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## Physics library

### Course: Physics library > Unit 4

Lesson 3: Newton's law of gravitation- Introduction to gravity
- Mass and weight clarification
- Gravity for astronauts in orbit
- Would a brick or feather fall faster?
- Acceleration due to gravity at the space station
- Space station speed in orbit
- Introduction to Newton's law of gravitation
- Gravitation (part 2)

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# Would a brick or feather fall faster?

What would fall faster on the moon, a brick or a feather? Created by Sal Khan.

## Want to join the conversation?

- I'm still convinced the brick would still, albeit imperceptably, "hit the surface of the moon" (the question posed in0:42) before the feather (even assuming that the moon is a true vacuum). F_f and F_b aren't only the forces of the moon pulling on the objects, but also the same force that the object pulls on the moon, so the moon is accelerating at G*m_f and G*m_b respectively, where the latter is larger and thus the brick collides first. My question is: am I wrong?(71 votes)
- Technically, you're correct.

When you drop the brick, it accelerates towards the moon at the same rate as a feather would, but the moon also accelerates towards the brick a tiny amount. This tiny amount is a slightly bigger tiny amount than the moon would accelerate toward a feather.

Of course, if you drop the brick and the feather at the same time, the moon will accelerate towards them both, so they will collide at the same time.(112 votes)

- the last experiment was ToTaLlY cool but why is it that the paper didnt float up but fell with the book???!!!(27 votes)
- Because of air resistance.

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If one object is placed above another, than air resistance is blocked out by the base object. So the resistance is not on the paper (which was on top...), rather it was on the book; the paper fell with the book.(10 votes)

- why does the moon have no atmosphere(22 votes)
- The gravitational field of the moon is too weak to sustain any significant atmosphere, as the moon isn't nearly as massive as needed to prevent most gases from escaping.(40 votes)

- Why is r(or distance) squared? Isn't it just distance from one object to another?(9 votes)
- Gravitational force follows the inverse square law. It is due to the spherical geometry of the force field. Intensity of the force twice as far from the source is spread over four times the area, hence one fourth the intensity.

http://hyperphysics.phy-astr.gsu.edu/hbase/forces/isq.html(9 votes)

- can you explain air friction to me? for example, when a skydiver is falling and reaches a terminal velocity of 120MPH after falling for 15 sec, what is the force of air friction on the skydiver as he falls at terminal speed?(7 votes)
- The reasoning for AndrewM's absolutely correct answer is this:

Terminal speed means that this speed no longer changes. In other words there is no accelleration happening. This means there is no net force acting on the diver. This means that all forces on the diver have to balance each other. There are only two relevant forces: friction and gravitational pull.(9 votes)

- what is the ratio g(earth)andg(moon)?(4 votes)
- The ratio would be about 6 to 1. For example if I weighed 100lb on earth and I were to travel to the moon. I would weigh about 16.6lb.(3 votes)

- If you would shoot a gun and drop a book would the bullet and book hit the ground at the same time(3 votes)
- it depends entirely on the direction you fired the gun. if you fired it perpendicular to the surface, and the bullet had enough velocity, it could potentially reach orbit. if you fired it straight into the ground, obviously it would be moving much faster than pure gravity could pull it, and therefore hit the ground before the book.(4 votes)

- a) We could calculate time by using kinematic formula: Vo=0 -> t=2X/g. From here we can see that the mass is irrelevant.

b) Air resistance is: F=p*A. Bigger the surface bigger the resistance force.

