Gravitational Fields
A-Level Physics Topic 13 12:15 English narration · English + 中文 subtitles burned in
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An apple falls from a tree.
一个苹果从树上落下。
The Moon sails around the Earth.
月亮绕着地球运行。
For thousands of years, no one saw any connection between them.
几千年来，没有人看出它们之间有任何联系。
Then Isaac Newton had one of the greatest ideas in all of science.
然后，艾萨克·牛顿产生了整个科学史上最伟大的想法之一。
What if it is the same force?
如果它们是同一种力呢？
The very same pull that drops the apple to the ground also reaches out, across four hundred thousand kilometres of empty space, to hold the Moon in its orbit.
让苹果落地的那同一个拉力，也伸展出去，穿过四十万公里的虚空，把月亮拉在它的轨道上。
Gravity is universal — it acts between every mass in the universe.
引力是普适的——它作用在宇宙中每一对质量之间。
Gravity shapes the whole cosmos, from a falling apple to the orbits of planets.
引力塑造着整个宇宙，从一个落下的苹果，到行星的轨道。
Today: gravitational fields, Newton's law of gravitation, orbits, and gravitational potential.
今天：引力场、万有引力定律、 轨道，以及引力势。
Let's begin.
让我们开始吧。
A gravitational field is a region where a mass feels a force.
引力场是一个区域，在其中质量会感受到力。
We measure its strength as the force on each kilogram — the gravitational field strength, g.
我们用每千克所受的力来量度它的强度—— 这就是引力场强度，g。
We draw it with field lines: for a planet, they point straight inward, toward the centre, all the way around.
我们用场线来画它：对于一颗行星，场线四面八方径直指向内部、指向中心。
Where the lines are close together, the field is strong; where they spread apart, it is weak.
场线密集的地方，场就强；场线稀疏的地方，场就弱。
Field lines point the way the force acts on a test mass, and their SPACING carries the information.
场线指出力作用在检验质量上的方向，而它们的疏密才是携带信息的东西。
Equally spaced parallel lines mean a constant field strength.
等间距的平行线表示场强恒定。
Lines spreading apart mean the field is getting weaker.
线向外散开表示场在变弱。
Lines converging mean it is getting stronger.
线向内会聚表示场在变强。
Closer lines, stronger field — that is the rule.
线越密，场越强——这就是规则。
Two patterns to know.
有两种图样要掌握。
Around a point mass, or a uniform sphere seen from outside, the lines are radial and point INWARDS, because gravity only attracts.
在一个质点周围， 或者从外部看一个均匀球体，场线是沿半径的，而且指向内侧，因为引力只有吸引。
And near the Earth's surface over a small area they are nearly parallel and equally spaced, pointing straight down — a uniform field.
而在地球表面附近的一小片区域内，它们几乎是等间距平行的，笔直向下——这是匀强场。
That second one is why g feels constant in a laboratory.
后面这一种，正是在实验室里 g 感觉起来是恒定的原因。
Here is the result that makes every calculation possible.
下面这个结论让所有的计算成为可能。
For a uniform sphere — a planet, a star — the field at any point OUTSIDE is exactly the same as that of a point mass equal to the total mass, sitting at the centre.
对一个均匀球体——一颗行星、一颗恒星—— 在它外部任何一点处的场，与把全部质量集中在球心的质点所产生的场完全相同。
Look at the two pictures: the radial pattern is identical.
看这两幅图：沿半径的图样一模一样。
So from above the surface you can treat the whole Earth as a point mass at its centre, and measure r from the CENTRE, not from the ground.
所以在地表以上， 你可以把整个地球当作位于球心的一个质点，而且 r 要从球心量起，不是从地面量起。
That last detail is where marks go.
最后这个细节正是丢分的地方。
Points INSIDE a sphere behave differently, and they are not in the syllabus.
球体内部各点的情形不同，而它们不在考纲范围内。
Newton captured it in one equation.
牛顿用一个方程抓住了它。
The force between two masses is the gravitational constant, times the two masses multiplied together, divided by the square of the distance between them.
两个质量之间的力，等于引力常量，乘以两个质量相乘， 再除以它们之间距离的平方。
Two things to notice.
有两点要注意。
The force grows with each mass.
力随每个质量的增大而增大。
And it obeys an inverse-square law: double the distance, and the force drops to a quarter.
而且它遵守平方反比定律：距离加倍，力就降到四分之一。
For a sphere, you measure the distance from its very centre.
