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## 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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# Space station speed in orbit

AP.PHYS:

INT‑3.A (EU)

, INT‑3.A.1 (EK)

, INT‑3.A.1.1 (LO)

, INT‑3.A.3 (EK)

, INT‑3.A.3.1 (LO)

Speed necessary for the space station to stay in orbit. Created by Sal Khan.

## Want to join the conversation?

- How does the craft in space get up to that speed without the jets and how does it maintain that speed?(41 votes)
- The craft uses a propulsive engine a bit like a jet to escape earth's atmosphere and get to the right height and then it lets gravity accelerate it back to earth. But it uses its engines to give it enough momentum to miss earth and so fall constantly in an orbit. Its speed is maintained purely by gravity, maybe using it's engines every so often if its speed deviates too much.(58 votes)

- Where the gravitational force is more at poles or equator ?(23 votes)
- Wow! That's a fascinating question!

This is what I found. It was posted in 2011:

Just a few weeks ago the European Space Agency released results from its satellite GOCE which answers this question - In general it seems that "Gravitational acceleration at Earth’s surface is about 9.8 m/s², varying from a minimum of 9.788 m/s² at the equator to a maximum of 9.838 m/s² at the poles." from the same site. But there is a lot of regional variation - some of which we cannot explain.

Really interesting- thank you for asking!!(44 votes)

- How does one derive centripetal acceleration =v^2/r(13 votes)
- That's a tricky question, and it is a bit beyond the scope of these videos. If you are comfortable with some calculus, though, a pretty nifty derivation can be found here:

http://blogs.msdn.com/b/matthew_van_eerde/archive/2010/01/24/deriving-the-centripetal-acceleration-formula.aspx

The author's notation is a little sloppy, going from vector notation to scalar notation at a whim, but it's still a pretty good derivation.

-RNS(2 votes)

- Why dont we need to cionsider the earth's orbit in these examples and in projectile motion? It doesnt matter right?(5 votes)
- I could be wrong about this, but I believe it's because we're calculating in relation to the earth.

If we introduced a third body then we'd probably have to factor in Earth's orbit and I imagine things would get much more complicated.(6 votes)

- How would you calculate the speed for an elliptical orbit?(4 votes)
- Speed is always distance per unit of time, regardless of the path.(5 votes)

- If you were at the center of the earth (assuming that the heat and pressure don't obliterate you and there is a space for you to stand), what would happen? Would you stay there? Would you be ripped apart as gravity pulls you in each direction?(3 votes)
- I was wondering, does a rocket actually need 11.2 km/sec of escape velocity initially to escape from the earth because that seems very crazy if so.(3 votes)
- No. 11.2 km/sec is needed to
*completely*escape Earth and start orbiting the Sun. Hence the term "escape velocity." Its the velocity you need to escape Earth's gravity well for good completely and then get into a heliocentric orbit.

Real rockets don't go that fast, they just get around 27,000 kmph to get into orbit, which is just around 7.6 km/s.

That is the speed where you are falling fast enough where you miss the Earth, or in other words you acheive orbit at that rate.

For instance, if you want to go to Mars, what they do first is they get into orbit, 7.6 km/s and*then*from there on do another burn to get their velocity to 11.2 km/sec to escape Earth's orbit and get onto the transfer orbit towards the planet.(4 votes)

- At0:42, How is a,c = v^2/r ? Is there a video to explain this concept?(3 votes)
- If I throw a hammer on the surface of the moon since there is no atmosphere all the potential energy will be converted to heat energy, contrary to what happens on earth where a part of energy gets converted to sound as well?(2 votes)
- Hello Krishnaonlyspam,

What a fun question!

Let's talk about the sound part.

If we bring a bell to the moon and strike it does it ring? Yes, it still rings - it you used a high speed camera you could see the vibrations. But, the energy does not get converted into "sound" as there is no atmosphere. Instead this vibrating energy is converted into heat inside the bell. A similar situation happens in the hammer. Like you said, all heat - assuming we ignore the PE change of dust as the hammer strikes the surface...

With regards to the bell, the atmosphere carries energy away in the form of sound. We could say the atmosphere dampens the bell. Or stated another way, the bell would vibrate longer on the moon than on the earth.

Regards,

APD(4 votes)

- Maybe it's an stupid question, but here we go.

In every radius (height from earth's surface, or massive any body) we have a different needed velocity for keep orbiting around. My question is, during the formation of our solar system, for example (assume the sun is completely formed), we only have billions of 'flying rocks' moving around with no sense. All of this 'rocks' have a different initial velocity, depending of the initial acceleration due to, i'd say, the previous supernova who gave all the mass to our solar system. Then, all of this 'rocks' would find his orbit radius, depending of them velocities. So, the fastest ones, will orbit much further than the slower ones. And this method will start to make a logic paths for every one.

