# 3. Newton’s Laws of Motion

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Professor Ramamurti

Shankar: Let me start again by reminding you what it is that

was done last time. If you say, “Can you summarize

for me in a few words the main ideas?”

I would like to do that. What I did last time was to

show you how to handle motion in more than one dimension.

I picked for it two dimensions as the standard way to explain

it. By the way, I would like to

make one recommendation. If you guys are coming in a

little late, don’t worry about submitting the homework.

Just come in and settle down, because it’s very hard to

lecture with this amount of traffic.

I do try to start a few minutes late, but I also have to finish

a few minutes early so you can go to your next class.

Also, people who come in early maybe should try to sit in the

middle part of the classroom, so the latecomers can come in

without too much disruption. So, summary of last time.

If you live in two dimensions or more, you’ve got to use

vectors to describe most things. The typical vector is called

V or A or B something with an arrow on it.

The most important vector is the position vector that tells

you where the object is. It’s got components which are

x and y that could vary with time.

I and J are unit vectors in the x and

y directions. You can deal with a vector in

one of two ways. You can either think of it as

an arrow and imagine the arrow, or you can reduce it to a pair

of numbers, x and y.

If you want to add two vectors, you can add the arrows by the

rule I gave you or just add the components of the two guys to

get the component of the sum and likewise for the y.

I mentioned something of increasing importance only

later, which is that you are free to pick another set of

axes, not in the traditional x

and y direction, but as an oblique direction.

If you do, the unit vectors are called I prime and

J prime, the components of the vector

change. You can imagine that if a

vector is viewed from an angle, then its components will vary

with the perspective. So the components of the vector

are not invariant characterizations of the vector.

That is, the vector itself has a life of its own.

The components come in the minute you pick your axis.

It’s not enough to say these are the components of the

vector; you’ve got to tell me,

“I am working with I and J, which I define in the

following manner.” I gave you a law of

transformation of the components;

namely, if the vector has components ax and

ay in one reference frame and ax prime and

ay prime in another reference frame,

how are the two related? They come from writing I

prime and J prime in terms of I and J,

then sticking it in an expression and identifying the

new representation. Somehow, when I told you to

invert the transformation, some of you had some

difficulty. Maybe you didn’t realize that

they are just simultaneous equations that you solve.

Normally, if I tell you 3x + 4y=6,

and 9x + 6y=14, you know how to solve it.

It’s like really the same kind of problem, except that the 3

and 4 will all be placed by sin φ and cos φ but

they are just the numbers. You eliminate them the same way

you eliminate them. I’ve given you a homework

problem where you can try your skills.

When you go today and look at the homework,

you’ll find the problem from the textbook,

one extra problem that deals with all of this.

Then, I gave you one other very important example of a particle

moving in the xy plane. x and y can be

whatever you like, but I picked a very special

example where x looked like this: r times cos

ωt. This times I + r

times sin ωt. You should go back and remember

what we did. If it took you a while to

digest it, I ask you to think harder.

This describes a particle that’s going around in a circle.

We know it’s going around in a circle because if I find the

length of this vector, which is the x-square

part, plus the y-square part, I just get r^(2) at

all times, because sine square plus cosine

square is one. So we know it’s moving on a

circle of radius r. Furthermore,

as time increases, the angle, ωt,

is increasing in this fashion. Omega is called the angle of

velocity. I related it to the time

period, which is the time it takes to go around a full

circle, by saying once you’ve done a

full circle, ωt better be 2π.

So this new quantity ω, which may be new to you,

is related to the time period. How long does it take to go one

round in this fashion? You’re also free to write it in

terms of the frequency. The time period and frequency

are reciprocals. If it takes you 1/60 of a

second to go around once, then you do it 60 times a

second. Omega is really very simple

quantity. It’s related to the frequency

with which you go around the circle but is multiplied by

2π. Why is that?

Frequencies, how many times you go around,

and 2π is the rate at which the angle is being

changed. But if every revolution is

worth 2π radians, then 2πf is the number

of radians per second. f is revolutions per

second and 2πf is radians per second.

It’s called the angle of velocity.

The most important result from last time was that if you took

this r, and you took two derivatives of

this to find the acceleration, d2r over

dt^(2), try to do this in your head.

If you took the two derivatives of this guy, first time it will

become -ω, sin ωt;

second time it will become -ω^(2) cos ωt.

In other words, it will become -ω^(2)

times itself. Same thing there.

The final result is the acceleration is -ω^(2)

times the position. That means the acceleration is

pointing towards the center of the circle and it has a

magnitude a. When I draw something without

an arrow, I’m talking about the magnitude.

It is just ω^(2)r. I have shown you yesterday that

the speed of the particle as it goes around the circle is this

(ωr). Again, you should make sure you

know how to derive this. You can do it any way you like.

You can take one full circle and realize the distance

traveled is 2πr, divide by the time and you will

get this. Another way,

take the derivative of this, get the velocity vector and you

notice his magnitude is a constant and the constant will

be ωr. Whichever way you do it,

you can then rewrite this as v^(2) over r.

This is called the centripetal acceleration. This is the acceleration

directed toward the center. I told you these are very

important results. You’ve got to get this in your

head. Whenever you see a particle

moving in a circle, even if it’s at a constant

speed, it has an acceleration,

v^(2) over r directed towards the center.

This formula doesn’t tell you which way it’s pointing,

because it’s a scaler; it’s not a vector equation.

If you want to write it as a vector equation,

you want to write it as v^(2) over r

minus–I want to say that it’s pointing in the direction toward

the center. So sometimes what we do is we

introduce a little vector here called e_r.

I’ll tell you more about it later.

e_r is a vector at each point of length

one pointing radially away from the center.

It’s like the unit vector I.

Unit vector I points away from the origin in the

x direction. J points away from the

origin in the y direction.

e_r is not a fixed vector.

At each point, e_r is a

different vector pointing in the radial direction of length one.

The advantage of introducing that guy is that if you like,

I can now write an equation for the acceleration as a vector.

The magnitude is v^(2) over r.

The direction is -e_r.

So e_r is a new entity I’ve introduced for

convenience. It plays a big role in

gravitation, in the Coulomb interaction.

It’s good to have a vector pointing in the radial direction

of length one. That’s what it is.

That’s really the heart of what I did last time.

Then we did some projectile problems.

You shoot something, you should know when it will

land, where it will land, with what speed it will land,

how high it will go. I assume that those problems

are not that difficult and I’ve given you a lot of practice.

