Acceleration Due to Gravity at the Space Station What is the acceleration due to gravity at the space station
Acceleration Due to Gravity at the Space Station
⇐ Use this menu to view and help create subtitles for this video in many different languages. You'll probably want to hide YouTube's captions if using these subtitles.
- Most physics books will tell you that acceleration due to gravity near the surface of the Earth is 9.81 m/s^2.
- This is an approximation, and what I want to do in this video is figure out if this is the value we get
- when we actually use **Newton's Law of Universal Gravitation**
- and that tells us that the force of gravity between two objects, (and let's just talk about the magnitude of the force of gravity between two objects)
- is equal to the universal gravitational constant times the mass of one of the bodies,
- *m1*, times the mass of the second body, divided by the distance between the center of masses of bodies square.
- So distance between the center of masses of the bodies squared.
- So let's use this, the Universal Law of Gravitation to figure out what acceleration due to gravity should be at the
- surface of the Earth.
- I have *G* over here, I have the mass of the Earth, which I have looked up over here,
- We also have the radius of the Earth and for the sake of this we are going to assume that the distance between the body at the surface of the
- Earth the distance between that and the center of the Earth is just going to be and the radius of the Earth.
- And so this will give us the magnitude of the force. If we want to figure out the magnitude of the acceleration
- (which this really is. I didn't write this as the vector, so this is just the magnitude of the acceleration.
- If you wanted the acceleration, which is a vector, you'd have to say downwards or towards the center of the Earth in this case)
- But if you want the acceleration we just have to remember that force
- is equal to mass times acceleration and if you wanted to solve for acceleration
- you just divide both sides times mass. So force divided by mass is equal to acceleration or, if you want,
- if you take the magnitude of the force and you divided by mass you're going to get the magnitude of the acceleration. This is a scalar
- quantity, this is a scalar quantity right over here. So if you want the acceleration due to gravity,
- let's try divide this in terms of the force of gravity on Earth. So the magnitude of the force of gravity on Earth.
- This one right over here, so this one being the case of Earth, one of these masses is going to be Earth.
- It's going to be this mass right over here. So if you want the acceleration due to gravity at the surface of the Earth
- you just have to divide by the mass that is being accelerated due to that force. And in this case it is the other
- mass. It's the mass sitting on the surface. So let's divide both sides by that mass,
- and this will give us the magnitude of the acceleration on that mass due to gravity.
- So this is equal to the magnitude of acceleration due to gravity.
- And the more reason that this is actually simplifying thing is that these two, these
- *m2* and *m2* cancel out, so the magnitude of our acceleration due to gravity using Newton's Universal Law
- of Gravitation is just going to be this expression right over here. It's going to be the Gravitational constant
- times the mass of the Earth divided by the distance between the object's center of mass and the center
- of the mass of the Earth. And we're going to assume the the object is right at the surface, and its center of mass
- is at the surface. So this is actually going to be the radius of the Earth squared.
- Sometimes this is also viewed as the gravitational field at the surface of the Earth, because if you it multiply by mass
- it tells you how much force is pulling on that mass. With that out of the way let's actually use a calculator to calculate what this value
- is and then what I want to do is, one, compare it to the value that the textbooks tell us and see if maybe why it may or
- may not be different, and then think about how it changes as we get further and further away from
- the surface of the Earth, and in particular, if we get to an altitude where the space shuttle or International Space Station
- might be at. This is an altitude of 400 km, give or take a little bit, depending on what it is up to. So
- let's figure out what this value is when we use Universal Law of Gravitation. So let's get my calculator out.
- We know what *G* is, 6.6738*10^-11. This "E" button means "times 10 to the-11"
- and I want to multiply that times the mass of the Earth, which is right over here,
- and it's 5.9722*10^24 and we want to divide that by radius of Earth squared.
- So this is in kilometers, and I want to make sure everything is in the same units. So 6371 km is the same thing as
- 6,371,000 m (you could just multiply that by a thousand). Or you could even write this as 6.371*10^6 m and we're going to square this.
- That's the radius of Earth, the distance from the center of mass of Earth, and the center of the mass of this object, which is sitting at the surface of the Earth.
- So let's get our drumroll, and we get... 9.8 and if we round we actually get something a
- little higher than what the textbooks give us. We get 9.82. Let's just round it, so we get 9.82 m/s^2. And so you might say,
- "What's going on here? Why do we get a discrepancy between what the Universal Law of Gravitation gives us and what is the average measured due to gravity at the surface of the Earth?"
- And the discrepancy here, the discrepancy between these two numbers is really because
- Earth is not a uniform sphere of uniform density, and that's what we have to assume here with Universal Law of Gravitation.
- It's actually a little bit flatter than a perfect sphere and definitely doesn't have uniform density.
- Different layers of the Earth have different densities, you have all sorts of different interactions,
- Also, if you measure effective gravity, there also a little bit of a buoyancy effect from the air,
- very, very negligible (I don't know if it would've been enough to change this), but there's other minor, minor effects, irregularities,
- Earth is not a perfect sphere, it is not of uniform density,
- and that is what accounts for the bulk of this. Now that out of the way, what I'm curious about is what is the acceleration
- due to gravity if we go up 400 km. So now, the main difference here
- *G* will stay the same,
- the mass of Earth will stay the same,
- but the radius is now going to be different because now we're placing the center of mass of our object
- whether it's the Space Station or someone sitting in the Space Station, they're going to
- be 400 km higher and I'm going to exaggerate what 400 km looks like. This is not drawn to scale.
- But now the radius is the radius of the earth plus 400 km. So now for the case of the Space Station,
- *r* going to be not 6371 km, we're going to add 400 km to this, 6771 km, which is the
- same thing as 6,771,000 m, which is the same thing as 6.771*10^6 m. So let's go back to our calculator
- We should be able to just second entry [that's the last entry we had] and instead of
- 6.371*10^6, let's add 400 km to that. So it was 371 and it's now 771 km.
- What do we get? We get We get 8.69 m/s^2.
- So now, the acceleration here, is 8.69 m/s^2. And you can verify that the units work out because
- Over there gravity is m^3/kg*s^2, you multiply that by mass of the Earth, which is kg. The kg cancels out with these kg.
- And then you're dividing by m^2. You're left with m/s^2. So the units work out as well.
- So there's an important thing to realise and this is a misconception.
- We do a whole video on it earlier when we talk about the Universal Law of Gravitation.
- There is gravity when you are in orbit up here. The only reason why it feels like there's no gravity or it looks like there's no gravity
- is that this Space Station is moving so fast that it's essentially in freefall but it moves so fast
- that it keeps missing the Earth. And in the next video we'll figure out how fast does it have to
- travel in order to stay in orbit, in order for it not to plummet to Earth due to the force of gravity, due to the acceleration that is occurring, this centripetal, center-seeking acceleration.
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
Have something that's not a question about this content?
This discussion area is not meant for answering homework questions.
Share a tip
When naming a variable, it is okay to use most letters, but some are reserved, like 'e', which represents the value 2.7831...
Have something that's not a tip or feedback about this content?
This discussion area is not meant for answering homework questions.
Discuss the site
For general discussions about Khan Academy, visit our Reddit discussion page.
Flag inappropriate posts
Here are posts to avoid making. If you do encounter them, flag them for attention from our Guardians.
- disrespectful or offensive
- an advertisement
- low quality
- not about the video topic
- soliciting votes or seeking badges
- a homework question
- a duplicate answer
- repeatedly making the same post
- a tip or feedback in Questions
- a question in Tips & Feedback
- an answer that should be its own question
about the site