Newton's law of gravitation · 牛顿万有引力定律
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| law of gravitation/lɔː ɒv ˌɡrævɪˈteɪʃn/ | 万有引力定律 | wàn yǒu yǐn lì dìng lǜ |
| attractive/əˈtræktɪv/ | 吸引 | xī yǐn |
| universal constant/ˌjuːnɪˈvɜːsl ˈkɒnstənt/ | 万有引力常数 | wàn yǒu yǐn lì cháng shù |
| inverse-square law/ɪnˈvɜːs skweə lɔː/ | 平方反比定律 | píng fāng fǎn bǐ dìng lǜ |
| point mass/pɔɪnt mæs/ | 质点 | zhì diǎn |
Every mass pulls every other
- The Earth pulls you down — and you pull the Earth up, just as hard.
- Every pair of masses attracts, anywhere in the universe.
- Newton captured it in one equation.
每个质量都吸引其他质量
- 地球把你向下拉——你也同样用力地把地球向上拉。
- 宇宙中任何地方,每一对 质量都相互吸引。
- 牛顿把它写进了一个方程。
The law of gravitation 万有引力定律
- $F = \dfrac{G m_1 m_2}{r^{2}}$ — an attractive 吸引 pull along the line joining the masses.
- $G = 6.67 \times 10^{-11}\ \dfrac{\text{N}\cdot\text{m}^2}{\text{kg}^2}$ is the universal constant 万有引力常数.
万有引力定律
- $F = \dfrac{G m_1 m_2}{r^{2}}$——沿两质量连线的吸引力。
- $G = 6.67 \times 10^{-11}\ \dfrac{\text{N}\cdot\text{m}^2}{\text{kg}^2}$ 是万有引力常数(universal constant)。
Newton's law of gravitation · 牛顿万有引力定律
F ∝ Mm / r²
Gravity pulls inward and weakens with the square of the distance. · 引力 向内 拉,并随距离的平方减弱。
Newton's law of gravitation gives the force between two masses as: · 牛顿万有引力定律给出两个质量之间的力是:
The pull is proportional to each mass and inversely proportional to the distance squared. · 拉力与每个质量成正比,与距离的平方成反比。
Gravity is always attractive. · 引力总是吸引的。
Yes — masses only ever pull together; there is no gravitational repulsion. · 是的——质量之间只会互相吸引;没有引力排斥。
An inverse-square law 平方反比定律
- The force falls off as $\dfrac{1}{r^{2}}$.
- Double the separation → the force drops to a quarter.
The International Space Station orbits Earth, held in its path by gravity
平方反比定律
- 力按 $\dfrac{1}{r^{2}}$ 减小。
- 距离加倍 → 力降为 四分之一。
国际空间站绕地球运行,被引力维持在它的轨道上
Two masses attract with $40\ \text{N}$ at separation $r$. What is the force at separation $2r$? · 两个质量在距离 $r$ 时以 $40\ \text{N}$ 互相吸引。在距离 $2r$ 时的力是多少?
Inverse-square: $\dfrac{40}{2^{2}} = \dfrac{40}{4} = 10\ \text{N}$. · 平方反比:$\dfrac{40}{2^{2}} = \dfrac{40}{4} = 10\ \text{N}$。
If both masses are doubled (same distance), the gravitational force becomes: · 如果两个质量都加倍(距离不变),引力变为:
$F \propto m_1 m_2$, so doubling each multiplies the force by $2 \times 2 = 4$. · $F \propto m_1 m_2$,所以各加倍使力乘以 $2 \times 2 = 4$。
Two masses attract with force $F$. Match each change to the new force. · 两个质量以力 $F$ 相互吸引。把每种变化与新的力配对。
$F \propto \dfrac{m_1 m_2}{r^{2}}$: masses scale the force directly, the separation scales it by the inverse square. · $F \propto \dfrac{m_1 m_2}{r^{2}}$:质量按正比改变力,距离按平方反比改变力。
Spheres act as points
- A uniform sphere pulls (from outside) exactly like a point mass 质点 at its centre.
- So you can treat the Earth as a point mass at its centre, and $r$ is always measured centre to centre.
