 $4$：$4^{3/2} = 8$。
Topic D1.4 · Gravitational Potential and Potential Energy HLTopic D1.4 · 引力势与引力势能 HL

Gravitational Potential and Potential Energy引力势与引力势能 HL D.1 HL

HL extensionHL 扩展 This whole section is HL-only. SL students may skip it; the SL core ends at D1.3.本节全部为 HL 专属。SL 学生可跳过；SL 核心止于 D1.3。
Potential = potential energy per unit mass. From the data booklet, V_g = -GM/r and E_p = -GMm/r: $$ V_{g} = -\frac{G M}{r}, \qquad E_{p} = m V_{g} = -\frac{G M m}{r}. $$
  • Both are negative and approach $0$ as $r \to \infty$ (the chosen zero point). Bringing a mass in from infinity releases energy, so $E_p$ drops below zero.
  • $V_g$ is a scalar (units $\mathrm{J\,kg^{-1}}$); add potentials from several sources by simple addition, no vectors.
  • $V_g \propto 1/r$ falls off more slowly than field strength $g \propto 1/r^{2}$.
Equipotentials are surfaces of constant $V_g$ — concentric spheres around a point mass. No work is done moving a mass along an equipotential. Field lines cross equipotentials at right angles.
势 = 单位质量的势能。由数据手册，V_g = -GM/r 与 E_p = -GMm/r： $$ V_{g} = -\frac{G M}{r}, \qquad E_{p} = m V_{g} = -\frac{G M m}{r}. $$
  • 二者均为负，当 $r \to \infty$ 时趋于 $0$（所选零点）。把质量从无穷远移入会释放能量，故 $E_p$ 降到零以下。
  • $V_g$ 是标量（单位 $\mathrm{J\,kg^{-1}}$）；多个源的势直接相加即可，无需矢量。
  • $V_g \propto 1/r$ 比场强 $g \propto 1/r^{2}$ 衰减得慢。
等势面是 $V_g$ 恒定的曲面——点质量周围的同心球面。沿等势面移动质量不做功。场线与等势面正交。
Worked Example D1.4 (potential and potential energy) HLD1.4 例题（势与势能）HL

At Earth's surface ($M = 6.0 \times 10^{24}\ \mathrm{kg}$, $R = 6.4 \times 10^{6}\ \mathrm{m}$), find the gravitational potential $V_g$, and the potential energy of a $1200\ \mathrm{kg}$ spacecraft sitting on the surface.在地球表面（$M = 6.0 \times 10^{24}\ \mathrm{kg}$，$R = 6.4 \times 10^{6}\ \mathrm{m}$），求引力势 $V_g$，以及停在表面的 $1200\ \mathrm{kg}$ 航天器的势能。

Identify. Use $V_g = -GM/r$ with $r = R$, then $E_p = m V_g$.

识别。用 $V_g = -GM/r$ 取 $r = R$，再用 $E_p = m V_g$。

Potential.

势。

$$ V_{g} = -\frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})}{6.4 \times 10^{6}} \approx -6.3 \times 10^{7}\ \mathrm{J\,kg^{-1}}. $$

Potential energy.

势能。

$$ E_{p} = m V_{g} = (1200)(-6.3 \times 10^{7}) \approx -7.5 \times 10^{10}\ \mathrm{J}. $$

Evaluate. Negative, as it must be: the spacecraft is bound. To free it completely (move it to infinity, $E_p = 0$) you must supply at least $+7.5 \times 10^{10}\ \mathrm{J}$.

评估。为负，必然如此：航天器处于束缚态。要彻底脱离（移到无穷远，$E_p = 0$）至少需供给 $+7.5 \times 10^{10}\ \mathrm{J}$。

Going deeper: why the zero of potential is at infinity深入：势的零点为何取在无穷远

Potential at a point is the work done per unit mass to bring a test mass from infinity to that point. Gravity does positive work as the mass falls inward, so the external agent does negative work, and the potential ends up negative. Choosing $V_g = 0$ at infinity is the natural reference because the force vanishes there.