对于一个球体， 你从它的正中心量起距离。
Combine that law with the definition of field strength, and the mass of a test object cancels out.
把这条定律和场强度的定义结合起来，测试物体的质量就消掉了。
What remains is beautifully simple: the field strength equals the gravitational constant, times the planet's mass, divided by the distance squared.
剩下的东西美妙而简单： 场强度等于引力常量，乘以行星的质量，再除以距离的平方。
Near the Earth's surface, the distance barely changes as you climb — so g stays almost constant, about nine point eight, whether you are on the ground or on the top of a tower.
在地球表面附近， 你往上爬时距离几乎不变——所以 g 几乎保持恒定，大约是九点八， 不管你是在地面上，还是在一座塔的顶端。
Find the gravitational field strength at the Earth's surface.
求地球表面的重力场强度。
The Earth's mass is six times ten to the twenty-four kilograms and its radius six point four times ten to the sixth metres.
地球质量为六乘以十的二十四次方千克， 半径为六点四乘以十的六次方米。
Use g equals G M over r squared, with r the radius because we want the value AT the surface.
用 g 等于 G M 除以 r 平方， 其中 r 取半径，因为我们要的是地表处的值。
Substituting gives about nine point eight newtons per kilogram — the value you have used since your first physics lesson, now derived rather than given.
代入后得到约九点八牛顿每千克—— 这正是你从第一节物理课起就一直在用的那个数值，如今是推导出来的，而不是给定的。
And note the unit: newtons per kilogram is the same thing as metres per second squared, which is why g is both a field strength and the acceleration of free fall.
再注意单位：牛顿每千克和米每二次方秒是同一回事， 这正是 g 既是场强、又是自由落体加速度的原因。
And g is a vector: it points the way the force acts, towards the source mass.
而且 g 是矢量：它指向力的方向，也就是指向源质量。
This graph shows g falling off as one over r squared.
这张图显示 g 按一除以 r 平方衰减。
So why does it feel constant in a laboratory?
那么为什么在实验室里它感觉是恒定的？
Because of the numbers.
因为数值的关系。
The Earth's radius is six point four MILLION metres.
地球半径是六百四十万米。
Rising to a height h changes the distance from the centre from R to R plus h.
升高到高度 h， 到地心的距离就从 R 变成 R 加 h。
For any building or mountain, h is utterly negligible next to R — going from five metres to ten metres high changes r by about one part in a million.
对任何建筑或高山来说， h 与 R 相比都完全微不足道——从五米升到十米， r 的变化大约只有百万分之一。
So g barely moves.
所以 g 几乎不动。
The lesson generalises: an inverse-square law looks constant whenever your change in distance is tiny compared with the distance you already are from the source.
这个道理可以推广：只要你在距离上的改变，相对于你与源之间已有的距离而言很小， 平方反比定律看上去就是恒定的。
Now, how does a satellite stay in orbit?
那么，卫星是怎么留在轨道上的？
Gravity provides exactly the centripetal force it needs to keep curving.
引力恰好提供它保持拐弯所需要的向心力。
Set the gravitational force equal to the centripetal force, and the satellite's mass cancels again.
让引力等于向心力，卫星的质量又一次消掉了。
This gives a direct link between the orbit's radius and its period: bigger orbits are slower.
这就给出了轨道半径和周期之间的直接联系： 越大的轨道越慢。
The Moon, far away, takes a month; a low satellite races around in ninety minutes.
遥远的月亮要走一个月；一颗低轨卫星九十分钟就绕一圈。
For a satellite in a circular orbit, gravity provides the centripetal force — that one sentence is the whole method, and every orbit question starts there.
对一个在圆轨道上运行的卫星来说，引力提供向心力—— 这一句话就是全部方法，每一道轨道题都从这里开始。
So set G M m over r squared equal to m v squared over r.
于是令 G M m 除以 r 平方，等于 m v 平方除以 r。
Now watch what cancels: the satellite's own mass m disappears from both sides.
现在看什么被约掉了： 卫星自身的质量 m 从两边消失了。
That is a real physical statement, not just algebra — the orbital speed does not depend on the satellite's mass at all.
这是一个真实的物理陈述，而不只是代数运算—— 轨道速度根本不取决于卫星的质量。
A bolt and a space station at the same radius travel at the same speed.
同一个半径上，一颗螺栓和一座空间站速度相同。
Cancelling one r as well leaves v equals the square root of G M over r.
