Is Our Weight Problem Worse than We Think?

This one came to me, I swear, in a dream. Okay, so I was taking some extra-strength cough medicine at the time, but still...

In my dream I saw a wet volleyball spinning so fast that droplets of water were being flung off in every direction. Waking up, I thought, gee, it's lucky the earth isn't spinning fast enough to fling us off like that water. But then I thought, wait a minute, I wonder if the spin of the earth, while not strong enough to eject me from the planet, still does have some (measurable?) effect on my attraction to said planet--that is, my weight? It seems to me it must.

Furthermore, I would guess that people at the equator, since they are experiencing the maximum acceleration, have the biggest differences between their apparent and actual weights; and the only completely accurate scales would be those on the north & south poles, and then, of course, you'd have to deduct for your parka. Can anyone say if this is so? Do I actually weigh more even than my scales currently indicate? And here I thought they were being cruelly honest already!

I'm pretty sure this is true. If I remember my physics, I think acceleration of an object travelling in a circle is velocity squared divided by the radius. If I'm crunching the numbers correctly, I think the effect is small--roughly (very roughly) 1/3 of 1% at the equator.😕

I don't think it matters. The acceleration one would experience on the surface of rotating object is called centripetal, i.e. "centre seeking" acceleration. In this case, gravity provides the centripetal acceleration. Friction also plays a part, keeping you from slipping along the surface of the Earth like a fat guy on a slip-n-slide. If gravity from the Earth were to suddenly stop acting on your body, you would continue moving in a straight line (simplified case) without curving, essentially flying off the face of the Earth (but not lifting off like a unanchored balloon).

The centripetal accelartion provided by gravity acts between the centres of mass of the two objects in question (the Earth and you), and depends only on the masses of the two objects and their distance apart, so it wouldn't change depending on how fast you're rotating. A variation in height (distance between centres of mass) would, so you would get different readings at the poles and the equator due to the shape of the Earth.

In the volleyball example, if it were real, the rotational speed of the ball would determine how fast the waterdroplets are moving before the friction/adhesive forces aren't enough to keep them on the ball.

We should be safe, even if Superman spins the Earth superfast to see the future, and then spins it backwards to save Lois Lane from her dirt grave.

Originally posted by PBE6
I don't think it matters. The acceleration one would experience on the surface of rotating object is called centripetal, i.e. "centre seeking" acceleration. In this case, gravity provides the centripetal acceleration. Friction also plays a part, keeping you from slipping along the surface of the Earth like a fat guy on a slip-n-slide. If gravity from the ...[text shortened]... g, essentially flying off the face of the Earth (but not lifting off like a unanchored balloon).

I think what you meant to say was "If the Earth suddenly stopped spinning, and gravity from the Earth were to suddenly stop having an effect on your body", because if the Earth kept spinning as normal, and gravity from the Earth stopped having an effect on your body, then you *would* seem to lift off like a balloon, since you would still be travelling as fast as the surface of the Earth.

Originally posted by Suzianne
I think what you meant to say was "If the Earth suddenly stopped spinning, and gravity from the Earth were to suddenly stop having an effect on your body", because if the Earth kept spinning as normal, and gravity from the Earth stopped having an effect on your body, then you *would* seem to lift off like a balloon, since you would still be travelling as fast as the surface of the Earth.

I agree that at first it would seem to ground level observer that you were float straight upwards as you both continue to travel in the forward-rotation direction, but after a short while it would look like you were travelling upwards and backwards away from the observer (like a balloon in the wind, I suppose. OK, it's pretty balloony, but definitely not straight up.)

Wasn't there a 99 Red Balloons thread around here somewhere last week?

Originally posted by Suzianne
I think what you meant to say was "If the Earth suddenly stopped spinning, and gravity from the Earth were to suddenly stop having an effect on your body", because if the Earth kept spinning as normal, and gravity from the Earth stopped having an effect on your body, then you *would* seem to lift off like a balloon, since you would still be travelling as fast as the surface of the Earth.

Actually, you would carry on moving in the direction that you were moving in when the forces were removed. ie at a complete tangent to the spot where you were standing. Due to the curvature of the Earth, it would fall beneath you, a bit like taking off in an aeroplane, but more subtle.

If gravity were switched off, nobody would have a "weight problem."

But we'd age faster...

Originally posted by PBE6
I agree that at first it would seem to ground level observer that you were float straight upwards as you both continue to travel in the forward-rotation direction, but after a short while it would look like you were travelling upwards and backwards away from the observer (like a balloon in the wind, I suppose. OK, it's pretty balloony, but definitely not straight up.)