某点的势是把单位测试质量从无穷远移到该点所做的功。质量内落时引力做正功，故外界做负功，势最终为负。把 $V_g = 0$ 取在无穷远是自然参考，因为那里力为零。

Near a surface, the familiar $E_p = mgh$ is the small-$\Delta r$ approximation of $-GMm/r$: over a height change $h \ll R$, $\Delta E_p \approx \dfrac{GMm}{R^2}h = mgh$. The IB formula is the exact, full-range version.

在表面附近，熟悉的 $E_p = mgh$ 是 $-GMm/r$ 在小 $\Delta r$ 下的近似：高度变化 $h \ll R$ 时，$\Delta E_p \approx \dfrac{GMm}{R^2}h = mgh$。IB 公式是精确的全程版本。

HL Which statement about gravitational potential $V_g$ is correct?HL 关于引力势 $V_g$ 的哪项陈述正确？
D1.4 · Q1
It is a vector pointing toward the mass它是指向质量的矢量
It is positive and increases with $r$它为正且随 $r$ 增大
It is zero at the surface and grows with height它在表面为零并随高度增大
It is a negative scalar that approaches zero as $r \to \infty$它是负的标量，当 $r \to \infty$ 时趋于零
$V_g = -GM/r$ is a scalar, negative everywhere, and rises toward $0$ as $r \to \infty$. It gets less negative (larger) as you move away from the mass.$V_g = -GM/r$ 是标量，处处为负，随 $r \to \infty$ 升向 $0$。远离质量时它变得不那么负（即变大）。
Potential is a scalar, not a vector. With the zero at infinity, $V_g = -GM/r$ is negative everywhere and tends to $0$ from below.势是标量而非矢量。零点取在无穷远时，$V_g = -GM/r$ 处处为负，从下方趋于 $0$。
HL A mass is moved from one point to another on the same equipotential surface. The work done by the gravitational field is:HL 一质量在同一等势面上从一点移到另一点。引力场所做的功为：
D1.4 · Q2
Zero为零
Equal to $m V_g$等于 $m V_g$
Always positive恒为正
Equal to $-GMm/r$等于 $-GMm/r$
Work done $= m\,\Delta V_g$. On one equipotential $\Delta V_g = 0$, so no work is done — motion is perpendicular to the field there.做功 $= m\,\Delta V_g$。同一等势面上 $\Delta V_g = 0$，故不做功——此时运动方向与场垂直。
Work depends on the change in potential, $W = m\,\Delta V_g$. Along one equipotential $\Delta V_g = 0$.功取决于势的变化 $W = m\,\Delta V_g$。沿同一等势面 $\Delta V_g = 0$。
Topic D1.5 · Field–Potential Relationship HLTopic D1.5 · 场与势的关系 HL

Field–Potential Relationship and Work Done场与势的关系及做功 HL D.1 HL

HL extensionHL 扩展 This whole section is HL-only.本节全部为 HL 专属。
Field is the negative potential gradient. From the data booklet, g = -ΔV_g/Δr: $$ g = -\frac{\Delta V_{g}}{\Delta r}. $$
  • The field points "downhill", from high (less negative) to low (more negative) potential. The minus sign encodes that.
  • Steeper potential change with distance $\Rightarrow$ stronger field. On a $V_g$-vs-$r$ graph, $g$ is minus the slope.
  • Closely spaced equipotentials mean a strong field; widely spaced ones mean a weak field.
Work done moving a mass. Between two points, the work done against the field equals the change in potential energy: $$ W = m\,\Delta V_{g} = G M m\!\left(\frac{1}{r_{1}} - \frac{1}{r_{2}}\right). $$
场是势的负梯度。由数据手册，g = -ΔV_g/Δr： $$ g = -\frac{\Delta V_{g}}{\Delta r}. $$
  • 场指向"下坡"方向，从高势（不那么负）指向低势（更负）。负号正是表达这一点。
  • 势随距离变化越陡 $\Rightarrow$ 场越强。在 $V_g$-$r$ 图上，$g$ 是斜率的相反数。
  • 等势面密集表示场强；稀疏表示场弱。
移动质量所做的功。在两点之间，克服场所做的功等于势能的变化： $$ W = m\,\Delta V_{g} = G M m\!\left(\frac{1}{r_{1}} - \frac{1}{r_{2}}\right). $$
Worked Example D1.5 (work to raise a satellite) HLD1.5 例题（抬升卫星所需的功）HL

