Yep. Most people don't realize this though, they think that space has microgravity or no gravity, and this is why astronauts are weightless. They don't even think about the "orbit" part. I posted this b/c I'm curious among my readers who has/how many have that misconception.
I am at the AAPT (follow my real-name twitter or Facebook for more details) and textbook author Hewitt reminded me that most of our students don't get this. See my comment above to spazzy444.
My guess is that the Apollo 11-17 astronauts experience a different sort of weightlessness as they left Earth orbit, traveled to the Moon, and entered Moon orbit. Am I right? Would they have been able to tell from one to the next?
Well keep in mind that the Moon orbits the Earth too, so even when orbiting the Moon you're experiencing gravity from the Earth.
"Free fall" or "weightlessness" is the same thing as saying that there's no normal force from being pressed against a surface. Weightlessness from orbiting Earth or the Moon, from moving in a projectile trajectory (either near a planet's surface or in space between bodies), or from hovering at the turnover/Lagrange point between two bodies, all of these would be perceived as the same in a closed box without windows by all instruments/sensors/human senses. Mathematically these are different (in a FBD for example), but there's no way to measure which situation you are in.
I had the impression that moving a certain distance from Earth reduces gravitational forces, but I don't know how by how much.
Examples that might help me understand this better would probably be the answers to the following problems (not that I know how to solve them)
1a) Assume a spaceship leaves Earth and travels 100,000 miles away, then fires retrorockets in order to come to a complete stop position relative to Earth. What sort of pull effect will the Earth have on it at this time (ignoring the effects of the Moon)? What sort of gravity will any astronauts feel?
1b) Assume the above, except the spaceship travels 1,000,000 miles away before stopping. What sort of pull effect will the Earth have? What sort of gravity will any astronauts feel?
2) Re-assume the existence of Moon, and the spaceship flies n miles away from Earth, between Earth and Moon. At about what n does the spaceship get pulled equally by Earth and Moon such that it stays more or less in that spot? I think, to make this make sense to me, we must assume the spaceship also keeps itself in line with the fraction of Earth-orbiting speed it needs to maintain a line between the Moon, itself, and Earth. I feel the necessary velocity is more or less the fractional speed of Moon rotation based on how big of an orbit the spaceship is tracing. Will astronauts notice any gravitational effects? If so, how much?
I also am not sure how difficult these problems are to solve, but if anyone feels like solving them and improving my understanding of mechanics, that would be awesome (:
To start off with, the formula to determine the force of gravity is F=GMm/r^2. F is Force, G is the gravitational constant (it's just a number, don't worry about it), M is the mass of one object (here the Earth), m is the mass of the other object (if you're looking at the acceleration due to gravity this term gets dropped), and r is the distance between the two objects' centers (or the radius of the orbit). The answers to all of your questions are well known, and should be calculable by anyone taking a calc-based Physics I course, or by most algebra-based Physics I students.
1a) [FYI: The speed of the spaceship when it crosses the 100,000 mi line is not relevant.] The radius of the Earth is around 4,000 mi, so this distance is 25 times further, so the strength of gravity will be 1/25^2 what it would be on the surface of the Earth. It will not be directly noticeable however unless there is a solid surface under the astronauts' feet, for example if the ship's thrusters were turned on just strongly enough to keep it at the same constant 100,000mi from Earth.
Note that the ISS is located only around 20 miles above the Earth's surface, so that its orbital radius is around 4,020mi, so the strength of gravity on the ISS is only a tiny bit weaker than it would be on the surface of the Earth [(4000/4020)^2 times weaker].
1b) Same as 1a, but now gravity is 1/250^2 weaker.
2) There are two different points in space that you are conflating. The first is the L1 Lagrange point, which is the spot in space between two objects where another thing (a spaceship) can orbit the bigger thing (the Earth) at the same rate as the smaller thing (the Moon). Some numbers related to this: Earth's radius ~6,400km, Moon's radius ~1,700km, Earth-Moon distance ~380,000km, L1 point ~60,000km from Moon (~320,000km from Earth).
