Gravitational potential · 引力势
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| escape velocity/eˈskeɪp vəˈlɒsɪti/ | 逃逸速度 | táo yì sù dù |
| gravitational potential/ˌɡrævɪˈteɪʃənl pəˈtenʃl/ | 引力势 | yǐn lì shì |
| infinity/ɪnˈfɪnɪti/ | 无穷远 | wú qióng yuǎn |
| gravitational potential energy/ˌɡrævɪˈteɪʃənl pəˈtenʃl ˈenədʒi/ | 重力势能 | zhòng lì shì néng |
| tightly bound/ˈtaɪtli baʊnd/ | 紧密束缚 | jǐn mì shù fù |
How fast to escape forever?
- Throw a ball up and it comes back. Throw it fast enough and it never returns.
- That "never returns" speed is the escape velocity 逃逸速度.
- To find it we need gravitational potential 引力势.
要多快才能永远逃离?
- 把球向上扔,它会落回来。扔得足够快,它就永远不回来。
- 这个"永不返回"的速率就是 逃逸速度(escape velocity)。
- 要求它,我们需要 引力势(gravitational potential)。
Gravitational potential
- $\phi$ is the work done per unit mass to bring a small test mass from infinity 无穷远 to a point: $\phi = \dfrac{W}{m}$. Unit: $\dfrac{\text{J}}{\text{kg}}$.
- Taken as zero at infinity, so $\phi = -\dfrac{GM}{r}$ is negative everywhere else.
引力势
- $\phi$ 是把一个小的试探质量从 无穷远(infinity) 移到某点、单位质量所做的功:$\phi = \dfrac{W}{m}$。单位:$\dfrac{\text{J}}{\text{kg}}$。
- 取无穷远处为 零,所以其他各处 $\phi = -\dfrac{GM}{r}$ 都是 负 的。
Gravitational potential · 引力势
V = −GM / r
Potential ∝ −1/r — deep near the mass, flattening with distance. · 势 ∝ −1/r——靠近质量处很深,随距离变平。
Gravitational potential is: · 引力势是:
Gravity does the work as a mass falls in, so $\phi = -\dfrac{GM}{r}$ is negative, reaching zero only at infinity. · 质量落入时引力做功,所以 $\phi = -\dfrac{GM}{r}$ 是负的,只有在无穷远处才到零。
Gravitational potential is taken to be zero at ____. · 引力势在 ____ 处被取为零。
That choice makes the potential negative everywhere a mass actually is. · 这样的选择使得在质量实际所在的任何地方,势都是负的。
Gravitational potential energy 重力势能
- A mass $m$ at potential $\phi$ has $E_{\text{P}} = m\phi = -\dfrac{GMm}{r}$.
- It is negative; closer masses are more negative — more tightly bound 紧密束缚.
Gravity provides the centripetal force that keeps a planet in a circular orbit
重力势能
- 势为 $\phi$ 处的质量 $m$ 具有 $E_{\text{P}} = m\phi = -\dfrac{GMm}{r}$。
- 它是 负 的;越靠近的质量越负——被 束缚(tightly bound) 得越紧。
引力提供向心力,使行星保持在圆轨道上
Two masses closer together have a more negative (lower) gravitational potential energy. · 相距更近的两个质量有更负(更低)的引力势能。
$E_{\text{P}} = -\dfrac{GMm}{r}$ — smaller $r$ gives a more negative value, i.e. a more tightly bound pair. · $E_{\text{P}} = -\dfrac{GMm}{r}$——更小的 $r$ 给出更负的值,即结合得更紧的一对。
Match each term to the definition the examiner marks. · 把每个术语与评分认可的定义配对。
Two of these are per unit mass and one is per unit mass of WORK. That single word is what separates potential from field strength. · 其中两条都是「单位质量的」,而一条是单位质量的功。就是这一个词把势与场强区分开。
Link with $mg\Delta h$
- For small height changes near the surface, $r$ barely changes, so $\Delta E_{\text{P}} \approx mg\Delta h$.
