Naked Science Forum
General Science => General Science => Topic started by: Alan McDougall on 09/08/2008 16:29:23

Centrifugal force.
If one took a hypothetical rope, half the diameter of the earth or radius, and swung it around at a 1000 miles per hour (the same speed the earth revolves around the celestial poles it must generate a force similar to gravity.
If the earth did not revolve, there would be no centrifugal force and it would seem logical that due to the absence of this action, our weight must differ minutely on the surfaces of revolving and nonrevolving earths.
Could the Naked Scientist come up with an answer please?
Regards
Alan

There is a difference, but it's small.
Imagine that you are at the north pole, the earth's spin is just turning you about a vertical axis, so there's no centrifugal force. You still weigh pretty much the same there as at the equator.
Actually this is complicated by the fact that, at the poles, you are slightly nearer to the centre of the earth so that also makes you weigh more.

In combination, the equatorial bulge and the effects of centrifugal force mean that sealevel gravitational acceleration increases from about 9.780 m·s2 at the equator to about 9.832 m·s2 at the poles, so an object will weigh about 0.5% more at the poles than at the equator.
http://en.wikipedia.org/wiki/Earth%27s_gravity
If the Earth stopped rotating I think the equatorial bulge would disappear and the Earth would become truly spherical,
rather than an oblatespheroid.

Bored Chemist and others
Thank you for your answers which cleared it up for me nicely.
There is a difference, but it's small.
Imagine that you are at the north pole, the earth's spin is just turning you about a vertical axis, so there's no centrifugal force. You still weigh pretty much the same there as at the equator.
Actually this is complicated by the fact that, at the poles, you are slightly nearer to the centre of the earth so that also makes you weigh more.
Just a point about your above quote, in another thread I asked how our weight would differ in a deep hole. The concensus was that one would weigh less. But you say at the poles we weight minutely more. This goes counter to the other explanation. RD calculation was an 0.5 difference which is considerable
Regards
Alan

This goes counter to the other explanation
No, the explanation is fine.
When you are down in a hole, there is some mass 'above you' or, more precisely, you are inside a thin shell of the Earth which has no net contribution to your weight  so you weigh less.
On the North Pole, all the mass is 'beneath your feet' and you are nearer the Centre of Mass  so you will weigh a bit more.

Sophie,
Nice explanation I really understand it now, thanks
Alan

In combination, the equatorial bulge and the effects of centrifugal force mean that sealevel gravitational acceleration increases from about 9.780 m·s2 at the equator to about 9.832 m·s2 at the poles, so an object will weigh about 0.5% more at the poles than at the equator.
http://en.wikipedia.org/wiki/Earth%27s_gravity
If the Earth stopped rotating I think the equatorial bulge would disappear and the Earth would become truly spherical,
rather than an oblatespheroid.
What force would create the sphere?

In combination, the equatorial bulge and the effects of centrifugal force mean that sealevel gravitational acceleration increases from about 9.780 m·s2 at the equator to about 9.832 m·s2 at the poles, so an object will weigh about 0.5% more at the poles than at the equator.
http://en.wikipedia.org/wiki/Earth%27s_gravity
If the Earth stopped rotating I think the equatorial bulge would disappear and the Earth would become truly spherical,
rather than an oblatespheroid.
What force would create the sphere?
The same one that currently (almost) does; gravity.