Now I’m going to move to the really important and central

topic. I guess you can guess what that

is. We’re going to talk about

Newton’s laws. This is a big day in your life.

This is when you learn the laws in terms of which you can

understand and explain a large number of phenomena.

In fact, until we do electricity and magnetism the

next semester, everything’s going to be based

on just the laws of Newton. It’s really amazing that

somebody could condense that much information into a few,

namely three, different laws.

That’s what we’re going to talk about.

Let’s start. Your reaction may be that

you’ve seen Newton’s laws, you applied them in school.

I’ve got to tell you that I realized fairly late in life

they are more subtle than I imagined the first time.

It’s one thing to plug in all the numbers and say,

“I know Newton’s laws and I know how they work.”

But as you get older and you have a lot of spare time,

you think about what you are doing,

which is something I have the luxury of doing right now,

and I realized this is more tricky.

I want to share some of that with you so you can fast forward

and get the understanding it took me much longer to get.

That’s what I’m going to emphasize, more than just

plugging in the numbers. Of course, we have to also know

how to plug in the numbers so we can pass all the tests,

but it’s good to understand the nature of the edifice set up by

Newton. First statement by Newton–I

don’t feel like writing it down. It’s too long and everybody

knows what the law is. It’s called the Law of Inertia.

Let me just say it and talk about it.

The Law of Inertia says that, “If a body has no forces acting

on it, then it will remain at rest if it was at rest to begin

with, or if it had a velocity to

begin with, it will maintain that velocity.”

One way to say it is, every body will continue to

remain in a state of rest or uniform motion in a straight

line. That’s another way of saying

maintaining velocity if it’s not acted upon by a force.

What makes the law surprising is that if I only gave you half

the law, namely every body will remain at rest if it’s not acted

upon by a force, you will say, “That’s fine.

I accept that, because here’s something.

You leave it there, it doesn’t move.

It’s not a big surprise.” People were used to that from

the time of Aristotle. But Aristotle used to think

that if you want something to move, there has to be some

agency making it move. That agency you could call

force. The great discovery that

Galileo and Newton made is that you don’t need a force for a

body to move at constant velocity.

It’s very clear you don’t need a force if something is doing

nothing, just sitting there. The fact that you don’t need a

force for it to move forever at a given speed in a given

direction, that’s not obvious,

because in daily life you don’t see that.

In daily life, everything seems to come to

rest unless you push it or you pull it or you exert some kind

of force. But we all know that the reason

things come to a halt when you push them is,

there eventually is some friction or drag or something

bringing them to rest. Somehow, if you could

manufacture a really smooth frictionless surface,

that if you took a hockey puck or something and an air cushion

and you give it a push, in some idealized world,

it’ll travel forever. So it’s hard to realize that in

the terrestrial situation. But Galileo already managed to

find examples where things would roll on for a very,

very long time. Nowadays, if you go to outer

space, you can check for yourself that if you throw

something out, it just goes on forever without

your intervention. It’s in the nature of things to

go at a constant velocity. They don’t need your help to do

that. You have to be careful that

this first law of Newton is not valid for everybody.

In fact, I’ll give an example in your own life where you will

find that this law doesn’t work. Here is the situation.

You go on an airplane and then after the usual delays,

the plane begins to accelerate down the runway.

At that instant, if you leave anything on the

floor, you know it’s no longer yours.

It’s going to slide down and the guy in the last row is going

to collect everything. Why is that?

Because we find in that plane, when objects are left at what

you think is at rest with no external agency acting on them,

they all slide backwards towards the rear end of the

plane. That happens during takeoff.

That doesn’t happen in flight, but it happens during takeoff.

That is an example of a person for whom the Law of Inertia does

not work. This is something you guys may

not have realized. Newton’s laws are not for

everybody. You have to be what’s called an

“inertial observer.” If you’re an inertial observer,

then in your system of reference, objects left at rest

will remain at rest. The plane that’s ready to take

off or is taking off is not such a system.

The Earth seems to be a pretty good inertial system,

because on the ground, you leave something,

it stays there. It depends on what you leave.

If you leave your iPod, it’s not going to stay there

for very long. But then you can trace it to

some external forces, which are carrying your iPod.

But if you don’t do anything, things stay.

Here is the main point. The point of Newton is,

two things in the Law of Inertia, which one may think is

trivial. First, free velocity,

constant velocity can be obtained for free without doing

anything. There are people for whom this

is true. For example,

in outer space, you’ve got an astronaut.

You send something, you’ll find it goes on forever.

Here’s another thing. If you find one inertial

observer, namely one person for whom this Law of Inertia works,

I can manufacture for you an infinite number of other people

for whom this is true. Who are these other people?

Do you know what I’m talking about?

If I give you one observer for whom the Law of Inertia is true,

I say that others for whom is also true.

Yes? Student:

[inaudible] Professor Ramamurti

Shankar: Did you hear that? Let me repeat that.

First of all, if the Law of Inertia is valid

for me, it’s valid for other people in the same room at rest

with respect to me. Because if I think it’s not

moving, you think it’s not moving.

That’s just fine. But suppose you are in a train

and you’re moving past me and you look at this piece of chalk.

Of course, everything in my room is going backwards for you.

But things which were at rest will move at a constant

velocity, opposite of the velocity that you have relative

to me. You will find that objects that

are at constant initial velocity maintain the velocity.

If I am an inertial observer, another person moving relative

to me at constant velocity will also be an inertial observer.

Why? Because any velocity I ascribe

to a particle or an object, you will add a certain constant

to it by the law of composition of velocity.

All velocities I see you will add a certain number to get the

velocities according to you. But adding a constant velocity

to objects does not change the fact that those which were

maintaining constant velocity still maintain a constant

velocity. It’s a different constant

velocity. In particular,

the things that I say are at rest, you will say are moving

backwards at the velocity that you have relative to me.

Things that I say are going at 50 miles per hour you may say

are going at 80. But 50 is a constant and 80 is

a constant. Therefore, it’s not that

there’s only one fortunate family of inertial observers.

There’s infinite number of them, but they’re all moving

relative to each other at constant velocity.

If the Earth is an inertial frame of reference,

if you go in a train relative to the Earth at constant

velocity, you’re also inertial. But if you go on a plane which

is accelerating, you’re no longer inertial.