- For large changes (raising a satellite), use $-\dfrac{GMm}{r}$ at each radius and subtract.
与 $mg\Delta h$ 的联系
- 对于表面附近的 小 高度变化,$r$ 几乎不变,所以 $\Delta E_{\text{P}} \approx mg\Delta h$。
- 对于 大 的变化(把卫星送上轨道),在每个半径处用 $-\dfrac{GMm}{r}$,然后相减。
The simple formula $\Delta E_{\text{P}} = mg\Delta h$ is valid when: · 简单公式 $\Delta E_{\text{P}} = mg\Delta h$ 在以下哪种情况下成立:
It assumes a uniform field. For big changes in $r$, use $-\dfrac{GMm}{r}$ at each radius instead. · 它假设场是均匀的。对于 $r$ 的大变化,改用每个半径处的 $-\dfrac{GMm}{r}$。
Why is gravitational potential always negative? Select all · 所有 that apply. · 引力势为什么总是负的?选出所有适用的。
"Because gravity attracts" on its own is not the marked answer. The mark needs the work argument: zero at infinity, and the field doing the work on the way in. · 只说"因为引力是吸引的"不是评分认可的答案。得分需要功的论证:无穷远为零,而质量移近途中是场在做功。
Worked example: the potential at the surface of Mars
Mars has mass $6.4 \times 10^{23}\ \text{kg}$ and radius $3.4 \times 10^{6}\ \text{m}$. Find the gravitational potential at its surface, and the energy needed to remove $1\ \text{kg}$ from the surface to infinity.
- Potential: $\phi = -\dfrac{GM}{R} = -\dfrac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{3.4 \times 10^{6}} = -1.3 \times 10^{7}\ \dfrac{\text{J}}{\text{kg}}$.
- The unit is asked for: joules per kilogram, because potential is energy per unit mass.
- Energy to escape per kilogram: from $\phi$ up to zero at infinity is $+1.3 \times 10^{7}\ \text{J}$ for each kilogram.
- Check: the minus sign is not optional. It says work must be done on the mass to remove it — the mass is in a "well" $1.3 \times 10^{7}\ \dfrac{\text{J}}{\text{kg}}$ deep.
例题:火星表面的引力势
火星质量 $6.4 \times 10^{23}\ \text{kg}$,半径 $3.4 \times 10^{6}\ \text{m}$。求其表面的引力势,以及把 $1\ \text{kg}$ 从表面移到无穷远所需的能量。
- 引力势: $\phi = -\dfrac{GM}{R} = -\dfrac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{3.4 \times 10^{6}} = -1.3 \times 10^{7}\ \dfrac{\text{J}}{\text{kg}}$。
- 题目要求单位: 焦耳每千克,因为势是 单位质量 的能量。
- 每千克的逃逸能量: 从 $\phi$ 升到无穷远的零,每千克需要 $+1.3 \times 10^{7}\ \text{J}$。
- 检查: 负号不是可有可无的。它说明必须 对 质量做功才能移走它——质量处在一个 $1.3 \times 10^{7}\ \dfrac{\text{J}}{\text{kg}}$ 深的"井"里。
The Moon has mass $7.35 \times 10^{22}\ \text{kg}$ and radius $1.74 \times 10^{6}\ \text{m}$. What is the gravitational potential at its surface, as a multiple of $10^{6}\ \dfrac{\text{J}}{\text{kg}}$? (Include the sign.) · 月球质量 $7.35 \times 10^{22}\ \text{kg}$,半径 $1.74 \times 10^{6}\ \text{m}$。其表面的引力势是多少(用 $10^{6}\ \dfrac{\text{J}}{\text{kg}}$ 的倍数表示,含符号)?