How much work is needed to raise a $500\ \mathrm{kg}$ satellite from Earth's surface ($r_{1} = 6.4 \times 10^{6}\ \mathrm{m}$) to an orbital radius $r_{2} = 1.3 \times 10^{7}\ \mathrm{m}$? Use $M = 6.0 \times 10^{24}\ \mathrm{kg}$. (Ignore kinetic energy; find the change in potential energy only.)把 $500\ \mathrm{kg}$ 的卫星从地球表面（$r_{1} = 6.4 \times 10^{6}\ \mathrm{m}$）抬升到轨道半径 $r_{2} = 1.3 \times 10^{7}\ \mathrm{m}$ 需做多少功？取 $M = 6.0 \times 10^{24}\ \mathrm{kg}$。（忽略动能，仅求势能变化。）

Identify. $W = \Delta E_p = GMm\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$, the work against the field.

识别。$W = \Delta E_p = GMm\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$，即克服场所做的功。

Set up. $GMm = (6.67 \times 10^{-11})(6.0 \times 10^{24})(500) \approx 2.0 \times 10^{17}$.

列式。$GMm = (6.67 \times 10^{-11})(6.0 \times 10^{24})(500) \approx 2.0 \times 10^{17}$。

Execute. $\dfrac{1}{r_1} - \dfrac{1}{r_2} = \dfrac{1}{6.4 \times 10^{6}} - \dfrac{1}{1.3 \times 10^{7}} \approx (1.56 - 0.77) \times 10^{-7} = 7.9 \times 10^{-8}$.

计算。$\dfrac{1}{r_1} - \dfrac{1}{r_2} = \dfrac{1}{6.4 \times 10^{6}} - \dfrac{1}{1.3 \times 10^{7}} \approx (1.56 - 0.77) \times 10^{-7} = 7.9 \times 10^{-8}$。

$$ W \approx (2.0 \times 10^{17})(7.9 \times 10^{-8}) \approx 1.6 \times 10^{10}\ \mathrm{J}. $$

Evaluate. Positive work, since the satellite moves to a higher (less negative) potential. Using the crude $mgh$ here would overestimate the cost, because $g$ weakens with altitude — the exact $1/r$ formula is needed.

评估。做正功，因为卫星移向更高（不那么负）的势。此处用粗糙的 $mgh$ 会高估，因为 $g$ 随高度减弱——需用精确的 $1/r$ 公式。

Going deeper: g as minus the slope of the potential深入：g 是势曲线斜率的相反数

In the continuous limit, $g = -\dfrac{dV_g}{dr}$. Differentiate $V_g = -GM/r$:

在连续极限下，$g = -\dfrac{dV_g}{dr}$。对 $V_g = -GM/r$ 求导：

$$ g = -\frac{d}{dr}\!\left(-\frac{GM}{r}\right) = -\frac{GM}{r^{2}}. $$

The magnitude is $GM/r^2$, exactly the field strength from D1.2, and the minus sign confirms the field points toward the mass (decreasing $r$). This is the single most important consistency check in the unit: field and potential are two views of the same thing, linked by a gradient.