Two masses attract along the line joining them
球体相当于质点
- 一个均匀球体(从外部看)的引力恰好像 位于球心的质点。
- 所以你可以把地球当作位于其中心的质点,而 $r$ 总是 从球心到球心 测量。
两个质量沿连线相互吸引
A uniform sphere attracts outside objects as if all its mass were at its centre. · 均匀的球吸引外部物体,就好像它的全部质量都集中在它的中心。
Yes — from outside, a uniform sphere behaves exactly like a point mass at its centre. · 是的——从外面看,均匀的球的行为与位于其中心的点质量完全相同。
Worked example: the Earth and the Moon
Earth's mass is $5.97 \times 10^{24}\ \text{kg}$, the Moon's is $7.35 \times 10^{22}\ \text{kg}$, and their centres are $3.84 \times 10^{8}\ \text{m}$ apart.
- Force: $F = \dfrac{G m_1 m_2}{r^{2}} = \dfrac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.35 \times 10^{22}}{(3.84 \times 10^{8})^{2}} = 2.0 \times 10^{20}\ \text{N}$.
- On which body? On both: the Moon pulls the Earth with exactly the same $2.0 \times 10^{20}\ \text{N}$ — a Newton's third law pair.
- Halve the distance: the force would be $4 \times 2.0 \times 10^{20} = 8.0 \times 10^{20}\ \text{N}$.
- Check: square the separation before dividing, and keep it in metres. An answer near $10^{28}\ \text{N}$ means $r$ was not squared.
例题:地球与月球
地球质量 $5.97 \times 10^{24}\ \text{kg}$,月球质量 $7.35 \times 10^{22}\ \text{kg}$,两者中心相距 $3.84 \times 10^{8}\ \text{m}$。
- 力: $F = \dfrac{G m_1 m_2}{r^{2}} = \dfrac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.35 \times 10^{22}}{(3.84 \times 10^{8})^{2}} = 2.0 \times 10^{20}\ \text{N}$。
- 作用在哪个天体上? 两个 都是:月球以完全相同的 $2.0 \times 10^{20}\ \text{N}$ 拉地球——一对牛顿第三定律的力。
- 距离减半: 力会变成 $4 \times 2.0 \times 10^{20} = 8.0 \times 10^{20}\ \text{N}$。
- 检查: 先把距离平方再相除,并且用米作单位。答案接近 $10^{28}\ \text{N}$ 说明 $r$ 没有平方。
Two $50\ \text{kg}$ people stand with their centres $1.0\ \text{m}$ apart. What is the gravitational force between them, as a multiple of $10^{-7}\ \text{N}$? · 两个 $50\ \text{kg}$ 的人站着,中心相距 $1.0\ \text{m}$。他们之间的引力是多少(用 $10^{-7}\ \text{N}$ 的倍数表示)?
$F = \dfrac{6.67 \times 10^{-11} \times 50 \times 50}{1.0^{2}} = 1.67 \times 10^{-7}\ \text{N}$ — less than the weight of a grain of dust, which is why $G$ is called tiny. · $F = \dfrac{6.67 \times 10^{-11} \times 50 \times 50}{1.0^{2}} = 1.67 \times 10^{-7}\ \text{N}$——比一粒灰尘的重力还小,这就是说 $G$ 极小的原因。
Weighing a planet
- You cannot put a planet on a balance, but its surface field gives its mass: $g = \dfrac{GM}{R^{2}}$, so $M = \dfrac{gR^{2}}{G}$.
- A moon's orbit works too: gravity supplies the centripetal force, $\dfrac{GMm}{r^{2}} = mr\omega^{2}$, so $M = \dfrac{4\pi^{2}r^{3}}{GT^{2}}$ — only the orbit's radius and period are needed.
给行星称重
- 你不能把行星放在天平上,但它的 表面场 能给出它的质量:$g = \dfrac{GM}{R^{2}}$,所以 $M = \dfrac{gR^{2}}{G}$。
- 卫星的轨道 也可以:引力提供向心力,$\dfrac{GMm}{r^{2}} = mr\omega^{2}$,所以 $M = \dfrac{4\pi^{2}r^{3}}{GT^{2}}$——只需要轨道的半径和周期。
Worked example: the mass of Mars
At the surface of Mars the field strength is $3.7\ \dfrac{\text{N}}{\text{kg}}$ and the radius is $3.4 \times 10^{6}\ \text{m}$. Show that the mass of Mars is about $6.4 \times 10^{23}\ \text{kg}$.