That’s the main point. The point is that there are

inertial frames of reference. You must know the Earth is not

precisely inertial. The Earth has an acceleration. Can you tell me why I’m sure

the Earth has an acceleration? Yes?

Student: Because it’s moving in a

circle. Professor Ramamurti

Shankar: It’s going around the Sun.

Let’s imagine it’s a circular orbit.

Then we’ve just shown here, it’s an accelerated frame of

reference. It just turns out that if you

put the v^(2) and you put the r,

and r is 93 million miles, you will find the

acceleration is small enough for us to ignore.

But there are effects of the Earth’s acceleration,

which we’ll demonstrate. The Focault pendulum is one

example where you can see that the Earth is rotating around its

own axis. Then, the fact that the Earth

is going around the Sun. All of them mean it’s really

not inertial, but it’s approximately

inertial. But if you go to outer space

nowadays, you can find truly inertial frames of reference.

That’s the first law. The first law,

if you want, if you want to say,

“Okay what’s the summary of all of this?”

The summary is that constant velocity doesn’t require

anything. The reason it looks like a

tautology, because you look around, nothing seems to have

its velocity forever. Then you say,

“Oh, that’s because there’s a force acting on it.”

It looks like a tautology because you’re never able to

show me something that moves forever at a constant velocity,

because every time you don’t find such a thing,

I give an excuse, namely, a force is acting.

But it’s not a big con, because you can set up

experiments in free space far from everything,

where objects will, in fact, maintain their

velocity forever. That’s a possibility.

It’s a useful concept on the Earth, because Earth is

approximately inertial. Now, we have come to the second

law, which is “the law.” This is the law that we all

memorize and learn. It says that,

“If a body has an acceleration, then you need a force and the

relation of the force to acceleration is this thing:

F=ma” Now, I have to say a few words about

units. Acceleration is measured in

meters per second squared. Mass is measured in kilograms.

So, the way to measure force is in kilogram meters per second

squared. But we get tired of saying that

long expression, so we’re going to call that a

Newton, right? If you invented it,

we’d call it whatever your name is, but this is the guy who

invented it, so it’s called a Newton, usually denoted by a

capital N. A typical problem that you may

have done in your first pass at Newton’s law,

someone tells you a force of 36 Newtons is acting on a mass of

whatever, 4 kilograms; what’s the acceleration?

You divide and you find it’s 9 and you say, “Okay,

I know what to do with Newton’s laws.”

That’s where I want to tell you that it’s actually more

complicated than that. Let’s really look at this

equation. Take yourself back to

1600-whatever, whenever Newton was inventing

these laws. You don’t know any of these

laws. You have an intuitive

definition of force. You sort of know what force is.

Somebody pushes you or pulls you.

That’s a force. Suddenly, you are told there is

a law. Are you better off in any way?

“Can you do anything with this law?”

is what I’m asking you. What can you do with this law?

I give you Newton’s law and say, “Good luck.”

What will you do? What does it help you predict?

Can you even tell if it’s true? Here’s a body that’s moving,

right? I want you to tell me,

is Newton right? How are we going to check that?

Well, you want to measure the left-hand side and you want to

measure the right-hand side. If they’re equal,

maybe you will say the law is working.

What can you measure in this equation?

Student: [inaudible]

Professor Ramamurti Shankar: Pardon me?

Student: Force and acceleration.

Professor Ramamurti Shankar: All right.

Let’s start with acceleration. What’s your plan for measuring

acceleration if some little thing is moving?

What do you need to measure it? Student:

The change in velocity over the change [inaudible]

Professor Ramamurti Shankar: Right,

but what instruments will you need to measure it?

You are supposed to really measure it.

What will you ask me for? Student: A watch.

Professor Ramamurti Shankar: A watch?

Yes. Student:

[inaudible] Professor Ramamurti

Shankar: And what else? Student:

[inaudible] Professor Ramamurti

Shankar: A ruler? Okay.

That’s right. You don’t mean Queen Elizabeth,

right? You mean– Very good.

What you really want–a ruler may not be enough,

but maybe it’s enough. So here’s the long ruler and

here’s this thing moving, right?

You ask for a Rolex so here’s your watch.

It’s telling time. Tell me exactly what you want

to do to measure acceleration. What do you have to do?

I want the acceleration now. What will you measure?

Okay, you go ahead. You can try.

Student: [inaudible]

Student: [inaudible]

I would start with the object at rest and then [inaudible]

Professor Ramamurti Shankar: It may not be at

rest to begin with. It’s doing its thing.

It’s going at some speed. Yes?

Student: Measure the distance it travels

over a constant interval of time?

Professor Ramamurti Shankar: That’ll give you

the velocity. Student:

Well, if you do distance after one second versus distance over

the next second versus distance over a third second,

you see how it increases. Professor Ramamurti

Shankar: Good, in principle.

Let me repeat what he said. He said, first,

let it go a little distance, take the distance over time.

That gives you the velocity now. Let it go a little more,

that gives you the velocity later.

Take the difference of the two velocities and divide by the

difference of the two times, and you’ve got the

acceleration. Of course, you have allowed it

to move a finite distance in a finite time.

What you should imagine doing is making these three

measurements more and more quickly.

You need three positional measurements.

Now, a little later, and a little later later,

because between the first and second,

you get a velocity, the second and third you get

another velocity. Their difference divided by the

difference in times is going to be the acceleration.

But if you imagine making these measurements more and more and

more quickly, in the end, you can measure

what you can say is the acceleration now.

That’s the meaning of the limit in calculus.

You take Δx and you take Δt.

What’s the meaning of Δx and Δt going

to 0? It means, measure them as

quickly as you can. In the real world,

no one’s going to measure it instantaneously,

but we can make the difference as small as we like.

Mathematically, we can make it 0.

In that limit, we can measure velocity right

now. That means we can also measure

velocity slightly later and make the slightly later come as close

to right now as we want. Then, that ratio will give you

acceleration. Acceleration is the easiest

thing to measure of these three quantities.

We all have a good intuitive feeling for what acceleration

is. You want to test if what Newton

told you is right. You see an object in motion,

you find a and you give a a certain numerical

value, 10 meters per second squared.

But that’s not yet testing the equation, because you’ve got to

find both sides. What about the mass?

What’s the mass of this object? Anybody want to try from this

section here? Yes?

Student: You could use some sort of a

standard unit of mass and then a balance to measure the mass of

something else, like put it at a certain

distance from the fulcrum on a scale and figure out the

relative mass. Professor Ramamurti

Shankar: Okay. His idea was the following.