大小为 $GM/r^2$，恰是 D1.2 的场强，负号确认场指向质量（$r$ 减小方向）。这是本单元最重要的一致性检验：场与势是同一事物的两种视角，由梯度相联。

HL On a graph of gravitational potential $V_g$ against distance $r$, the field strength $g$ at a point equals:HL 在引力势 $V_g$ 对距离 $r$ 的图上，某点的场强 $g$ 等于：
D1.5 · Q1
The value of $V_g$ at that point该点 $V_g$ 的值
The area under the curve up to that point到该点为止曲线下的面积
Minus the gradient (slope) of the curve at that point该点曲线斜率的相反数
The gradient of the curve, with no sign change曲线斜率，不变号
$g = -\Delta V_g/\Delta r$, so on a $V_g$-$r$ graph the field strength is minus the local slope. The minus sign makes $g$ point toward decreasing potential.$g = -\Delta V_g/\Delta r$，故在 $V_g$-$r$ 图上场强为局部斜率的相反数。负号使 $g$ 指向势减小方向。
Field is the negative potential gradient, $g = -\Delta V_g/\Delta r$ — minus the slope, not the value or the area.场是势的负梯度，$g = -\Delta V_g/\Delta r$——是斜率的相反数，不是值也不是面积。
HL A $2.0\ \mathrm{kg}$ mass is moved from a point where $V_g = -40\ \mathrm{MJ\,kg^{-1}}$ to a point where $V_g = -25\ \mathrm{MJ\,kg^{-1}}$. The work done against the field is:HL 一 $2.0\ \mathrm{kg}$ 质量从 $V_g = -40\ \mathrm{MJ\,kg^{-1}}$ 处移到 $V_g = -25\ \mathrm{MJ\,kg^{-1}}$ 处。克服场所做的功为：
D1.5 · Q2
$-30\ \mathrm{MJ}$$-30\ \mathrm{MJ}$
$+30\ \mathrm{MJ}$$+30\ \mathrm{MJ}$
$+130\ \mathrm{MJ}$$+130\ \mathrm{MJ}$
$+15\ \mathrm{MJ}$$+15\ \mathrm{MJ}$
$W = m\,\Delta V_g = (2.0)\big[(-25) - (-40)\big] = (2.0)(+15) = +30\ \mathrm{MJ}$. Positive because the mass moves to a higher (less negative) potential.$W = m\,\Delta V_g = (2.0)\big[(-25) - (-40)\big] = (2.0)(+15) = +30\ \mathrm{MJ}$。为正，因为质量移向更高（不那么负）的势。
$W = m\,\Delta V_g$ with $\Delta V_g = V_2 - V_1 = (-25) - (-40) = +15\ \mathrm{MJ\,kg^{-1}}$. Multiply by $2.0\ \mathrm{kg}$.$W = m\,\Delta V_g$，其中 $\Delta V_g = V_2 - V_1 = (-25) - (-40) = +15\ \mathrm{MJ\,kg^{-1}}$。乘以 $2.0\ \mathrm{kg}$。
Topic D1.6 · Escape Speed and Orbit Energy HLTopic D1.6 · 逃逸速度与轨道能量 HL

Escape Speed, Bound vs Unbound, Orbit Energy逃逸速度、束缚与非束缚、轨道能量 HL D.1 HL

HL extensionHL 扩展 This whole section is HL-only.本节全部为 HL 专属。
Escape speed. The minimum launch speed to reach infinity with zero leftover kinetic energy. Set total energy to zero, $\tfrac{1}{2} m v_{esc}^{2} - \dfrac{GMm}{r} = 0$. From the data booklet, v_esc = √(2GM/r): $$ v_{esc} = \sqrt{\frac{2 G M}{r}}. $$
  • Independent of the escaping mass $m$. For Earth's surface, $v_{esc} \approx 1.1 \times 10^{4}\ \mathrm{m\,s^{-1}}$ ($11\ \mathrm{km\,s^{-1}}$).
  • Note $v_{esc} = \sqrt{2}\, v_{orbit}$: escape speed is $\sqrt2$ times the circular orbital speed at the same $r$.
Bound vs unbound. Total energy $E = E_k + E_p$. If $E < 0$ the object is bound (closed orbit); if $E \ge 0$ it is unbound and escapes. Orbit energy. For a circular orbit, $E_k = \dfrac{GMm}{2r}$ and the total is $$ E = E_{k} + E_{p} = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}. $$
逃逸速度。恰好到达无穷远且剩余动能为零所需的最小发射速度。令总能量为零，$\tfrac{1}{2} m v_{esc}^{2} - \dfrac{GMm}{r} = 0$。由数据手册，v_esc = √(2GM/r)： $$ v_{esc} = \sqrt{\frac{2 G M}{r}}. $$
  • 与逃逸物体质量 $m$ 无关。地球表面 $v_{esc} \approx 1.1 \times 10^{4}\ \mathrm{m\,s^{-1}}$（$11\ \mathrm{km\,s^{-1}}$）。
  • 注意 $v_{esc} = \sqrt{2}\, v_{orbit}$：逃逸速度是同一 $r$ 处圆轨道速度的 $\sqrt2$ 倍。
束缚与非束缚。总能量 $E = E_k + E_p$。若 $E < 0$ 则束缚（闭合轨道）；若 $E \ge 0$ 则非束缚并逃逸。 轨道能量。对圆轨道，$E_k = \dfrac{GMm}{2r}$，总能量为 $$ E = E_{k} + E_{p} = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}. $$
Worked Example D1.6 (escape speed from a moon) HLD1.6 例题（从卫星上逃逸的速度）HL