Hi all
So Alan
Centrifugal force.
If one took a hypothetical rope, half the diameter of the earth or radius, and swung it around at a 1000 miles per hour (the same speed the earth revolves around the celestial poles it must generate a force similar to gravity.
If the earth did not revolve, there would be no centrifugal force and it would seem logical that due to the absence of this action, our weight must differ minutely on the surfaces of revolving and nonrevolving earths.
Could the Naked Scientist come up with an answer please?
Regards
Alan
Yes your correct and its quite a straight forward calculation.
Acceleration (centripetal) = Velocity squared /Radius or a(c) = V^2/R
plug in the numbers in SI units (metres and seconds )
Now the rotational velocity is obviously greatest at the equator/middle, which when you do the calculation gives a value of
a(c) = 33.69 x 10 3 ms^2 or 33.69 mm^s2 (read that as, metres per second squared or millimetres per second squared )
and given the value of earths gravitational acceleration is 9. 81 ms^2 it is not an inconsiderate value, indeed the strength of the suns gravity field at earths distance is approximately
6.0 x10^3 m s^2 ( 6 millimetres per second squared.)
("its per second squared because its acceleration" due to the circular/centripetal motion is towards the center )
Where I am in the middle of England at an altitude of 250 m my velocity is reduced to 278.3 ms^1 which gives;
a(c) = 20.24 mms^2
so your weight difference at the equator (purely due to centripetal acceleration or no centripetal acceleration )
would be
Delta = 1Kg x 9.776 /1Kg x 9.810x 100%
Delta = 0.99656.....% or 0.3434 % less
So if you want to lose weight the easy way, move towards the equator ;)

Pedantically, rotation might not affect your weight!
Weight can be defined the force exerted by gravitation on a body. This doesn't change, but rotation provides an opposing force.
Or you can define it as the total force exerted by gravitation and acceleration, which does change.
0.34% reduction in weight is important when reviewing athletic records for throwing or jumping!

Hi all
So Alan
Centrifugal force.
If one took a hypothetical rope, half the diameter of the earth or radius, and swung it around at a 1000 miles per hour (the same speed the earth revolves around the celestial poles it must generate a force similar to gravity.
If the earth did not revolve, there would be no centrifugal force and it would seem logical that due to the absence of this action, our weight must differ minutely on the surfaces of revolving and nonrevolving earths.
Could the Naked Scientist come up with an answer please?
Regards
Alan
Yes your correct and its quite a straight forward calculation.
Acceleration (centripetal) = Velocity squared /Radius or a(c) = V^2/R
plug in the numbers in SI units (metres and seconds )
Now the rotational velocity is obviously greatest at the equator/middle, which when you do the calculation gives a value of
a(c) = 33.69 x 10 3 ms^2 or 33.69 mm^s2 (read that as, metres per second squared or millimetres per second squared )
and given the value of earths gravitational acceleration is 9. 81 ms^2 it is not an inconsiderate value, indeed the strength of the suns gravity field at earths distance is approximately
6.0 x10^3 m s^2 ( 6 millimetres per second squared.)
("its per second squared because its acceleration" due to the circular/centripetal motion is towards the center )
Where I am in the middle of England at an altitude of 250 m my velocity is reduced to 278.3 ms^1 which gives;
a(c) = 20.24 mms^2
so your weight difference at the equator (purely due to centripetal acceleration or no centripetal acceleration )
would be
Delta = 1Kg x 9.776 /1Kg x 9.810x 100%
Delta = 0.99656.....% or 0.3434 % less
So if you want to lose weight the easy way, move towards the equator ;)
You've forgotten that centripetal acceleration acts perpendicular to the axis of rotation, not the earths surface, which means that a plumb line doesn't hang vertically, except at the poles and equator. At worst (at 45° latitude), it's about 0.1° off, or 4mm on a 2.4m wall.

Hi all,
So vhfpmr
because the rotational motion affects both the surface of the Earth as well as
the bob, then the plumb Bob remains perpendicular to the Earth’s surface
4 mm out of plumb in 2.4 m on every wall built would have a good bricklayer wondering wtf that’s an extra perp joint for every floor up

Hi all,
So vhfpmr
because the rotational motion affects both the surface of the Earth as well as
the bob, then the plumb Bob remains perpendicular to the Earth’s surface
4 mm out of plumb in 2.4 m on every wall built would have a good bricklayer wondering wtf that’s an extra perp joint for every floor up
Er, no it doesn't. Gravity (G, in black) acts perpendicular to the earths surface, centripetal acceleration (C in green) acts perpendicular to the axis of rotation. The centripetal acceleration can then be resolved into two orthogonal components (red), one Cv acting vertically which makes no difference to the angle of the plumb line, and the other, Ch, acting horizontally, which swings the plumb line away from the vertical. The plumb line only hangs vertically at the equator, where the horizontal component is zero, and at the poles where the centripetal acceleration is zero.
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Hi all,
So vhfpmr I was quoting directly from
https://www.sfu.ca/~boal/211lecs/211lec12.pdf
Bottom of page 2
Quote:
(Lastly, because the rotational motion affects both the surface of the Earth as well as
the bob, then S remains perpendicular to the Earth’s surface)
What I believe this means is if you did the plumb Bob test over say an area of water 💧 on earth’s surface the plumb line would be vertical/perpendicular at every latitude.
So 🧱bricklayers rest easy, you cannot blame the latitude and north/south elevation of the wall
one on top of two