- Start from the field of a sphere: $g = \dfrac{GM}{R^{2}}$.
- Rearrange: $M = \dfrac{gR^{2}}{G} = \dfrac{3.7 \times (3.4 \times 10^{6})^{2}}{6.67 \times 10^{-11}}$.
- Answer: $M = 6.4 \times 10^{23}\ \text{kg}$, as required — about a tenth of the Earth's mass.
- Check: a "show that" is marked on the working, so write the equation, the rearrangement and the substitution, and let the number fall out.
例题:火星的质量
火星表面的场强为 $3.7\ \dfrac{\text{N}}{\text{kg}}$,半径为 $3.4 \times 10^{6}\ \text{m}$。证明火星的质量约为 $6.4 \times 10^{23}\ \text{kg}$。
- 从球体的场出发: $g = \dfrac{GM}{R^{2}}$。
- 整理: $M = \dfrac{gR^{2}}{G} = \dfrac{3.7 \times (3.4 \times 10^{6})^{2}}{6.67 \times 10^{-11}}$。
- 答案: $M = 6.4 \times 10^{23}\ \text{kg}$,与要求相符——约为地球质量的十分之一。
- 检查: "证明"题按过程给分,所以写出方程、整理和代入,让数字自然得出。
A planet has surface field strength $25\ \dfrac{\text{N}}{\text{kg}}$ and radius $7.0 \times 10^{7}\ \text{m}$. What is its mass, as a multiple of $10^{27}\ \text{kg}$? · 一颗行星的表面场强为 $25\ \dfrac{\text{N}}{\text{kg}}$,半径为 $7.0 \times 10^{7}\ \text{m}$。它的质量是多少(用 $10^{27}\ \text{kg}$ 的倍数表示)?
$M = \dfrac{gR^{2}}{G} = \dfrac{25 \times (7.0 \times 10^{7})^{2}}{6.67 \times 10^{-11}} = 1.84 \times 10^{27}\ \text{kg}$ — close to Jupiter. · $M = \dfrac{gR^{2}}{G} = \dfrac{25 \times (7.0 \times 10^{7})^{2}}{6.67 \times 10^{-11}} = 1.84 \times 10^{27}\ \text{kg}$——接近木星。
$r$ is the distance between the centres — never the gap between the surfaces, and never a height above the ground on its own. $G$ is tiny, which is why gravity between everyday objects is negligible; only a planet-sized mass makes it noticeable. And the two forces in the pair are equal, however different the masses — the apple pulls the Earth as hard as the Earth pulls the apple.
$r$ 是两 中心 之间的距离——绝不是两表面之间的间隙,也绝不是单独的离地高度。$G$ 极小,所以日常物体之间的引力可以忽略;只有行星大小的质量才让它显著。而且这一对力 相等,无论质量相差多大——苹果拉地球的力和地球拉苹果的力一样大。
In $F = \dfrac{G m_1 m_2}{r^{2}}$, the distance $r$ is measured between the ____ of the two masses. · 在 $F = \dfrac{G m_1 m_2}{r^{2}}$ 中,距离 $r$ 是在两个质量的 ____ 之间测量的。
A uniform sphere acts as a point mass at its centre, so for a satellite $r$ is the planet's radius plus the height above the surface. · 均匀球体相当于位于球心的质点,所以对卫星来说 $r$ 是行星半径加上离地高度。
You've got it
- $F = \dfrac{G m_1 m_2}{r^{2}}$ — always attractive, along the joining line, equal on both masses
- it is an inverse-square law: double $r$ → quarter the force
- a uniform sphere acts as a point mass at its centre, so $r$ is centre-to-centre; $M = \dfrac{gR^{2}}{G}$ weighs a planet
你掌握了
- $F = \dfrac{G m_1 m_2}{r^{2}}$——总是吸引力,沿连线,对两个质量相等
- 它是 平方反比 定律:$r$ 加倍 → 力变为四分之一
- 均匀球体相当于 位于球心的质点,所以 $r$ 是中心到中心;$M = \dfrac{gR^{2}}{G}$ 可以给行星称重