Wasn't there a 99 Red Balloons thread around here somewhere last week?

Firstly, I don't believe anyone said "straight up", and

secondly, yes, I agree that the net effect would be to appear to rise and float backwards towards the west (away from the motion of rotation).

Originally posted by jimslyp69
Actually, you would carry on moving in the direction that you were moving in when the forces were removed. ie at a complete tangent to the spot where you were standing. Due to the curvature of the Earth, it would fall beneath you, a bit like taking off in an aeroplane, but more subtle.

Like.... a balloon, perhaps?

Firstly, I don't believe anyone said "straight up"

Touchy, touchy. But I beg to differ. Paula Abdul said it many times on her groundbreaking middle-of-the-road dance pop extravaganza "Straight Up" from her innocuous debut smash sensation "Forever Your Girl".

Originally posted by Suzianne
Like.... a balloon, perhaps?

An idealized balloon would float straight up from the Earth's surface in the radial direction. A person floating off the Earth due to lack of gravity would move in a direction tangential to the Earth's surface. Those directions are perpendicular to each other.

Originally posted by PBE6
[b]Firstly, I don't believe anyone said "straight up"

Touchy, touchy. But I beg to differ. Paula Abdul said it many times on her groundbreaking middle-of-the-road dance pop extravaganza "Straight Up" from her innocuous debut smash sensation "Forever Your Girl".[/b]

LOL

Originally posted by PBE6
An idealized balloon would float straight up from the Earth's surface in the radial direction. A person floating off the Earth due to lack of gravity would move in a direction tangential to the Earth's surface. Those directions are perpendicular to each other.

Oh my... but since the surface of the Earth is, for all purposes, a circle, wouldn't it have an infinite number of tangents?

Ok, sorry... I know what you meant, I just thought I'd take a stab at making a "Bowmann"... I was far too verbose, tho... and not nearly irritating enough... 😞

And I did say "seem to lift off like a balloon" in my first post and "appear to rise up and float backwards" in my second. I fully agree that the path would be tangential.

EDIT: That is, minus any external forces like wind, or coriolis effect.

Okay, I dusted off the old Physics book (cough, cough).

The acceleration due to gravity at the earth's surface is roughly 9.8 meters per second squared and the radius (r) is roughly 6380000 meters. If we assumed a uniform radius and density and the planet was not rotating, then then this acceleration would apply everywhere on the planet. A person standing on the surface does not undergo this acceleration, but rather the force of gravity is countered by the force exerted upwards against their feet and hence the sensation of weight.

The earth rotates 360 degrees every 86164 seconds (t, had to google this one). Thus for a person standing on the surface of the earth (not at a pole) a small portion of the force of gravity is necessary to accelerate them around in a circle. At the equator the velocity is roughly 465 m/sec (2*pi*r/t). As a sanity check we can convert this to 1047 mph, which sounds about right. The centripetal acceleration undergone by an object travelling in circular motion is given by v^2/r, or at the earth's equator 0.0339 m/sec^2. This would act to reduce a person's weight by 0.346%.

Perhaps a more interesting result would be to calculate how fast the earth would have to rotate in order for you to be weightless at the equator. Essentially you could stand on the ground at the equator and be in a geosynchronous orbit--how cool would that be! I believe the answer is 1 hour 24.5 minutes. Of course this all assumes that the shape of the earth is not alterred by these same forces--which is certainly not true.

For those who's curiosity truly knows no bounds, consider the following. The force of gravity acts towards the center of the earth. The centripetal acceleration is normal to the axis of rotation--aligned with gravity only at the equator. Thus, if you could (very gently) stop the planet's rotation, you would find that buildings which had previously been standing straight up and down (relative to the vector sum of the gravitational and centripetal accelerations), would now be slightly tilted relative only to gravity. What is the angle of this tilt as a function of lattitude?

Russ: An equation editor would be a really nice feature in the forums... For the truly nerdy! 🙂

Originally posted by leisurelysloth
Okay, I dusted off the old Physics book (cough, cough).

The acceleration due to gravity at the earth's surface is roughly 9.8 meters per second squared and the radius (r) is roughly 6380000 meters. If we assumed a uniform radius and density and the planet was not rotating, then then this acceleration would apply everywhere on the planet. A perso ...[text shortened]... : An equation editor would be a really nice feature in the forums... For the truly nerdy! 🙂

It would have to be inversely proportional to the latitude.

As a function of the latitude, tho, I'm not at all sure about the actual angle of the tilt.

EDIT: (And yes, I did take trig in school, but I forget most of it, hehe)