Is this correct?(3 votes)- Yes, that would work for any object on Earth(5 votes)

- How do we prove that the acceleration due to free fall is the same as Earth's gravitational field strength?(1 vote)
- Hello! I have a question regarding Newton's second law and law of gravitation. If, through reconciling F=ma and F=G(Mm)/r^2, we see that the acceleration due to gravity on Earth is equivalent to the mass of the earth multiplied by the universal gravitational constant, divided by the square of the distance between the object and the Earth, shouldn't we see a massive decrease in acceleration as distance increases? Why don't we see it in real-life? Please let me know if I am mistaken. Thank you very much!(2 votes)
- Do you have an example of where the inverse square law is not seen?(2 votes)

## Video transcript

Let's say we were to take a
little excursion to the moon. And so here we are
sitting on the surface of the moon-- that's the
surface of the moon right there. And with us to our excursion to
the moon we brought two things. We brought ourself
a concrete brick. So that's my brick
right over there, although it's orange-- we'll say
it's an orange concrete brick. And I also bought a
bird feather with us. So this is the bird feather. And then my question
to you is, if I were to hold both the
brick and the bird feather at the same time, and I were
to let go of both of them at the same time and ask
you, which one of them would hit the surface of the
moon first, what would you say? Well, if you based it on
your experience on Earth-- on Earth if you were to
take a break and a feather, a brick would just
go straight down. A brick would just
immediately fall to the Earth, and it would do
it quite quickly. It would accelerate
quite quickly. While a feather would
kind of float around. If you had a feather on Earth,
it would just float around. It would go that way,
then it would go that way, and it would slowly make
its way down to the ground. So on Earth, at least
in the presence of air, it looks like the brick
will hit the ground first. But what would
happen at the moon? And what's interesting about the
moon is we have no atmosphere. We have no air to
provide resistance for either the brick
or the feather. So what do you think
is going to happen? So your first temptation
would say, well, let's just use the
universal law of gravity. So what is the force of
gravity on the brick? Well, you could
calculate that out. The force of
gravity on the brick is going to be equal to big G
times the mass of the moon-- I'll say that's m
for mass and then the subscript is lowercase
m for moon-- the mass of the moon times the
mass of the brick divided by the distance
between the brick and the center of
the moon squared. Fair enough. That's the force on the brick. What's going to be the force
due to gravity on the feather? Or another way to
think about it, the weight of the
feather on the moon? We'll do the same calculation. The force on the
feather is going to be equal to big G times
the mass of the moon times the mass of the feather
divided by the distance between the center
of this feather and the center of
the moon squared. That's the distance,
and then we square it. So if you look at both
of these expressions, they both have
this quantity right over here-- G times the
mass of the moon divided by the distance
between this height and the center of
the moon squared. So they both have this
exact expression on it. So let's replace
that expression. Let's just call that
the gravitational field on the moon. So if you apply this
number by any mass, it will tell you the weight
of that object on the moon, or the gravitational
force acting downward on that object on the moon. So this is the gravitational
field of the moon. So I'll just call it g sub m. And all it is, is all of
these quantities combined. So if we simplify
that way, the force on the brick due to
the moon is going to be equal to that lowercase
g on the moon-- normally we use this lowercase g for
the gravitational constant on Earth, or the
gravitational field on Earth, or sometimes the acceleration
of gravity on Earth, but now we're
referring to the moon. That's what this lowercase
subscript m is doing for us. So it's equal to that times
the mass of the brick. For the case of the feather,
the force on the feather is equal to all
of this business. So that's the g sub m times
the mass of the feather. So we're going to assume,
which is a reasonable thing to assume, that the
mass of the brick is greater than the
mass of the feather. What's going to be
their relative forces? Well, here you have a greater
mass times the same quantity. Here you have a smaller mass
times the same quantity. So if the mass of
the brick is greater than the mass of the feather,
it's completely reasonable to say that the force
of gravity on the brick is going to be greater
than the force of gravity on the feather. So if you do all
this, and everything we've done to this point
is correct, you might say, hey, there's going to be
more force due to gravity on the brick, and that's
why the brick will be accelerated
down more quickly. But what you need to
remember is that there is more gravitational