The second is sometimes called the turnover point or the neutral gravity point, which is the spot in space at which the net gravitational force from the Earth and Moon are balanced. It is called the turnover because when traveling between the two objects as the Apollo ships did, when you pass through that point you transition from working against Earth's gravity, to instead working against the Moon's gravity. This is located approximately ~340,000km from Earth.
i put ‘yes’, because that's why it's orbiting and because gravity is everywhere, but i think people who say ‘no’ probably mainly just mean that, relative to a frame of reference that moves with the ISS, gravity is not a force that's easily observed in the behavior of normal human-scale objects, as reflected in colloquial terms like ‘zero gravity’.
i think people who say ‘no’ probably mainly just mean that, relative to a frame of reference that moves with the ISS, gravity is not a force that's easily observed in the behavior of normal human-scale objects
I don't think most people who say "no" think that much about it. Your speculation is that they realize that the force of gravity exists but is not directly measurable by way of a normal force. If you are correct, then if these people were able to draw a Free Body Diagram, they would include an arrow for gravity. It is my belief that the people who say "no" think that there is no force of gravity at all. If I am correct, they wouldn't have an arrow for gravity at all.
it's not that it's not measurable by normal force - it's that it's not there at at all in the frame that takes the orbiting station to be stationary. if you adopt such a frame, i'm pretty sure you can do all the computations for the mechanics of objects in the station in a way that does not once mention gravity (or, at least, does not mention the Earth's gravitational pull - the objects bouncing about the station are of course interacting with each other gravitationally, but i assume it's customary to ignore this). and it ought to be a pretty well-behaved inertial frame in other respects, if i remember this right (it has, admittedly, been a while).
of course, i don't think most people are thinking about it like this, but the point is that, for the movements of objects aboard the station relative to the station, if we do all our math as if there is no gravity involved, we ought to get the right results, and even people who aren't up to doing the math probably have an intuitive grasp of that.
for the movements of objects aboard the station relative to the station, if we do all our math as if there is no gravity involved, we ought to get the right results,
Well, technically it would still be a non-inertial reference frame, so if the station is large enough then an assumption of no gravity and an inertial reference frame would not be correct.
If you care about the details, if the station is large enough in the radial direction, objects floating in the station closer to or farther from the Earth would appear to drift forward or backwards in orbit (respectively, I think). If the station is large enough in the other direction perpendicular to travel (i.e., for an East-West orbit, long in the North-South direction) you would be able to measure the Coriolis force.
well, i mean, you're clearly entitled to risk talking down if i seem not to know what i'm talking about. in some cases, when i perceive things as talking down, i get snippy, but even in those cases that's an issue with me, but a sign that you've done anything wrong.
It sounds like you're mixing up Special and General Relativity. Special Relativity can be invoked when everything can be done assuming something is stationary in an inertial reference frame, i.e. there is no acceleration involved relative to any other inertial reference frame. It's usually used for objects moving at significant fractions of the speed of light relative to each other. Gravity, however, being force-like (and therefore appearing to cause acceleration) is necessarily non-inertial. In General Relativity, to deal with gravity as a not-force, rather than treating an object as stationary in some frame, we treat spacetime as warped and an object as moving along a geodesic of spacetime's new geometry (it's been a long time and I forget whether we pick the largest object or the center of mass of the entire system of objects to be the stationary reference frame, probably the latter). A geodesic through spacetime can only be stationary in space when spacetime is completely flat--that is, there is nothing with mass around (so in that case, it reduces back to Special Relativity).
you're right, of course, that i was mixing things up and/or being sloppy.
i was mainly remembering being taught to think of an elevator in free-fall as a canonical inertial frame, which is, i'm sure, wrong in the details, but which is really good enough a lot of the time.
Well, it is true that if you decide to invoke curvature of spacetime rather than gravity, you can redefine inertial to mean anything moving along a geodesic in your curved spacetime, and it won't contradict the Special Relativity definition of inertial (just expand on it). In that case, an elevator in free-fall is an inertial frame, yes. It's an either-or thing, though. If you call it "gravity" instead, then it's an accelerating frame, which is not inertial. But the basic argument in favor of General Relativity (ignoring, of course, that now that the framework has been built up, GR was shown to predict new phenomena which have since been discovered) is that an object in freefall looks, in a lot of ways, more like an object in an inertial frame than like an object with a force applied to it.
Based on what you said about people getting this wrong, I think a lot of people parse "gravity" as "that sensation I feel when I'm falling to the ground", rather than "a force between any two objects". I'm not sure how you'd fix that impression, actually.
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My guess is that the Apollo 11-17 astronauts experience a different sort of weightlessness as they left Earth orbit, traveled to the Moon, and entered Moon orbit. Am I right? Would they have been able to tell from one to the next?