Is that correct?(2 votes)- It would be that simple if the "rocks" didn't interact through collisions and mutual gravitational attraction. The initial velocities would have the rocks in all sorts of orbital distances and amounts of eccentricities that would have them crossing each other's orbits so over time some material gets ejected from the solar system and the stuff that is left settles down into regular orbits like we see in the current solar system.(2 votes)

## Video transcript

Now that we know the
magnitude of the acceleration due to gravity at 400 kilometers
above the surface of the earth, where the space
station might hang out, what I want to do in this video
is think about how fast does the space station need to be
moving in order to keep missing the earth as it's falling, or
another way to think about it, in order to stay
in orbit, in order to maintain its circular
motion around the earth. So we know from our
studies of circular motion so far, what's keeping it going
in circular motion, assuming that it has a constant
speed, is some type of centripetal acceleration. And that centripetal
acceleration is the acceleration
due to gravity. And we figured out what
it was at 400 kilometers. And so we know that that
centripetal acceleration-- let me write it here
in pink-- we know that the magnitude of that
centripetal acceleration has to be equal to the speed or
the magnitude of the velocity squared divided by
divided r, where r is the radius of
the circular path. So in this case,
it'd be the radius of the orbit, which
would be the radius of the Earth, plus the altitude. So that, we already figured
out in the last video, is 6,771 kilometers. Let's just solve this
for v, and then we can put in the numbers
in our calculators. So you multiply both sides by
r, and you flip the two sides. You get v squared is
equal to the magnitude of our acceleration
times the radius. The magnitude of our
velocity or speed is equal to the square root
of our acceleration times-- or the magnitude of
our acceleration-- times the radius. And so let's get
our calculator out, and you can verify that
the units work out. This is meters per second
squared times meters, which gives us meters squared
per second squared. Take the square root
of that, and you get meters per second, which
is the appropriate units. But let's get our calculator
out and actually calculate this. Let me see. My calculator's sitting
on my other screen. There you go. And then we want to calculate
the principal square root of the acceleration due to
gravity at this altitude, the magnitude of
the acceleration due to gravity at that
altitude is 8.69 meters per second squared times the
radius of our circular path. That's going to be the
radius of Earth, which is 6,371 kilometers plus the
400 kilometers of altitude that we have in this scenario. So that gives us 6.-- and we
did this in the last video-- 6.771 times 10 to
the sixth meters. And it's important that
everything here is in meters. Our acceleration is in
meters per second squared. This right over
here is in meters. So the units don't
do anything strange. And then we get a drum
roll for how fast-- and this is going to be
in meters per second. We already thought about
how the units will work out. We get 7,670-- I could say 71. But actually I'm
just going to stick to three significant digits. 7,670 meters per second. So let me write that down. The necessary velocity
to stay in orbit is 7,670 meters per second. So let's just conceptualize
that for a second. Every second, it's
going over 7,000 meters. Or every second, it's
going over 7 kilometers. Every second. It's going at this super-- if
we assume that's the direction it's traveling, it's going at
this super-, superfast speed. And if we want to translate
that into kilometers per hour, you just take 7,670
meters per second. If you want to know
how many meters it's going to do in an
hour, you just say, well, there's 3,600
seconds per hour. And so if you
multiply that, that's how many meters it
will travel in an hour. But if you want
that in kilometers, you just divide by 1,000. You have 1 kilometer
for every 1,000 meters. Meters will cancel out. Seconds will cancel out. And you are left with
kilometers per hour. So let's do that. So that was our previous answer. We multiply by 3,600 and
then divide by 1,000. So we really could have
just multiplied by 3.6. And then we get 27,000, roughly
27,600 kilometers per hour. So this is really an
unfathomable speed. And you might be wondering, how
does such a big thing maintain that type of speed, because even
a jet plane, which is nowhere near this fast, has to have
these huge engines to maintain its speed. How does this thing maintain it? And the difference between
this and a jet plane is that a jet or a car or if
I throw a ball or anything like that-- but a jet plane. Let's focus on a jet
plane so we don't have to worry
about other things. A jet plane has to
travel through the air. It has to travel
through the air. And actually, it uses the air as
kind of its form of propulsion. It sucks in the air, and then it
spits out the air really fast. But it has all of
this air resistance. So if the engines were
to just shut down, all the air would
bump into the plane and provide essentially
friction to slow down the plane. What the space station or the
space shuttle or something in space has going for
itself is that it's traveling in an almost
complete vacuum. Not 100% complete vacuum,
but almost complete vacuum. So it has pretty much
no air resistance, negligible air resistance
to have to deal with. So we know from Newton's
laws, an object in motion tends to stay in motion. So once this thing
gets going, it doesn't have air
to slow it down. It'll keep staying that speed. In fact, if it did
not have gravity, which is causing this
centripetal acceleration right over here, it would
just go in a straight path. It would go in a straight
path forever and ever. And that brings up
an interesting point. Because if you are
in orbit like this, traveling at this very,
very, very fast speed, you have to make sure that
you don't vary from this speed too much. If you slow down, you will
slowly spiral into the Earth. And if you speed up a
lot beyond that speed, you will slowly spiral
away from the Earth. Because then the centripetal
acceleration due to gravity won't be enough to keep you
in a perfect circular path. So you really have to stay
pretty close to that speed right there.