force on this brick. But it also has greater mass. And we remember the
larger something's mass is, the less acceleration it'll
experience for a given force. So what really determines how
quickly either of these things will fall is their
accelerations. And let's figure out
their accelerations. I'll do this in a neutral color. We know that force is equal
to mass times acceleration. So if we want to figure out
the acceleration of the brick-- or we could write
it the other way. If we divide both
sides by mass, we get acceleration is equal
to force divided by mass. And acceleration is
a vector quantity, and force is also
a vector quantity. And in this situation, we're
not using any actual value. But if I were using
actual values, I would use negative numbers for
downwards and positive values for upwards. But we're not using
any signs here. But you could assume that
the direction is implicitly being given. So what's the
acceleration of the brick? That's a lowercase
b I was writing. The acceleration of
the brick is going to be equal to the force
applied to the brick divided by the mass of the brick. But the force applied to
the brick, we already know, is this business
right over here. It is little g on the moon,
the gravitational field on the moon, times
the mass of the brick, and we're dividing that
by the mass of the brick. So the acceleration on
the brick on the moon-- the acceleration that the
brick will experience-- is the same thing as that
gravitational field expression. It is g sub m. This is how quickly it would
accelerate on the moon. Now let's do the same
thing for the feather. I think you see
where this is going. The acceleration
of the feather is going to be the force
on the feather divided by the mass of the feather. The force on the
feather is g sub m-- g with the subscript m--
times the mass of the feather, and then we're
going to divide that by the mass of the feather. And so, once again,
its acceleration is going to be
the same quantity. So they are both going to
accelerate at the same rate downwards, which
tells us that they'll both hit the ground
at the same rate. They'll both accelerate from
the same point at the same time, and they'll both have
the same velocity when they hit the ground. And they'll both hit it
at the exact same time, despite one having
a larger mass. So the reality is, because
it has a larger mass, it has a larger gravitational
attraction to the moon. But because of its
mass, that attraction gives it the same acceleration
as something with a smaller mass. So any mass at the same level
on the surface of the moon would experience the
same acceleration. So now the quite
natural question is, wait, Sal, if that's true
on the moon it should also be true on Earth. And it would be true on Earth. If you did this
exact same experiment and you evacuated all
the air from the room, so that you didn't
have air resistance, and you took a
brick and a feather and let them go
at the same time, they will both hit the ground
at the exact same time, which is a little unintuitive,
to imagine a feather just plummeting the same
way a brick would. But it would if you
evacuated all the air. And so the reason why
we see this over here, and I think you get the
sense because I already talked about
evacuating the air, is that the difference between
the brick and the feather is all due to air resistance. If you took the same brick,
or if you took something that had the same
mass as the brick, and you were to flatten it out
so it has more air resistance-- but let's say it has the same
mass, let's say this thing and this thing have
the same mass-- this thing would
fall slower than that because it'll have
more air resistance. It has more air to collide
into, to provide resistance as it falls. And if you took a feather and
if you compacted it really, really, really, really, really,
really small-- the same mass as a feather, but you made
it so small that it could cut through the air-- you'll see
that it will drop a lot faster. So the real difference between
how things fall on Earth-- if you had no air,
they would all fall at the exact same rate. It's only because of air that
they fall at different rates. And the air does two things. For constant
pressure-- so if you have two objects that
have the same shape, the object that is heavier,
that has more weight, will fall faster
because it'll overcome-- it'll be able to provide more
net force against the air pressure. If you have something
that has the same weight, the object that is
more aerodynamic will fall faster-- the
one that cuts through, the one that has the
least air resistance. And as a little
experiment that you can try in the comfort of
your own room right now, take a book like this. And you could drop it. And then you could take
another piece of paper, or even a little postcard or
something, and you drop it. And you'll see, of
course, a postcard will fall much slower
than this book. But what you do is put the
postcard on top of the book so that the book is
essentially breaking all of the air resistance
for the postcard. And what you'll see is, if
you put it on top of the book and you were to
drop it, you'll see that they fall at
the exact same rate.