The Moon has mass $M = 7.3 \times 10^{22}\ \mathrm{kg}$ and radius $R = 1.7 \times 10^{6}\ \mathrm{m}$. Find the escape speed from its surface, and state whether a probe launched at $2.0\ \mathrm{km\,s^{-1}}$ would escape.月球质量 $M = 7.3 \times 10^{22}\ \mathrm{kg}$，半径 $R = 1.7 \times 10^{6}\ \mathrm{m}$。求从其表面逃逸的速度，并判断以 $2.0\ \mathrm{km\,s^{-1}}$ 发射的探测器能否逃逸。

Identify. Use $v_{esc} = \sqrt{2GM/r}$ with $r = R$.

识别。用 $v_{esc} = \sqrt{2GM/r}$ 取 $r = R$。

Set up.

列式。

$$ v_{esc} = \sqrt{\frac{2 (6.67 \times 10^{-11})(7.3 \times 10^{22})}{1.7 \times 10^{6}}}. $$

Execute. Inside the root: $\dfrac{9.74 \times 10^{12}}{1.7 \times 10^{6}} \approx 5.7 \times 10^{6}$, so $v_{esc} \approx 2.4 \times 10^{3}\ \mathrm{m\,s^{-1}} = 2.4\ \mathrm{km\,s^{-1}}$.

计算。根号内：$\dfrac{9.74 \times 10^{12}}{1.7 \times 10^{6}} \approx 5.7 \times 10^{6}$，故 $v_{esc} \approx 2.4 \times 10^{3}\ \mathrm{m\,s^{-1}} = 2.4\ \mathrm{km\,s^{-1}}$。

Evaluate. A launch at $2.0\ \mathrm{km\,s^{-1}}$ is below $2.4\ \mathrm{km\,s^{-1}}$, so the probe is still bound ($E < 0$): it rises, slows, and falls back. It does not escape.

评估。$2.0\ \mathrm{km\,s^{-1}}$ 的发射速度低于 $2.4\ \mathrm{km\,s^{-1}}$，故探测器仍束缚（$E < 0$）：上升、减速、落回。无法逃逸。

Going deeper: deriving escape speed and orbit energy深入：推导逃逸速度与轨道能量

Escape speed. "Just escaping" means reaching $r \to \infty$ with $v \to 0$, where both $E_k$ and $E_p$ vanish, so total energy $E = 0$. By conservation, at launch:

逃逸速度。"恰好逃逸"指到达 $r \to \infty$ 时 $v \to 0$，此处 $E_k$ 与 $E_p$ 都为零，故总能量 $E = 0$。由守恒，在发射时：

$$ \frac{1}{2} m v_{esc}^{2} - \frac{GMm}{r} = 0 \;\Rightarrow\; v_{esc} = \sqrt{\frac{2GM}{r}}. $$

Orbit energy. For a circular orbit, gravity supplies the centripetal force: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$, so $mv^2 = \dfrac{GMm}{r}$ and $E_k = \tfrac12 mv^2 = \dfrac{GMm}{2r}$. Adding $E_p = -\dfrac{GMm}{r}$:

轨道能量。圆轨道中引力提供向心力：$\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$，故 $mv^2 = \dfrac{GMm}{r}$，$E_k = \tfrac12 mv^2 = \dfrac{GMm}{2r}$。加上 $E_p = -\dfrac{GMm}{r}$：

$$ E = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}. $$

The total is negative (bound) and equals minus the kinetic energy. A neat consequence: $E_p = 2E$ and $E_k = -E$. To raise a satellite to a higher orbit you must add energy, which paradoxically reduces its orbital speed.