再约掉一个 r，就剩下 v 等于 G M 除以 r 的平方根。
A satellite orbits the Earth in a circular orbit of radius seven times ten to the sixth metres.
一颗卫星在半径为七乘以十的六次方米的圆轨道上绕地球运行。
Find its orbital speed, taking G M for the Earth as four times ten to the fourteenth.
求它的轨道速度， 取地球的 G M 为四乘以十的十四次方。
Note the question gave you the PRODUCT G M rather than the two separately — that is common, and it saves a step.
注意题目给的是乘积 G M， 而不是分别给出两个量——这很常见，而且省了一步。
So v equals the square root of four times ten to the fourteenth over seven times ten to the sixth, which is about seven point six times ten cubed metres per second.
于是 v 等于四乘以十的十四次方除以七乘以十的六次方，再开平方根， 约为七点六乘以十的三次方米每秒。
Sanity-check it: seven point six kilometres per second is about twenty-five times the speed of sound, and that is genuinely how fast low-orbit satellites move.
做个合理性检验： 每秒七点六千米大约是声速的二十五倍，而低轨卫星确实就是这么快。
The period follows from T equals two pi r over v.
周期由 T 等于二 pi r 除以 v 得出。
Substitute the speed we just found and rearrange, and you get T squared equals four pi squared over G M, times r cubed.
把刚才求出的速度代进去并整理， 就得到 T 平方等于四 pi 平方除以 G M，再乘以 r 的三次方。
That is Kepler's third law for circular orbits: T squared is proportional to r cubed.
这就是圆轨道情形下的开普勒第三定律：T 平方与 r 的三次方成正比。
Now look at what the graph does with it.
再看图上是怎么用它的。
Plot T squared against r cubed and orbital data lies on a straight line through the origin, with gradient four pi squared over G M.
把 T 平方对 r 的三次方画出来， 轨道数据就落在一条过原点的直线上，斜率是四 pi 平方除以 G M。
So measuring the periods and radii of a planet's moons gives you the gradient — and from the gradient, the mass of the planet.
所以测出一颗行星各卫星的周期和半径，你就得到了这个斜率—— 再由斜率得到那颗行星的质量。
That is how the masses of distant bodies are actually known.
遥远天体的质量，实际上正是这样知道的。
A geostationary satellite stays directly above the same point on the Earth, so a fixed dish can always point at it.
地球同步卫星始终停在地球上同一点的正上方，所以固定的碟形天线可以一直对准它。
Four conditions make that happen, and an exam wants all four.
有四个条件使这成为可能，而考试四个都要。
It must have a period of twenty-four hours — the same angular speed as the Earth — matching the Earth's rotation.
它的周期必须是二十四小时——与地球有相同的角速度—— 与地球自转相同。
It must orbit west to east, the same way the Earth turns — the same period going the wrong way would not help.
它必须自西向东运行，与地球转动的方向一致—— 周期相同但方向反了并没有用。
It must be directly above the equator, because an orbit tilted out of the equatorial plane would carry the satellite north and south during the day even with the right period.
它必须正好在赤道上空， 因为轨道面若偏离赤道面，即使周期正确，卫星在一天之内也会南北往复移动。
And putting a twenty-four hour period into Kepler's law, the radius works out at about four point two times ten to the seventh metres — roughly thirty-six thousand kilometres above the surface.
而把二十四小时的周期代入开普勒定律，算出的半径约为 四点二乘以十的七次方米——大约在地表以上三万六千千米。
One orbit is special.
有一条轨道很特别。
Choose exactly the right radius, and the period becomes twenty-four hours — the same as the Earth's spin.
选取恰好合适的半径，周期就变成二十四小时——和地球自转一样。
Place the satellite above the equator, moving west to east, and it stays fixed above one single point on the ground.
把卫星放在赤道上空，自西向东运动，它就固定地停在地面上某一个点的正上方。
This is a geostationary orbit.
这就是地球同步轨道。
It is why a satellite dish can point at one spot in the sky, and never move.
这也是为什么卫星天线可以对准天上的一个位置，永远不动。
Finally, gravitational potential: the energy per kilogram to bring a mass from infinitely far away, to a point in the field.
最后，引力势：把一个质量从无穷远处带到场中某一点，每千克所做的功。
Here is the strange part — it is always negative.
奇怪的地方在这里——它总是负的。
Why?
为什么？
Because gravity does the work for you as the mass falls inward, so you end up with less than nothing.
