I’ll use 9.8 m/s^2 for g[sub]0[/sub] and 6.310^6m for the radius of the earth. I’ll also use 110^5 for h, since that’s about 100km above the surface of the earth, which is generally used as the border for space. Working that out, I get a g[sub]h[/sub] of 9.5 m/s^2. So I have nearly the same weight at the border of space as I do at sea level? I thought I was supposed to be weightless. Where have I erred?
I assumed as much, but the post I was responding to (Chronos’) referred specifically to a balance scale. Unlike scales based on material elasticity (springs, load cells, etc.), balance scales measure mass correctly no matter what the local acceleration is (as long as its non-zero, of course). That is, aside from buoyancy effects, or cases where the local approximation doesn’t hold.
It is a misnomer that there is “zero gravity” in free fall conditions such as at the apex of a parabolic trajectory or in orbit; instead, you are experiencing no net force because everything around you is being accelerated at the same speed. Just as your apparent weight is reduced when you are riding down in an express elevator, when you are in a spacecraft or airplane that is being allowed to fall freely it can impart no force upon you. However, when you are on solid earth the ground doesn’t move and you experience the reaction to gravitational acceleration on your body through your feet.
When you are in orbit what is happening is that you are falling around the planet so fast that you literally fall and keep falling above the horizon. It is critically important to not thing too much about this because it could suddenly stop happeni…oh, no, sorry, that’s Douglas Adams. In fact, you have a stable condition in which your outward acceleration is, on average, matched by gravity and your instantaneous velocity vector is always pointing above the horizon so you keep going indefinitely unless drag from the upper atmosphere or some other factor slows you down. Just being up at altitude, however, doesn’t help you; you could be even beyond the orbit of the Moon, but if you didn’t have any velocity tangent to Earth orbit (and weren’t ejected into solar orbit) you’d just eventually fall back down to Earth. The only exceptions to this are points of gravitational equilibrium between the Earth and the Moon called the libration or Lagrange points (after the mathematician who discovered how to find them and determine their stability) in which you would remain suspended above the Earth orbiting at only the same rate as the Moon.
Stranger
Pretty much as soon as I posted, I realized that it must be that the weightlessness is caused by the free-fall effect, but that was a fantastic explanation. Thanks, Stranger.
Which, once you get to the level of general relativity, is exactly what “zero gravity” means. Given a phrase that’s accurate at the layman’s level, and which is also accurate at the nitty-gritty technical level, I’m inclined to just say that the phrase is accurate.
Substitute “mass” for weight.
Picking this up, not only will my sugar not be sugar at the border of a black hole, but wouldn’t what remained necessarily be at a scale where quantum effects dominate?
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Meaning yeah, you have massy ingredients for a pound cake, but after measuring them, when it comes to pouring them you can expect only some level statistical assurance that what you’re pouring is (was) measured accurately?
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[hijack territory further exploded:] And does that statistical accuracy depend on which subatomic particle is being measured?
But that is the definition of zero gravity! There can be no situation where there is “no gravity”. Zero Gs means you experience no G force.
Addendum to above:
I also wouldn’t know if the ingredient I poured was the actual stuff I measured, but as long as I know it’s defining properties one unit-mass of sugar is the same as any other unit-mass. Right?
- Is even knowing that the damned ingredient is correct–is what I think it is–can be verified only within a statistical boundary?
So does it make a difference if I buy my flour in cloves or quintals? I have a kilderkin of milk I need to use before it goes bad.
This is some hilarious thread, an apparent pound cake recipe that veers to the edge of space and back, quibbles about whether zero G means no gravity (there is no no gravity) or uniform acceleration of one’s surroundings. It even makes a side trip to Mars. Meanwhile, Leo keeps plugging away, talking about I’m honestly not sure what.
Yes, although only slowly until you start to get really, really close. Nevertheless Mercury’s velocity at perihelion influences its mass measurably enough that its orbital precession can’t be calculated correctly without taking relativistic effects into account.
Are you sure that’s the reason?
Kilogram is a base unit of mass not a weight.
It doesn’t matter WHERE you are in the Galaxy, it’s ALWAYS the SAME.
Now, if you’d say: How many cups (volume measurement) that’s a different story.
If you are on the moon, you’re still having the same kilograms but your muscles are used to work against a much higher gravitational force, therefore you can lift heavier equipment and jump higher and further, but you still can’t bent steel like superman.
Ingredients. Measurements. You know, science.
True, though I think Stranger was referring to the misconception by some that there is “no gravity” on space missions because the spacecraft is “far away” from earth. The point being that a spacecraft in orbit is well within the earth’s gravity well. But you’re right, of course, from a relativistic standpoint an astronaut with no external reference could not tell whether he was in a gravity well or not.
Actually, the interesting thing is that orbits are just a special case of the fact that zero gravity is experienced by any object in a pure ballistic trajectory, which really means that all objects in the universe that are not being acted upon by a net force are in zero G. There’s absolutely no difference between an object in a stable orbit and one that is hurtling unpowered (and therefore on a ballistic trajectory) through deep space, even though everything in the universe “feels” the gravitational attraction of everything else within the relativistic limits of its surrounding observable universe.
Well, I’ve made my contribution. I await a revised recipe for pound cake. 