$\phi = -\dfrac{GM}{R} = -\dfrac{6.67 \times 10^{-11} \times 7.35 \times 10^{22}}{1.74 \times 10^{6}} = -2.8 \times 10^{6}\ \dfrac{\text{J}}{\text{kg}}$ — about a twentieth of the Earth's, which is why leaving the Moon is easy. · $\phi = -\dfrac{GM}{R} = -\dfrac{6.67 \times 10^{-11} \times 7.35 \times 10^{22}}{1.74 \times 10^{6}} = -2.8 \times 10^{6}\ \dfrac{\text{J}}{\text{kg}}$——约为地球的二十分之一,所以离开月球很容易。
Escape velocity
- To reach infinity, kinetic energy must match the depth of the well: $\tfrac{1}{2}mv_{\text{esc}}^{2} = \dfrac{GMm}{r}$.
- $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}} \approx 11\ \dfrac{\text{km}}{\text{s}}$ at Earth's surface — independent of the object's mass.
逃逸速度
- 要到达无穷远,动能必须与井的深度相当:$\tfrac{1}{2}mv_{\text{esc}}^{2} = \dfrac{GMm}{r}$。
- $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}} \approx 11\ \dfrac{\text{km}}{\text{s}}$(地球表面)——与物体质量无关。
Roughly, what is the escape velocity from the Earth's surface, in km/s? · 大致而言,从地球表面逃逸的速度是多少(用 km/s)?
$v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}} \approx 11\ \dfrac{\text{km}}{\text{s}}$ for Earth. · 地球的 $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}} \approx 11\ \dfrac{\text{km}}{\text{s}}$。
Escape velocity depends on the mass of the escaping object. · 逃逸速度取决于逃逸物体的质量。
The object mass cancels: $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$ depends only on the planet and the distance. · 物体质量抵消:$v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$ 只取决于行星和距离。
A planet has GM/r = 6.3e7 J/kg at its surface. What is its escape speed, in km/s? · 某行星表面的 GM/r = 6.3e7 J/kg。它的逃逸速率是多少 km/s?
v = sqrt(2GM/r) = sqrt(1.26e8) = 1.12e4 m/s = 11.2 km/s, the Earth's value. Note the factor of 2 and that the mass of the escaping object cancels out entirely. · v = sqrt(2GM/r) = sqrt(1.26e8) = 1.12e4 m/s = 11.2 km/s,正是地球的值。注意那个 2,以及逃逸物体的质量完全约掉了。
Worked example: lifting a satellite
A $500\ \text{kg}$ satellite is raised from the Earth's surface ($R = 6.37 \times 10^{6}\ \text{m}$) to an orbit of radius $7.0 \times 10^{6}\ \text{m}$. Earth's mass is $5.97 \times 10^{24}\ \text{kg}$. Find the gain in potential energy, and compare it with $mg\Delta h$.
- Two potentials, subtract: $\Delta E_{\text{P}} = GMm\left(\dfrac{1}{R} - \dfrac{1}{r}\right)$.
- Substitute: $6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 500 \times \left(\dfrac{1}{6.37 \times 10^{6}} - \dfrac{1}{7.0 \times 10^{6}}\right) = 2.8 \times 10^{9}\ \text{J}$.
- The rough way: $mg\Delta h = 500 \times 9.81 \times 6.3 \times 10^{5} = 3.1 \times 10^{9}\ \text{J}$ — about $10\%$ too big, because $g$ falls with height and $mg\Delta h$ assumes it does not.
- Check: the gain is positive — the satellite ends less negative, less tightly bound — even though both potentials are negative numbers.