总能量为负（束缚），且等于动能的相反数。一个漂亮的推论：$E_p = 2E$、$E_k = -E$。把卫星抬到更高轨道必须补充能量，但这反而降低其轨道速度。

HL Escape speed from a planet's surface does NOT depend on:HL 从行星表面逃逸的速度不取决于：
D1.6 · Q1
The mass of the escaping object逃逸物体的质量
The mass of the planet行星的质量
The radius of the planet行星的半径
The gravitational constant $G$引力常量 $G$
$v_{esc} = \sqrt{2GM/r}$ contains $G$, $M$ (planet) and $r$, but not $m$ (the escaping object). Its mass cancels because both $E_k$ and $E_p$ are proportional to $m$.$v_{esc} = \sqrt{2GM/r}$ 含 $G$、$M$（行星）与 $r$，但不含 $m$（逃逸物体）。其质量约去，因为 $E_k$ 与 $E_p$ 都正比于 $m$。
Set $\tfrac12 m v^2 = GMm/r$; the $m$ cancels, leaving $v_{esc} = \sqrt{2GM/r}$, which depends on $G$, $M$ and $r$ only.令 $\tfrac12 m v^2 = GMm/r$，$m$ 约去，得 $v_{esc} = \sqrt{2GM/r}$，只依赖 $G$、$M$ 与 $r$。
HL A satellite in a circular orbit of radius $r$ has total energy $E = -GMm/2r$. To move it to a larger orbit, an engineer must:HL 半径为 $r$ 的圆轨道卫星总能量 $E = -GMm/2r$。要把它移到更大轨道，工程师必须：
D1.6 · Q2
Remove energy; the satellite then speeds up移除能量；卫星随后加速
Add energy; the satellite then speeds up补充能量；卫星随后加速
Add energy; the satellite then slows down补充能量；卫星随后减速
Do nothing; energy is unchanged in a larger orbit什么都不做；更大轨道能量不变
A larger $r$ makes $E = -GMm/2r$ less negative, i.e. larger — so energy must be added. But $v = \sqrt{GM/r}$ falls with $r$, so the satellite ends up slower. Higher orbit, more total energy, lower speed.更大的 $r$ 使 $E = -GMm/2r$ 不那么负（即更大）——故必须补充能量。但 $v = \sqrt{GM/r}$ 随 $r$ 减小，故卫星最终更慢。轨道更高、总能量更大、速度更低。
$E = -GMm/2r$ increases (becomes less negative) with $r$, so energy is added. Yet $v \propto 1/\sqrt{r}$ decreases — the higher orbit is slower.$E = -GMm/2r$ 随 $r$ 增大（变得不那么负），故补充能量。但 $v \propto 1/\sqrt{r}$ 减小——更高轨道更慢。
Exam Tactics考试技巧