因为质量往里落时，引力替你做了功， 所以你最后得到的比零还少。
The potential is minus the gravitational constant, times the mass, over the distance.
势等于负的引力常量，乘以质量，再除以距离。
And the energy of two masses together follows the same shape: minus big G, big M, little m, over r.
而两个质量在一起的能量遵循同样的形式：负的大 G，大 M，小 m，除以 r。
Gravitational potential at a point is the work done per unit mass in bringing a small test mass from infinity to that point, phi equals W over m, in joules per kilogram.
某一点的引力势，是把一个小检验质量从无穷远处移到该点、每单位质量所做的功， phi 等于 W 除以 m，单位是焦耳每千克。
Zero is taken at infinity.
零点取在无穷远。
And here is why it is negative everywhere else: as the test mass falls inwards, GRAVITY does the work for you, so you do negative work — which is why the graph sits entirely below the axis.
而它在别处为什么处处为负，原因在这里：当检验质量向内落时， 是引力替你做了功，所以你做的是负功——这正是这条曲线整个位于横轴下方的原因。
For a point mass, phi equals minus G M over r.
对一个质点，phi 等于负的 G M 除以 r。
Note the r, not r squared, just as with the electric case.
注意分母是 r，不是 r 平方， 这和电场的情形一样。
The picture is a potential WELL: deepest near the surface, rising towards zero far away.
这幅图是一口势阱：在地表附近最深， 向远处升向零。
And phi is a scalar, so for several masses you simply add the potentials, with no directions to resolve.
而且 phi 是标量，所以有多个质量时， 你只要把各自的势相加，没有方向需要分解。
A test mass m sitting where the potential is phi has gravitational potential energy E P equals m phi, which is minus G M m over r.
一个位于势为 phi 处的检验质量 m，具有引力势能 E P 等于 m phi， 也就是负的 G M m 除以 r。
Like the potential it is negative, reaching zero only at infinite separation, so closer masses are more tightly bound — more negative means harder to pull apart.
和势一样它是负的，只有在无限远处才为零， 所以靠得越近的质量束缚得越紧——越负就越难分开。
Now the two regimes.
现在看两种情形。
For small height changes near the surface, r barely changes, and the whole thing reduces to the familiar delta E P equals m g delta h.
对地表附近的小高度变化，r 几乎不变，整个式子退化成熟悉的 delta E P 等于 m g delta h。
For a large change — a satellite moving to a higher orbit — that formula is wrong, because g is not constant over that distance.
而对大的变化——比如卫星升到更高的轨道—— 那个公式就错了，因为在那么长的距离上 g 并不恒定。
Use minus G M m over r at each radius and take the difference.
要在两个半径上分别用负的 G M m 除以 r，再取差值。
The answer comes out positive, which is right: energy must be supplied to raise the satellite.
算出来的结果是正的，这是对的：把卫星举高必须提供能量。
To escape from radius r all the way to infinity, an object's kinetic energy must equal the SIZE of its gravitational potential energy — you have to climb the whole well.
要从半径 r 处一路逃逸到无穷远，物体的动能必须等于它引力势能的大小—— 你得爬完整口势阱。
So a half m v squared equals G M m over r.
于是二分之一 m v 平方等于 G M m 除以 r。
The mass cancels again, and rearranging gives the escape velocity: v escape equals the square root of two G M over r.
质量又一次被约掉了，整理后得到逃逸速度等于二 G M 除以 r 的平方根。
Put in the Earth's numbers and you get about one point one times ten to the fourth metres per second — about eleven kilometres per second.
代入地球的数据，得到约一点一乘以十的四次方米每秒——大约是每秒十一千米。
And say the consequence out loud, because it is worth a mark: it does not depend on the object's mass.
而且要把这个推论说出来，因为它值一分：它不取决于物体的质量。
A pebble and a rocket need exactly the same escape speed.
一颗石子和一枚火箭，需要的逃逸速度完全相同。
Three marks to secure.
三个要拿稳的分。
First, gravity is an inverse-square law — double the distance, quarter the force.
第一，引力是平方反比定律——距离加倍，力变四分之一。
Second, for an orbit, set the gravitational force equal to the centripetal force.
第二，对于轨道，让引力等于向心力。
Third, gravitational potential and potential energy are always negative, and zero only at infinity.
第三，引力势和引力势能总是负的， 只有在无穷远处才为零。
Master these, and gravity is yours.
掌握这些，引力就是你的了。