I was quoting directly from
https://www.sfu.ca/~boal/211lecs/211lec12.pdf
Hmm, OK.
I follow the maths which concludes with
"ε ~ 0.0017 x 57.3 = 0.1 of a degree"
which is the figure I got. But then having arrived at that, they throw in
"because the rotational motion affects both the surface of the Earth as well as the bob, then S remains perpendicular to the Earth’s surface"
without further explanation, which is the bit I don't follow.

The plumb line only hangs vertically at the equator,
For a particular definition of vertical. (essentially: "points to the middle of the Earth").
But if your definition is "perpendicular to the surface of the Earth" then the plumb line is always vertical.
This definition also means that things fall vertically and you can build a vertical tower of bricks without needing cement.
Er, no it doesn't. Gravity (G, in black) acts perpendicular to the earths surface
You have assumed a spherical Earth often a reasonable start, but not in this thread about the Earth not being a sphere.

Here's a spindle with a radius arm attached. Hanging by a piece of string from the radius arm is a weight, which is pulled toward the centre of the spindle by a force provided by a piece of elastic. If I take it up to the space station where there's little/no gravity, the string and elastic will be colinear whist the spindle is stationary, but when the spindle rotates, centrifugal force will move the weight away from the spindle like the bobs on a steam engine regulator, no? What changes if I replace the elastic with gravity?
Globe 2.png (3.8 kB . 594x689  viewed 3073 times)

The plumb line only hangs vertically at the equator,
For a particular definition of vertical. (essentially: "points to the middle of the Earth").
But if your definition is "perpendicular to the surface of the Earth" then the plumb line is always vertical.
This definition also means that things fall vertically and you can build a vertical tower of bricks without needing cement.
Er, no it doesn't. Gravity (G, in black) acts perpendicular to the earths surface
You have assumed a spherical Earth often a reasonable start, but not in this thread about the Earth not being a sphere.
I've assumed a spherical earth because I'm arguing about the effect of rotation on a plumb line, not variations in gravity or the effect of rotation on the shape of the earth. If the earth is spherical, 'points at the centre' is the same as 'perpendicular to the surface'.

[ Invalid Attachment ] Here is a picture of a section through the planet Zog it's very clearly elliptical.
If you look you will see that a line perpendicular to the surface does not go through the centre of Zog.
The reason Zog is that shape is because it's spinning fast.
The effect of the combination of gravity and spin ensures that the pendulum always hangs perpendicular to the surface of Zog, but it does not go through the centre unless you are at the equator or a pole.

the effect of rotation on a plumb line, not variations in gravity or the effect of rotation on the shape of the earth.
The shape of the earth is the same thing as the effect of rotation.

I can see that the distortion of the earth's shape moves the vertical in the same direction as centrifugal force moves the pendulum, but it wasn't obvious that they necessarily cancel out to zero. It would depend on the degree to which the earth behaves like a liquid and not like an elastic solid, but I suppose the earth's solid crust is too thin to have any significant effect.

It would depend on the degree to which the earth behaves like a liquid and not like an elastic solid
In fact, it behaves like a liquid it even has tides and viscosity.
If the net effect wasn't zero then, over the aeons, the sand grains and the sea would shift until it did.
It seems Mr Newton was cleverer than me. :)
When I first considered this problem I thought about it in terms of the forces acting on a grain of sand on the surface of the Earth. That's a nasty calculus problem, and I gave up on it without even starting.
He noticed that the question was equivalent to having a water filled pipe from the pole to the equator which is a rather easier problem to address.
I'm unlikely to get a knighthood, run the royal Mint and be remembered centuries after my death. :)