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

Measure r from the centre (every paper)r 从中心起算（每张试卷）
  • For orbits and fields, $r$ is centre-to-centre, not the altitude above the surface. An orbit "$600\ \mathrm{km}$ up" has $r = R_{\text{Earth}} + 600\ \mathrm{km}$.
  • 对轨道与场，$r$ 是中心到中心，不是距表面的高度。"$600\ \mathrm{km}$ 高"的轨道 $r = R_{\text{地}} + 600\ \mathrm{km}$。
  • Surface values set $r = R$ (the planet radius). Mixing altitude and $r$ is the single most common D.1 slip.
  • 表面值取 $r = R$（行星半径）。混淆高度与 $r$ 是 D.1 最常见的失误。
Know your scalings: 1/r vs 1/r²记牢标度：1/r 与 1/r²
  • Force and field strength go as $1/r^{2}$; potential and potential energy go as $1/r$. Quote the right power before substituting.
  • 力与场强按 $1/r^{2}$；势与势能按 $1/r$。代入前先确认正确的幂次。
  • Orbital speed $v \propto r^{-1/2}$ and period $T \propto r^{3/2}$. Most ratio questions reduce to one of these four scalings.
  • 轨道速度 $v \propto r^{-1/2}$，周期 $T \propto r^{3/2}$。多数比例题都归结为这四个标度之一。
Signs of potential and energy (HL, Paper 2)势与能量的符号（HL，Paper 2）
  • $V_g$ and $E_p$ are negative; do not drop the minus sign. Moving outward makes them less negative (larger), so $\Delta V_g > 0$ and work done is positive.
  • $V_g$ 与 $E_p$ 为负，不要丢负号。向外移动使其不那么负（变大），故 $\Delta V_g > 0$，做正功。
  • Total orbit energy $E = -GMm/2r$ is negative for bound orbits. $E \ge 0$ means escape.
  • 束缚轨道的总能量 $E = -GMm/2r$ 为负。$E \ge 0$ 表示逃逸。
State the modelling assumption写明建模假设
  • "Spherically symmetric body, treated as a point mass at its centre." Examiners credit this when the question says "explain why we can use $F = GMm/r^2$".
  • "球对称物体，视为中心处的点质量。"当题目要求"解释为何可用 $F = GMm/r^2$"时，写明此句可得分。
  • For escape-speed problems, state "ignore air resistance and the planet's rotation". Otherwise the energy-conservation argument is not clean.
  • 逃逸速度题应写明"忽略空气阻力与行星自转"。否则能量守恒论证不够严谨。
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Newton's law of gravitation?万有引力定律？
$$F = \frac{GMm}{r^{2}}$$
Value of $G$?$G$ 的值？
$$6.67 \times 10^{-11}\ \mathrm{N\,m^{2}\,kg^{-2}}$$
Gravitational field strength?引力场强度？
$$g = \frac{F}{m} = \frac{GM}{r^{2}}$$
Units of $g$?$g$ 的单位？
$\mathrm{N\,kg^{-1}}$, equal to $\mathrm{m\,s^{-2}}$.$\mathrm{N\,kg^{-1}}$，等于 $\mathrm{m\,s^{-2}}$。
Circular orbital speed?圆轨道速度？
$$v = \sqrt{\frac{GM}{r}}$$
Kepler's third law?开普勒第三定律？
$$T^{2} = \frac{4\pi^{2}}{GM}\,r^{3} \;\Rightarrow\; T^{2} \propto r^{3}$$
Gravity = centripetal force gives? 引力=向心力给出？
$$\frac{GMm}{r^{2}} = \frac{mv^{2}}{r}$$
Gravitational potential? HL引力势？HL
$$V_{g} = -\frac{GM}{r}$$Scalar; negative; $\to 0$ at infinity.标量；为负；无穷远 $\to 0$。
Gravitational potential energy? HL引力势能？HL
$$E_{p} = -\frac{GMm}{r}$$
Field–potential relationship? HL场与势的关系？HL
$$g = -\frac{\Delta V_{g}}{\Delta r}$$
Escape speed? HL逃逸速度？HL
$$v_{esc} = \sqrt{\frac{2GM}{r}}$$
Total energy of a circular orbit? HL圆轨道总能量？HL
$$E = -\frac{GMm}{2r}$$Negative ⇒ bound.为负 ⇒ 束缚。
Mixed Practice综合练习