You take a standard mass and you maybe go to a seesaw.

You put the standard mass here and you put some other fellow at

the mass you’re looking at there.

You add just the lengths and when it balances,

you can sort of tell what this mass is, right?

But suppose you were in outer space.

There’s no gravity. Then the seesaw will balance,

even if you put a potato on one side and an elephant on the

other side. You cannot tell the mass,

because what you are doing now is appealing to the notion of

mass as something that’s related to the pull of the Earth on the

object. But Newton’s law is–You see,

you’ve got to go back and wipe out everything you know.

If this is what you have, there is no mention of the

Earth in these equations. Yet, the notion of mass is

defined. It’s not talking about the

gravitational pull of the Earth on an object.

Yes? Student:

If you had an object that you say has a certain mass that you

don’t know and then you took one that’s the same density,

but say twice the size and you can see if it slows the force.

Professor Ramamurti Shankar: But do you know

what the force on the body is? We don’t know how to measure

that either; do you agree?

We don’t know what the force is, because we are hoping to say

force is going to be m times a.

We’re just getting on to measuring m.

It looks like a circular definition right now.

Yes? Student:

Do you mean what material this object is made of and its

density [inaudible] Professor Ramamurti

Shankar: Ah, but density is mass over volume.

But we don’t know the mass, right?

Student: But if you use the density of

that [inaudible] Professor Ramamurti

Shankar: How is anyone going to give you a density of

anything? We’re just asking what’s the

mass of any object? We have not yet found a

satisfactory answer to what’s the mass of an object?

Yes? Student:

Are we still operating in outer space?

Professor Ramamurti Shankar: Yes.

This cannot depend on the planet Earth.

Student: Could we have,

by any chance, a spring?

Professor Ramamurti Shankar: Yes.

Student: Whose spring constant and the

definition [inaudible] Professor Ramamurti

Shankar: I’m sorry. What did you just say?

You can have a spring. Student:

Yeah, you could have a spring and you could compare how fast

that objects will travel when the spring is compressed and

they’re placed against it and released, but you’d only have a

comparison then. Professor Ramamurti

Shankar: Okay. Let me repeat what he said.

He said, take a spring to outer space and we’ll hook up some

objects to them and see how fast they move and do a comparison.

That’s fairly close to what I had in mind.

But it’s not the word perfect answer.

I just want to take some time thinking about it.

Yes? Student:

[inaudible] Professor Ramamurti

Shankar: What do I learn from the period?

Student: [inaudible]

Professor Ramamurti Shankar: How?

Student: [inaudible]

Professor Ramamurti Shankar: Remember,

you cannot peek into chapters 6 and 7, because you’ve seen it

before. I’m asking you if somebody

wrote apparently Shakespeare’s plays for him;

it’s one of the rumors, right? Suppose Newton comes to you and

says, “I have this great law, but I don’t want to publish it

under my name. I’m going to give it to you.”

You have got this new law, but how are you going to sell

this? You’ve got to tell people how

to use it. You realize it’s very subtle,

because the very first thing in that equation,

which is m, has not yet been defined.

He gave an answer which is fairly close.

It doesn’t rely on gravity. It doesn’t rely on the planet.

You cannot say to me, “Take a force,

due to a spring, and see what force it applies

and divide by the acceleration and get the mass,”

because we haven’t defined force either.

You’ve got to realize that. Let me ask you something.

How do we decide how long a meter is?

Can you tell me, how do you know how long a

meter is? Student:

It’s just sort of arbitrary. Professor Ramamurti

Shankar: Right. Student:

You can just pick an arbitrary mass, as well,

like one object, which is an arbitrary mass.

Professor Ramamurti Shankar: And we’ll call it.

First thing you’ve got to do is, realize that some of these

things are not God-given. A meter, for example,

is not deduced from anything. Napoleon or somebody said,

“The size of my ego is one meter.”

That’s a new unit of length. You take a material like silver

and put it in a glass case and that’s the definition of a

meter. It’s not right or wrong.

Then I ask you, “What is two meters and what is

three meters?” We have ways of doing that.

I take the meter and put it next to the meter,

that’s two meters. I cut it in half,

I’ll use some protractors and dividers and compasses,

you can split the meter into any fraction you like.

Likewise for mass, we will take a chunk of some

material and we will call it a kilogram.

I should give you some hint. That kilogram,

I don’t expect you to deduce. That is a matter of convention.

Just like one second is some convention we use and one meter

is some convention we use. I’m going to give you a little

help. I’m going to give you a glass

case and in the glass case is an entity with this as one

kilogram, by definition. Then I give you another object,

an elephant. Here’s an elephant.

I’m telling you, “What’s the mass of the

elephant?” How do you find this mass?

You got to take the hint he gave.

I give you a spring and an elephant.

What should we do? Yes?

Student: Measure how far,

if you hung an elephant from a spring [inaudible]

Professor Ramamurti Shankar: No.

Remember, when we hang the elephant in outer space,

nothing is going to happen. Student: Okay.

You push the elephant and measure how far– what the

distance from the unstretched spring is that the elephant

travels in either direction. Professor Ramamurti

Shankar: Okay. Student:

Then, you also do that with one kilogram.

It should be proportional to how far from the resting state

of the spring. That should be the proportion

that the one kilogram object is to the other.

Professor Ramamurti Shankar: When you say,

“push the elephant,” you want to push it in a particular way?

What do you want to do? Student:

Towards the [inaudible] to compress the spring or to

[inaudible] Professor Ramamurti

Shankar: That is correct, but you don’t want to give it a

definite push. Yes?

Student: Do you push it radially along

the axis of the spring, because you know that the

centripetal force is mv^(2) is over t.

Professor Ramamurti Shankar: Yeah.

Maybe that’s an interesting thing.

That’s correct. You can do that, too.

Let me now put you out of your suspense.

I think I’ve heard bits and pieces of the answer everywhere,

so I don’t want to wait until we get it word-perfect.

The point is, the one kilogram is a matter of

convention. We want to know what is the

mass of the elephant. We can do the seesaw

experiment, you suggested, but the seesaw experiment

requires gravity, so we don’t want to do that.

A spring will, on the other hand,

exert a force. We don’t know what the force of

the spring is. If you assume that,

you are not playing by the rule, because we don’t know what

force it exerts. We do know it exerts a force,

so here’s what you do. You hook one end of the spring

to a wall and you pull it from rest by some amount and you

attach the one kilogram mass to it.