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"Free fall" or "weightlessness" is the same thing as saying that there's no normal force from being pressed against a surface. Weightlessness from orbiting Earth or the Moon, from moving in a projectile trajectory (either near a planet's surface or in space between bodies), or from hovering at the turnover/Lagrange point between two bodies, all of these would be perceived as the same in a closed box without windows by all instruments/sensors/human senses. Mathematically these are different (in a FBD for example), but there's no way to measure which situation you are in.
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Examples that might help me understand this better would probably be the answers to the following problems (not that I know how to solve them)
1a) Assume a spaceship leaves Earth and travels 100,000 miles away, then fires retrorockets in order to come to a complete stop position relative to Earth. What sort of pull effect will the Earth have on it at this time (ignoring the effects of the Moon)? What sort of gravity will any astronauts feel?
1b) Assume the above, except the spaceship travels 1,000,000 miles away before stopping. What sort of pull effect will the Earth have? What sort of gravity will any astronauts feel?
2) Re-assume the existence of Moon, and the spaceship flies n miles away from Earth, between Earth and Moon. At about what n does the spaceship get pulled equally by Earth and Moon such that it stays more or less in that spot? I think, to make this make sense to me, we must assume the spaceship also keeps itself in line with the fraction of Earth-orbiting speed it needs to maintain a line between the Moon, itself, and Earth. I feel the necessary velocity is more or less the fractional speed of Moon rotation based on how big of an orbit the spaceship is tracing. Will astronauts notice any gravitational effects? If so, how much?
I also am not sure how difficult these problems are to solve, but if anyone feels like solving them and improving my understanding of mechanics, that would be awesome (:
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1a) [FYI: The speed of the spaceship when it crosses the 100,000 mi line is not relevant.] The radius of the Earth is around 4,000 mi, so this distance is 25 times further, so the strength of gravity will be 1/25^2 what it would be on the surface of the Earth. It will not be directly noticeable however unless there is a solid surface under the astronauts' feet, for example if the ship's thrusters were turned on just strongly enough to keep it at the same constant 100,000mi from Earth.
Note that the ISS is located only around 20 miles above the Earth's surface, so that its orbital radius is around 4,020mi, so the strength of gravity on the ISS is only a tiny bit weaker than it would be on the surface of the Earth [(4000/4020)^2 times weaker].
1b) Same as 1a, but now gravity is 1/250^2 weaker.
2) There are two different points in space that you are conflating. The first is the L1 Lagrange point, which is the spot in space between two objects where another thing (a spaceship) can orbit the bigger thing (the Earth) at the same rate as the smaller thing (the Moon). Some numbers related to this: Earth's radius ~6,400km, Moon's radius ~1,700km, Earth-Moon distance ~380,000km, L1 point ~60,000km from Moon (~320,000km from Earth).
The second is sometimes called the turnover point or the neutral gravity point, which is the spot in space at which the net gravitational force from the Earth and Moon are balanced. It is called the turnover because when traveling between the two objects as the Apollo ships did, when you pass through that point you transition from working against Earth's gravity, to instead working against the Moon's gravity. This is located approximately ~340,000km from Earth.
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I don't think most people who say "no" think that much about it. Your speculation is that they realize that the force of gravity exists but is not directly measurable by way of a normal force. If you are correct, then if these people were able to draw a Free Body Diagram, they would include an arrow for gravity. It is my belief that the people who say "no" think that there is no force of gravity at all. If I am correct, they wouldn't have an arrow for gravity at all.
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of course, i don't think most people are thinking about it like this, but the point is that, for the movements of objects aboard the station relative to the station, if we do all our math as if there is no gravity involved, we ought to get the right results, and even people who aren't up to doing the math probably have an intuitive grasp of that.
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Well, technically it would still be a non-inertial reference frame, so if the station is large enough then an assumption of no gravity and an inertial reference frame would not be correct.
If you care about the details, if the station is large enough in the radial direction, objects floating in the station closer to or farther from the Earth would appear to drift forward or backwards in orbit (respectively, I think). If the station is large enough in the other direction perpendicular to travel (i.e., for an East-West orbit, long in the North-South direction) you would be able to measure the Coriolis force.
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i was mainly remembering being taught to think of an elevator in free-fall as a canonical inertial frame, which is, i'm sure, wrong in the details, but which is really good enough a lot of the time.
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