Unit D.1 Practice Quiz单元 D.1 练习测验

The gravitational field strength at a planet's surface is $g$. At a distance of $3$ planet-radii from the centre, the field strength is:某行星表面引力场强为 $g$。在距中心 $3$ 个行星半径处，场强为：
Q1
$g/3$
$g/6$
$g/9$
$3g$
$g \propto 1/r^{2}$. Surface is $r = R$; the point is $r = 3R$. Factor $(R/3R)^{2} = 1/9$, so the field is $g/9$.$g \propto 1/r^{2}$。表面 $r = R$；该点 $r = 3R$。系数 $(R/3R)^{2} = 1/9$，故场强为 $g/9$。
Inverse-square: tripling $r$ divides $g$ by $3^{2} = 9$.反平方：$r$ 变 $3$ 倍使 $g$ 除以 $3^{2} = 9$。
A satellite orbits a planet of mass $M$ at radius $r$. Which quantity is independent of the satellite's own mass?卫星以半径 $r$ 绕质量 $M$ 的行星运动。下列哪个量与卫星自身质量无关？
Q2
Its gravitational potential energy其引力势能
Its orbital speed and period其轨道速度与周期
The gravitational force on it它所受的引力
Its total orbital energy其总轨道能量
$v = \sqrt{GM/r}$ and $T^2 = 4\pi^2 r^3/(GM)$ contain no $m$. Force ($GMm/r^2$), $E_p$ ($-GMm/r$) and total energy ($-GMm/2r$) all scale with $m$.$v = \sqrt{GM/r}$ 与 $T^2 = 4\pi^2 r^3/(GM)$ 都不含 $m$。力（$GMm/r^2$）、$E_p$（$-GMm/r$）与总能量（$-GMm/2r$）都正比于 $m$。
When you set gravity equal to the centripetal force, $m$ cancels — so speed and period depend only on $M$ and $r$.令引力等于向心力时 $m$ 约去——故速度与周期只依赖 $M$ 与 $r$。
A geostationary satellite has period $24\ \mathrm{h}$. A second satellite of the same planet has period $3\ \mathrm{h}$. The ratio of their orbital radii $r_{\text{geo}}/r_{2}$ is closest to:同步卫星周期 $24\ \mathrm{h}$。同一行星另一卫星周期 $3\ \mathrm{h}$。它们轨道半径之比 $r_{\text{geo}}/r_{2}$ 最接近：
Q3
$2$
$8$
$16$
$4$
$T^{2} \propto r^{3}$ so $r \propto T^{2/3}$. The period ratio is $24/3 = 8$, giving $r$ ratio $8^{2/3} = (2^{3})^{2/3} = 2^{2} = 4$.$T^{2} \propto r^{3}$ 故 $r \propto T^{2/3}$。周期比 $24/3 = 8$，得 $r$ 比 $8^{2/3} = 2^{2} = 4$。
From $T^2 \propto r^3$, $r \propto T^{2/3}$. With period ratio $8$: $8^{2/3} = 4$.由 $T^2 \propto r^3$，$r \propto T^{2/3}$。周期比 $8$：$8^{2/3} = 4$。
HL Two equal point masses $M$ are separated by $2d$. At the midpoint between them, the gravitational potential and field strength are respectively:HL 两个相等点质量 $M$ 相距 $2d$。在它们连线中点处，引力势与场强分别为：
Q4
$V_g = -2GM/d$ (nonzero), field $= 0$$V_g = -2GM/d$（非零），场 $= 0$
Both zero二者皆为零
$V_g = 0$, field $= 0$$V_g = 0$，场 $= 0$
$V_g = -GM/d$, field nonzero$V_g = -GM/d$，场非零
Potentials are scalars and add: $V_g = -GM/d - GM/d = -2GM/d$ (nonzero). Fields are vectors of equal magnitude pointing opposite ways, so they cancel: field $= 0$.势是标量，相加：$V_g = -GM/d - GM/d = -2GM/d$（非零）。场是等大反向的矢量，相互抵消：场 $= 0$。
Add scalar potentials (they reinforce, giving $-2GM/d$) but vector fields cancel by symmetry (giving zero). They behave differently.标量势相加（相互加强，得 $-2GM/d$），但矢量场因对称相消（得零）。二者表现不同。
HL The escape speed from a planet's surface is $v_{esc}$. The orbital speed of a satellite skimming just above the same surface (radius $\approx R$) is:HL 某行星表面逃逸速度为 $v_{esc}$。紧贴同一表面（半径 $\approx R$）掠行卫星的轨道速度为：
Q5
$2 v_{esc}$$2 v_{esc}$
$v_{esc}$$v_{esc}$
$v_{esc}/\sqrt{2}$$v_{esc}/\sqrt{2}$
$v_{esc}/2$$v_{esc}/2$
$v_{orbit} = \sqrt{GM/R}$ and $v_{esc} = \sqrt{2GM/R} = \sqrt{2}\,v_{orbit}$. So $v_{orbit} = v_{esc}/\sqrt{2}$.$v_{orbit} = \sqrt{GM/R}$，$v_{esc} = \sqrt{2GM/R} = \sqrt{2}\,v_{orbit}$。故 $v_{orbit} = v_{esc}/\sqrt{2}$。
Escape speed is $\sqrt2$ times the orbital speed at the same radius, so the orbital speed is $v_{esc}/\sqrt2$.逃逸速度是同半径处轨道速度的 $\sqrt2$ 倍，故轨道速度为 $v_{esc}/\sqrt2$。
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IB Physics HL · Unit D.1 Gravitational fields · v1 · First Assessment 2025IB 物理 HL · 单元 D.1 引力场 · v1 · 首次考核 2025