Yes? Student:

One problem that we’d have, though, is if we’re in outer

space, where are we going to find a wall that won’t move?

Professor Ramamurti Shankar: A wall that will

not move? Student:

Actually, if you hook it up to any wall in outer space

[inaudible] Professor Ramamurti

Shankar: You’re right. What can actually happen is

that if you’re in an enormous laboratory, which is made up of

an enormous object, then you will find that you

can, in fact, attach it to the wall.

You can go in outer space– Student:

It won’t move very much? Professor Ramamurti

Shankar: No, it won’t move very much.

Once we know enough dynamics, we can answer your question.

Does outer space even rob you of something to which you can

anchor a spring? The answer is, “no.”

You can anchor things to objects in outer space.

They just won’t act the way gravity does.

But you can nail it to the wall and pull one end.

So you pull one end. We don’t know what force it

exerts. But it exerts some force.

Now, I tie this one kilogram mass to it and let it go.

I find the acceleration. That is the force which I do

not know in magnitude. But this is the acceleration of

the one kilogram mass. Then, I bring the elephant and

I pull the spring by the same amount and I find the

acceleration of the elephant and the denominator is obviously the

mass of the elephant. The force is not known,

but it’s the same force. So, when I divide these two

numbers, I’m going to find a_1 over

a_E is equal to m_E over

m, which is the one kilogram mass.

What we needed was some mechanism of exerting some fixed

force. We didn’t have to know its

magnitude. But the acceleration it

produces on the elephant and on the mass, are in an inverse

ratio of their masses. If you knew this was one

kilogram, then the acceleration of the elephant,

which will be some tiny number, maybe 100^(th) of what this guy

did; the mass of the elephant is

then 100 kilograms. Note there are,

again, subtleties even here. If you think harder,

you can get worried about other things.

For example, how do I know that when I pull

the spring the first time for the mass,

it exerted the same force when I pulled the spring the second

time for the elephant? After all, springs wear out.

That’s why you change your shock absorbers in your car.

After a while, they don’t do the same thing.

First, we got to make sure the spring exerts a fixed force

every time. You can say,

“How am I going to check that? I don’t have the definition of

force yet.” But we do know the following.

If I pull the one kilogram mass and I let it go,

it does something, some acceleration.

Then, I pull it again by the same amount and let it go;

I do it 10 times. If every time I get the same

acceleration, I’m convinced this is a

reliable spring that is somehow producing the same force under

the same condition. On the eleventh time,

I pull the mass. I will put the elephant in.

With some degree of confidence, I’m working with a reliable

spring and then I will get the mass of the elephant.

Why is it so important? It’s important for you guys to

know that everything you write down in the notebook or

blackboard as a symbol is actually a measured quantity.

You should know at all times how you measure anything.

If you don’t know how to measure anything,

you are doing algebra and trigonometry.

You are not doing physics. This also tells you that the

mass of an object has nothing to do with gravitation.

Mass of an object is how much it hates to accelerate in

response to a force. Newton tells you forces cause

acceleration. But the acceleration is not the

same on different objects. Certain objects resist it more

than others. They are said to have a bigger

mass. We can be precise about how

much bigger by saying, “If the acceleration of a body

to a given force is ten times that of a one kilogram mass,

then this mass is one-tenth of one kilogram.”

This is how masses can be tabulated using a spring.

Imagine then from now on, we can find the mass of any

object, right? We know now with the same

spring, by this comparison, we will find.

All objects now can be attributed a mass.

Then we may, from this equation,

say a certain force is acting in a given situation by

multiplying the m times the a.

Then, here is what we actually do.

Now, we go back to the spring. We go back to the spring and we

want to learn something about the spring.

We want to know how much force it exerts when I pull it by a

certain amount. Now I can measure that,

because I pull it by one centimeter and I find the

acceleration it exerts on a known mass.

That m times a is the force the spring is

exerting. Then I pull it by 1.1 cm,

and I find ma. I find 1.2, I find ma.

I draw a graph here of the amount by which I pull the

spring versus the force it exerts.

It will typically look like this and the formula we say is

F=-kx, where k is called “the

force constant.” You got to understand what the

minus sign is doing here. This is the force exerted by

the spring on the mass. It says, if you pull it to the

right, so that x is positive, the spring will exert

a force which is in the negative direction;

that’s why you have a minus sign.

Then, all springs will do something like this.

Further out they can do various things.

The force may taper off, the force may not be given by a

straight line, but for modest deformations,

every spring will have a linear regime in which the force is

linearly proportional to the stretching.

It also tells you that if you compress the spring,

compress it means x is now–x is measured from

this position, where the spring is neither

compressed nor expanded. So x is not really the

coordinate of the end point. You’ve got to understand that.

Springs have a natural length; x is measured from that

length. If it’s positive,

it means you’ve stretched it, if it’s negative,

it means compressed. This equation is telling you if

you compress it, namely if x is negative,

F will be then positive, because it’s pushing you

outwards. Therefore, what we have done

now is, we can take all kinds of springs and we can calibrate the

force they will exert under various conditions.

Namely, if you pull it by so much, that’s the force it will

exert. I want you to think for a

second about two equations. One equation says F=ma.

Other equation says F=-kx. What’s going on?

Is one of them Newton’s law? Then what’s the other one?

Maybe F=-kx should be called Newton’s law?

Why is F=ma called Newton’s law?

Then, what’s the meaning of this expression? What’s the difference between

saying F=-kx and F=ma?

They are saying very different things.

Yes? Student:

F=ma is a universal law.

Professor Ramamurti Shankar: Right.

Let me repeat. F=ma is universally

true, independent of the nature of the force acting on a body.

Then? Student:

F=-kx is only describing how the spring is.

Professor Ramamurti Shankar: Very good.

That’s the whole point. The cycle of Newtonian dynamics

has two parts. First one says,

if you knew the force acting on any body, without going into

what caused the force, then you may set that force

equal the mass times acceleration of the body.

We think of force as the cause and a as the result or

the effect. Force causes acceleration and

this is a precise statement. There, Newton doesn’t tell you

what forces are going to be acting on a body in a given

situation. If you leave the body alone,

maybe there’s no force acting on it.

If you connect the body to a spring, which is neither

compressed nor extended, there’s no force acting on it.

If you pull the spring, there is a force acting on it.

Newton is not going to come and tell you what force the spring

will exert when it’s pulled by some amount.