例题:把卫星送上去
一颗 $500\ \text{kg}$ 的卫星从地球表面($R = 6.37 \times 10^{6}\ \text{m}$)被送到半径 $7.0 \times 10^{6}\ \text{m}$ 的轨道。地球质量 $5.97 \times 10^{24}\ \text{kg}$。求势能的增加,并与 $mg\Delta h$ 比较。
- 两个势相减: $\Delta E_{\text{P}} = GMm\left(\dfrac{1}{R} - \dfrac{1}{r}\right)$。
- 代入: $6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 500 \times \left(\dfrac{1}{6.37 \times 10^{6}} - \dfrac{1}{7.0 \times 10^{6}}\right) = 2.8 \times 10^{9}\ \text{J}$。
- 粗略的方法: $mg\Delta h = 500 \times 9.81 \times 6.3 \times 10^{5} = 3.1 \times 10^{9}\ \text{J}$——大了约 $10\%$,因为 $g$ 随高度减小,而 $mg\Delta h$ 假设它不变。
- 检查: 增量是 正 的——卫星最终的势能负得少了,束缚得没那么紧了——尽管两个势都是负数。
Potential is negative and increases towards zero as you move away. "Energy gained" is final minus initial with the signs kept: $-1.3 \times 10^{7} \to -0.9 \times 10^{7}$ is a gain of $0.4 \times 10^{7}$ per kilogram. $mg\Delta h$ is only for $\Delta h \ll R$. And the unit of $\phi$ is $\dfrac{\text{J}}{\text{kg}}$ — writing joules loses the mark.
引力势是 负 的,离开时 向零增大。"获得的能量"是末减初、保留符号:$-1.3 \times 10^{7} \to -0.9 \times 10^{7}$ 是每千克增加 $0.4 \times 10^{7}$。$mg\Delta h$ 只适用于 $\Delta h \ll R$。而且 $\phi$ 的单位是 $\dfrac{\text{J}}{\text{kg}}$——写成焦耳会丢分。
Which statements about gravitational potential are correct? Select all that apply. · 关于引力势的哪些说法正确?选出所有正确的。
Potential is energy per unit mass, so its unit is joules per kilogram. It is negative, rises towards zero with distance, and its gradient gives the field: $g = -\dfrac{\Delta\phi}{\Delta r}$. · 势是单位质量的能量,所以单位是焦耳每千克。它为负,随距离向零增大,它的梯度给出场:$g = -\dfrac{\Delta\phi}{\Delta r}$。
Field strength from the potential graph
- The field strength is the gradient of the potential–distance graph: $g = -\dfrac{\Delta\phi}{\Delta r}$.
- Where the $\phi$ graph is steep (close to the mass) the field is strong; far away the graph flattens and the field fades.
- The minus sign says the field points down the potential slope — towards the mass, where $\phi$ is most negative.
从势的图得到场强
- 场强是势–距离图的 斜率(potential gradient):$g = -\dfrac{\Delta\phi}{\Delta r}$。
- $\phi$ 图陡的地方(靠近质量)场强;远处图变平,场减弱。
- 负号说明场指向势 下降 的方向——朝向质量,那里 $\phi$ 最负。
You've got it
- gravitational potential $\phi = -\dfrac{GM}{r}$ in $\dfrac{\text{J}}{\text{kg}}$ (negative, zero at infinity); $g = -\dfrac{\Delta\phi}{\Delta r}$
- gravitational PE $E_{\text{P}} = -\dfrac{GMm}{r}$ — more negative = more tightly bound; large changes use $GMm\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$
- escape velocity $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$ (mass-independent, $\approx 11\ \dfrac{\text{km}}{\text{s}}$ for Earth)
你掌握了
- 引力势 $\phi = -\dfrac{GM}{r}$,单位 $\dfrac{\text{J}}{\text{kg}}$(负,无穷远处为零);$g = -\dfrac{\Delta\phi}{\Delta r}$
- 重力势能 $E_{\text{P}} = -\dfrac{GMm}{r}$——越负 = 束缚越紧;大的变化用 $GMm\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$
- 逃逸速度 $v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}$(与质量无关,地球约 $11\ \dfrac{\text{km}}{\text{s}}$)