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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.

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  • blobby green style avatar for user Marcus
    How does the craft in space get up to that speed without the jets and how does it maintain that speed?
    (41 votes)
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    • leafers sapling style avatar for user Peter Collingridge
      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)
  • male robot hal style avatar for user varghesesherwin
    Where the gravitational force is more at poles or equator ?
    (23 votes)
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    • male robot hal style avatar for user Yisrael F
      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)
  • hopper jumping style avatar for user Ravi
    How does one derive centripetal acceleration =v^2/r
    (13 votes)
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  • blobby green style avatar for user Sabrina Wu
    Why dont we need to cionsider the earth's orbit in these examples and in projectile motion? It doesnt matter right?
    (5 votes)
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    • piceratops ultimate style avatar for user Ivan Occam
      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)
  • female robot ada style avatar for user Cece C
    How would you calculate the speed for an elliptical orbit?
    (4 votes)
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  • female robot grace style avatar for user cjddowd
    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)
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  • starky sapling style avatar for user 👑💎 Professor Sykes 💎👑
    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)
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    • aqualine ultimate style avatar for user Inspiron13
      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)
  • piceratops ultimate style avatar for user Rickston Pinto
    At , How is a,c = v^2/r ? Is there a video to explain this concept?
    (3 votes)
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  • female robot ada style avatar for user krishnaonlyspam
    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)
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    • purple pi purple style avatar for user APDahlen
      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)
  • orange juice squid orange style avatar for user BLDR
    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)
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    • male robot hal style avatar for user Charles LaCour
      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.