That is another part of your assignment.

The physicist has to constantly find out what forces act on

bodies. That’s a separate exercise.

In every context in which I place a body,

I’ll have to know what are the forces acting on it.

I’ve got to find them by experimenting,

by putting other bodies and seeing how they react and then

finding out what’s the force that acts on a body when it’s

placed in this or that situation.

Once you’ve got that, then you come back.

In the case of a spring, this is the law that you will

deduce. If it’s something else,

you will have to deduce another law.

For example, we know that if a body is near

the surface of the Earth, the force of gravity and that

object seems to be m times g,

where g is 9.8. That’s something you find out

by experiment. Every time you are finding out

a different force that’s acting on a body with different origin.

One says, leave any body near the Earth, it yields a force.

I know this is the right answer, because if I now find

the acceleration, I find it’s mg divided

by m and I get -g as the answer for all bodies.

By the way, that’s a very remarkable property of the

gravitational force–the cancellation of the two

ms. If you look at the electrical

force, the force of electricity, on the proton and electron or

something, it’s not proportional to the

mass of either object. It’s proportional to the

electric charge of either object.

Therefore, when you divide by the mass to get the

acceleration, the response of different

bodies is inverse to the mass. But gravity has a remarkable

property that the pull of the Earth is itself proportional to

the inertia of the object. So, when you divide by

m, m cancels and everything falls at the same

rate on the surface of the Earth.

In fact, there’s a property of gravitational fields anywhere,

even in outer space, but there is some residual

field between all the planets and all the stars in the

universe, that the force on a body is

proportional to the mass of the body.

So, when you divide by the mass to get the acceleration,

you get the same answer. Everything — gold,

silver, diamonds, particles — everything

accelerates the same way in a gravitational field,

due to this remarkable fact. This was known for a long time,

but it took Mr. Einstein to figure out why

nature is behaving in that fashion.

If I have some time, I’ll tell you later.

But there are two qualities which happen to be equal.

One is inertial mass, which is how much you hate your

velocity to change, how hard you resist

acceleration. That exists far from planets,

far from everything. Other is gravitational mass,

which is the measure of how much you’re attracted to the

Earth. There’s no reason why these two

attributes had to be proportional,

but they are proportional and they are equal by choice of

units and you can ask, “Is this just an accident or is

it part of a big picture?” It turns out,

it’s part of a big picture and all of general relativity is

based on this one great equivalence of two quantities

which are very different attributes.

Why should the amount by which you’re attracted to the Earth be

also a measure of how much you hate acceleration?

Two different features, right? But they happen to be the same.

Anyway, what physicists do is they put bodies in various

circumstances and they deduce various forces.

This is the force of gravity. This is the force of the spring.

Here’s another force you might find.

You put a chunk of wood on a table and you try to move it at

constant speed. Then you find that you have to

apply minimum force. We are moving at constant

velocity. That means the force you’re

applying is cancelled by another force, which has got to be the

force of friction. So force of friction is yet

another force. Then, there are other forces.

You guys know there is the electrical force.

If you bring a plus charge near a plus charge,

if my body m, has a plus charge and another

plus charge is there, it’ll feel a force due to that.

That’s not going to be given by Newton.

So, Newton did not ever tell you what the expression for

force is in a given context. That is a constant study.

Coulomb discovered the Coulomb’s law,

which is a repulsion between charges.

Nowadays we know if you go into the neutron or the proton,

there are quarks. There are forces between the

quarks. You can ask what the force that

this quark will exert and that quark at a certain separation.

That was obviously not known to Newton.

Remember, Newton said F=ma, but didn’t tell you what

value F has in a given context.

He just said whenever there’s an acceleration,

it’s going to be due to some forces and it’s your job to find

what the forces are. To find the force,

what you will do is, suppose somebody says,

“Hey, I’ve got a new force. Every time I go near the

podium, I find I’m drawn to it.” Okay, that’s a new force.

The word gets around and we want to measure the force.

What do I do? I stand near this podium.

I’m drawn to it. I cannot stop.

I tie a spring to my back and I anchor it to the wall and see

how much the spring stretches before the two forces balance.

Then I know that kx is equal to the force this is

exerting at this separation. I move a little closer and I

find the stretching is a different number.

Maybe the force is getting stronger.

That’s how by either balancing the unknown force with a known

force or by simply measuring the acceleration as I fall towards

this podium and multiplying by mass,

you can find the force that exerts on me.

It’s not a cyclical and useless definition.

It’s a very interesting interplay and that’s the

foundation of all of mechanics. We are constantly looking for

values of F and we’re constantly looking for responses

or bodies to a known force. Here’s a simple example of a

complete Newtonian problem. A mass is attached to a spring.

It is pulled by a certain amount x,

and is released. What is it going to do?

We go to Newton. Newton says F=ma,

so to make it a useful result of this problem,

we know the mass of this guy. We did the comparison with the

elephant or something; a is the second

derivative of x and for this problem,

when F is due to a spring,

we know the force is that by studying the spring.

Suddenly, you have a mathematically complete problem.

Mathematically complete problem is that you can find the

function x(t) by saying that the second derivative of

the function is equal to -k over m times

the function. Then, you go to the Math

Department and say, “Please tell me what’s the

answer to this equation?” We don’t have to worry about

how you solve it, but it’s problem in mathematics

and the answer will be–surprise,

it’s going to be oscillating back and forth and that’ll come

out of the wash. This is how you formulate

problems. You can formulate another

problem. Later on, we know about gravity.

Newton finds out there’s a force of gravity acting on

everything. Here’s the Sun.

Here’s your planet. At this instant,

the planet may be moving at that speed.

Then the acceleration of the planet is the force of gravity

between the planet and the Sun, which Newton will tell you is

directed towards the Sun and it depends on how far you are.

Depending on how far the planet is from the Sun and where it’s

located, you will get the left-hand side.

That’s another law. That’s the Law of Universal

Gravitation. Then again, you will find the

evolution of the planetary motion, because the rate of

change of the position is connected to the position.

Again, go to the math guys and say, “What’s the answer to

that?” and they tell you the answer,

which will be some elliptical motion.

Okay. By the way, Mr.

Newton did not have math guys he could go to.

Not only did he formulate laws of gravitation,

he also invented calculus and he also learned how to solve the

differential equation for calculus.

He probably felt that nobody around was doing any work,

because all the thing was given to this one person.

It’s really amazing that what Newton did in the case of

gravity was to find the expression for this.

A few years earlier he had also gotten this law.

By putting the two together, out comes the elliptical motion

of the planets. We’ll come back to that,

but you have to understand the structure of Newtonian

mechanics. Generally, any mechanics will

require knowledge of the force. Now, I’m going to add one more

amendment. You don’t have to write in your

notebook, but you’ve got to remember.

Maybe you’ll but a little T and circle it.

Let me write it here. F_T means the

total force on a body. You’ve got many forces acting

on a body. The acceleration is controlled

by the sum. If I’m now working in one

dimension, it’s obvious because I’m not using any vectors.

Then, you may have forces to the right, forces to the left

pushing, pulling. You add them all up

algebraically, keeping track of their sign,

and that’s the total force. That’s connected to mass times

acceleration. Now, I’ll give you the third

law. The third law says that if

there are two bodies, called one and two,

force of one on two is minus the force of the second on the

first one. This is the thing about action

and reaction. All the laws that anybody knows

have this property. What does it require to be a

successful mechanic, to do all the mechanics

problems? You got to be good at writing

down the forces acting on a body.

That’s what it’s all going to boil down to.

Here is my advice to you. Do not forget the existing

forces and do not make up your own forces.

I’ve seen both happen. Right now at this point in our

course, whenever you have a problem where there is some body

and someone says, “Write all the forces on it”,

what you have to do is very simple.

Every force, with one exception,

can be seen as a force due to direct contact with the body.

Either a rope is pulling, a rope is pushing it,

you are pushing it, you are pulling it.

That’s a contact on the body. If nothing is touching the

body, there are no forces on it, with one exception which is,

of course, gravity. Gravity is one force that acts

on a body without the source of the force actually touching it.

That’s it. Do not draw any more forces.

People do draw other forces. When a body is going around a

circle, they say that’s some centrifugal force acting.

There is no such thing. Be careful.

Whenever there is a force, it can be traced back to a

tangible material cause, which is all the time a force

of contact, with the exception of gravity.

Okay, so with that, if you write the right forces,

you will be just fine. You will be able to solve all

the problems we have in mechanics.

I’m going to now start doing simple problems in mechanics.

They will start out simple and, as usual, they will get

progressively more difficult. Let’s start with our first

triumph will be motion in 1D. Here is some object,

it’s 5 kilograms and I apply 10 Newtons.

Someone says, “What’s the acceleration?”

Everyone knows it is 10 over 5 equals 2.

Now we know how we got all the numbers that go into the very

question. How do we know 10 Newtons is

acting? I think you people know how we

can say that with confidence. How do we know the mass of this

is 5 kilograms? We know how you got that from

an earlier experiment. Now, we know how the numbers

come in. The algebra is,

of course, very trivial here. Then, the next problem is a

little more interesting. Here I got 3 kg and I got 2 kg

and I’m pushing with 10 Newtons and I want to know what happens. One way is to just use your

common sense and realize that if you push it this way,

these two guys are going to move together.

And know intuitively that if they move together,

they will behave like an object of mass 5 and the acceleration

will again be 2. But there’s another way to do

this and I’m going to give you now the simplest example of the

other way, which is to draw free-body diagrams.

By the way, when I say there’s 10 Newtons acting this way,

you might say, “What about gravity?

What about the table?” Imagine that this is in outer

space where there is no gravity for now.

The motion is just along the x axis.

The free-body diagram, it says you can pick any one

body that you like and apply F=ma to it,

provided you identify all the forces acting on that body.

We’ll first pick the body, with mass 3.

Here’s the body of mass 3. What are the forces on it?

This is certainly acting on it. Then, you have to ask,

“What other force is acting?” Here is where you have to think.

Anybody want to guess from the last row what force?

Yes? Student:

The reactive force of 2 on 3? Professor Ramamurti

Shankar: Right. So let’s give it a name.

Let’s point it that way and call it F2 on 3.

That’s the end of this guy. Let’s look at the other fellow.

Maybe you should complete the force acting on this one.

Can you tell me what it is? Same person in the last row.

Student: [inaudible]

Professor Ramamurti Shankar: And how big is

that? Student:

The same as F2, 3.

Professor Ramamurti Shankar: Right.

I don’t want to give it too many different names,

because F2, 3 and F3,

2 are equal and opposite. I’m already showing F2,

3 acting to the left. Let me give it some other name

like f. Then you agree this will be the

same f but pointing that way.

Here is the mistake some people make.

They add to that the 10 Newtons. It’s the 10 Newtons acting on

it, because I know the 10 Newton is pushing me and I’m going to

feel it even if I’m here. That will be a mistake.

That’s an example of adding a force that you really shouldn’t

be adding. The only force acting on this

guy is this little f. That’s, in turn,

because this guy’s being pushed by the 10 Newtons,

but that’s not your problem. Your problem is to only look at

the forces of contact on you, and that is just this f.

Then we do F=ma for these two guys.

For this guy, F=ma will be 10 minus

f is 3 times a. The other one,

it’ll be f=2a. Notice I’m using the same

acceleration for both. I know that if the second mass

moved faster than the first one, then the picture is completely

wrong. If it moved slower than the

first one, it means it’s rammed into this one.

That also cannot happen, so they’re moving with the same

acceleration. There’s only one unknown

a. Once you got this,

you realize what you got to do. This equation is begging you to

be added to this equation. You got a plus f and a

minus f, so you got to do what you got

to do. You add the two,

you get 10=5a and you get a=2.

Once you got a=2, you can go back and realize

here that f=4, so you got really 4 Newtons.

Now, we know the full story. 4 Newtons acting on 2 kg,

gives you an acceleration of 2. On this guy,

I have 10 from the left and I have 4–10 acting this way,

4 acting that way. That is 6 Newtons divided by 3

kg is also an acceleration of 2. This is a simpler example.

A simple example of free body diagrams.

Very simple. If you can do this,

you can do most of the problems you will run into.

Just don’t add stuff that’s not there.

That’s all you have to be careful about.

The stuff that people tend to add sometimes is to keep drawing

the 10 Newtons that’s acting on that. Now, here’s another variation.

The variation looks like this. I got 3 kg and I have a rope.

I got 2 kg and I pull this guy with 10 Newtons. What’s going to happen?

Again, your common sense tells you, “Look, you are pulling

something whose effective mass seems to be 5,

the answer is 2.” Let’s get that systematically

by using free-body diagrams. Now, there are really three

bodies here. Block one and block two and the

rope connecting them. In all these examples,

this rope is assumed to be massless.

We know there is no thing called a massless rope,

but most ropes have a mass, but maybe negligible compared

to the two blocks you are pulling, so we’ll take the

idealized limit where the mass of the rope is 0.

Here is the deal. 3 kg is being pulled by the

rope on the right with a force that I’m going to call T,

which stands for tension. The rope is being pulled

backwards by this guy, the T.

What is the force on the other side?

What should that be? Who said T?

Why is it definitely T and not something else?

What would happen if it’s something else?

Student: If it was greater,

then the rope would snap. Well, if it was greater

[inaudible] Professor Ramamurti

Shankar: It won’t snap, but something else will be a

problem. Yes?

Student: [inaudible]

Professor Ramamurti Shankar: Not only faster,

but what will be its acceleration?

If the two forces don’t cancel, you have a net force.

What are you going to divide by to get the acceleration?

Student: [inaudible]

Professor Ramamurti Shankar: Zero,

right? So, a massless body cannot have

a net force on it, because the acceleration of the

rope cannot be infinite. In fact, it has to be some

finite number, which is the acceleration of

either of these two guys. So, massless bodies will always

have, like a massless rope, equal and opposite forces on

the two ends. That is called the tension on

the rope. When you say the rope is under

tension being pulled from both sides by a certain force,

the tension is not 0 just because this T and that

T cancel. It’s true the net force is 0,

but it doesn’t mean you can ignore it.

Suppose you are being pulled by my favorite animals — the

elephants — from both sides by equal force.

You don’t find any consolation in the fact that these forces

add up to 0. You feel the pain.

That pain is what–This gentleman doesn’t agree.

One of you guys nodding your head.

Do you feel the pain? Do you agree?

Okay. That force is called a tension.

Whenever you’re asked on a problem, “What’s the tension on

the rope?” you’re looking for that equal

and opposite forces acting at two ends of the rope.

We don’t know what it is. We’ll give it a name,

but by linking that T to that T,

we can also figure out the same T must be exerted on this

one by Newton’s third law, because if this block is the

only one that could be pulling this rope with T.

Therefore, the ropes will be pulling the block with T

in the other direction, then I got 10 Newtons here.

Now, you can do F=ma for the three different objects.

There’s nothing to do here, because the forces are 0,

the mass is 0. It doesn’t tell you anything.

The first one tells you T=3a.

The other one tells you 10 – T=2a.

Again, we know what to do. We add these numbers and we get

10=5a. Therefore, we find a=2.

Once you find a=2, you find T=6.

So, tension on the rope is 6 Newtons.

This is very important, because when you buy a rope,

they will tell you how much tension it can take before it’ll

snap. If your plan is to accelerate a

3 kg mass with an acceleration of 2 meters per second,

you better have a rope that can furnish that force and it can

take the tension of 6 Newtons. Now, for the- whoa! I’m going to give the last

class a problem which is pretty interesting, which is what

happens to you when you have an elevator.

Here is a weighing machine and that’s you standing on the

elevator. We’re going to ask,

“What’s the needle showing at different times?”

First, take the case in the elevator is on the ground floor

of some building and completely addressed.

Then, let’s look at the spring. The spring is getting squashed

because you are pushing down and the floor is pushing up.

You are pushing down with the weight mg,

and the floor has got to be pushing up with the md,

because the spring is not going anywhere.

So the spring is being pushed by mg here and mg

here. Therefore, it’ll compress by an

unknown x, which is equal to mg

divided by the force constant of the spring.

By the way, that is a subtle thing people may not have

realized. Even in the case of this

spring, when you pull it, if you pull it to the right by

some force. Remember, the wall is pulling

to the left with the same force. So springs are always pushed or

pulled on either side with the same force.

We focus on one because we are paying for it,

but the wall is doing the opposite.

We don’t pay any attention to that.

You cannot have a spring pulled only at one side,

because then it will then accelerate with infinite

acceleration in that direction. This spring is getting squashed

on either side, and it’ll squash by certain

amount x, that depends on your mass,

and that x will be turned into a motion of a needle

and that’ll read your mass. Now, what happens if the

elevator is accelerating upwards with an acceleration a?

That’s the question. The way I analyze it is,

I say, if I look at me and I write the forces on me,

that is mg acting down, then we use w as the

symbol to represent the force exerted by the spring;

w – mg=ma.

That is, F=ma. When I was not accelerating and

everything was at rest, ma was 0,

w was mg, and w was the reading on

the needle. But if I’m accelerating,

the force exerted by the spring and therefore,

the needle the weighing machine reads is m times g

plus a. It means, when you are

accelerating upwards, as the elevator picks up speed,

the reading on the spring will be more and you will feel heavy.

You feel heavy and it reads more because the poor spring not

only has to support you from falling through the floor,

but also accelerate you counter to what gravity wants to do.

That’s why it is g plus a.

So, you picked up some speed, then you’re coasting along at a

steady speed. Then, a drops out and

you weigh your normal self. As you come to the top of the

building, the elevator has to decelerate, so that it loses its

positive velocity and comes to rest.

So a will be negative and w,

in fact, will be less than mg.

So, you will feel weightless for a short time or you’ll feel

your weight is little. Then, you come to rest and the

opposite happens on the way down.

Let me just briefly look at the ride on the way down.

As you start on the top and go down, your acceleration is

negative. Remember, you’ve got to keep

track of the sign of acceleration,

so if you’re picking up speed towards the ground,

a is negative, so it will be g plus

a, but a is a negative number.

Let me write it as g minus the absolute value of

a. You can see that if a

was equal to g, your downward acceleration is

that of gravity, namely the cable has snapped in

the elevator, then you don’t feel any weight.

You don’t feel any weight because your weight is the

opposition you get to falling through the floor;

but if the floor is giving way and you’re just falling freely,

you feel weightless. It’s wrong to think that you

feel weightless because you escaped the pull of gravity.

We all know that in a falling elevator, you definitely do not

escape the pull of gravity. It’s going to catch up with you

in a few seconds. Likewise, when in outer space,

when you are orbiting the Earth, people are always

floating around in these space stations.

They have not escaped the pull of gravity either.

They have just stopped fighting it.

If you escape the pull of gravity, your spaceship will be

off, won’t be orbiting the Earth.

Anyway, I will return to this next time.

Do your problems. There’s quite a few problems.

You should start them right away.