Question

[Johnny5]

yes it rotates (relative to the distant stars) about once every 27.3 days

so there would be a slight centrifugal force lifting you up

which would be some tiny fraction of the weight you would feel without the rotation

 

Doesn't only one side of the moon face us at all times?

 

So to visualize the 27.3 day rotation' date=' I have to stand on the moon? Then, the "fixed stars" will orbit me once every 27.3 days, do I have that right?

 

Assuming I do, you are saying that if I were at the equator of the moon, there is a slight "centrifugal force" reducing my weight? I have read that centrifugal force is a fictious force. Can a fictitous force reduce my weight?

 

Is there an action reaction pair, for the centrifugal force?

 

Without the rotation say my weight is W. I could measure W by standing upon a scale on the moon. Now, let the moon be spinning with respect to the fixed stars. In this case, the moon is buldging at the equator. Let my scale be located on the equator of the moon. So I am looking up at the stars, and they are orbiting me once every 27.3 days. But because the moon is spinning, my weight is now [i']reduced [/i] from W? How do you compute my new weight?

 

I thought the centrifugal force wasn't a real force, how can it reduce my weight. If a force acts on me, there is an action reaction pair, what is the reaction force called?

 

This is highly interesting.

[Johnny5]

what about that centrifugal force for a person standing on the earth's equator?

 

my book says the earth's equatorial radius is 6378 km.

 

if you were standing at equator' date=' by what fraction would your weight-force be diminished by the centrifugal business?

 

my book says the net gee at the equator is 9.78 meters per second^2

(this takes into account both the bulge, being farther from center, and also the centrif effect, so 9.78 is the acceleration that the person actually feels at sealevel at equator) what I want is a comparison. how big is the centrif effect compared to that

 

the centrif effect on the moon would be extremely less than that, of course because slower rotation and smaller radius.

 

do you know the V^2/R formula? have no idea of where you are age and schoolwise.[/quote']

 

I need to review centrifugal force, in order to answer that, so in this post that's exactly what I'm gonna do.

 

I'm going to start off by reading this article Centrifugal Force in wikipedia. If that's not a good article, let me know.

 

Years ago, i derived some very complex formulas using calculus, which had a lot of omegas in them, and omega-dots... dw/dt, dots denoting differentiation with respect to time. The book was classical mechanics, by Kleppner Kolenkow (sp) I don't remember if I still have the book.

 

I want to re-derive those formulas again, but I cannot remember the starting point. I do remember that centripetal acceleration was a special case of a more general derivation. Also, about a year ago, I saw the perfect geometrical argument to derive centripetal acceleration = v^2/R, but I don't remember what site it was.

 

Ok this article on centripetal acceleration is helping me to remember.

 

You start off with:

 

[math] x(t) = R(t) cos(\omega t) [/math]

 

[math] y(t) = R(t) sin (\omega t) [/math]

 

No, that isn't how the argument started. The argument started off like this I think:

 

Let r(t) denote the position of something in some two dimensional reference frame as a function of time.

 

[math] r(t) = x(t) \hat i + y(t) \hat j [/math]

 

I can't remember what they did, it's been too long, so I will just try to re-work it all out myself. I should have the answer I want within the hour. Somewhere, one of the terms in my formula, will be related to the centrifugal force, if I am remembering correctly. And if not, re-deriving those formulas can't hurt anything.

 

So I will start off in three dimensions instead of two, here is what I start off knowing:

 

Spherical Coordinates

 

The position of an arbitrary particle in the reference frame is given by (x,y,z), and the following equations express that location in terms of spherical coordinates:

 

[math] x = r cos \theta sin \phi [/math]

[math] y = r sin \theta sin \phi [/math]

[math] z = r cos \phi [/math]

 

Let us restrict our analysis to the motion of this particle constrained to move in a plane. Thus, the polar angle phi is ninety degrees, and we can work out the motion of the particle in the xy plane, using polar coordinates. The cosine of pi/2 is equal to zero, and the sine of pi/2=1 so that we have this at some arbitrary moment in time:

 

[math] x = r cos \theta[/math]

[math] y = r sin \theta [/math]

 

The position vector of this particle in this frame of reference is defined to be:

 

[math] r = x \hat i + y \hat j [/math]

 

If the particle is permanently at rest in this frame, the equation above gives its position at all moments in time in the future. But let the particle be free to move in the plane. Thus, its position can change in time, so that r is a function of a variable t, which stands for time, so that we have:

 

[math] r(t) = x(t) \hat i + y(t) \hat j [/math]

 

I have paused to read this article:

 

Nasa article on centrifugal force

 

Ok, the problem I have with their derivation, is that they make use of the following coordinate transformation, but I have no idea as to where they came up with it:

 

[math] (\frac{d}{dt})_i = (\frac{d}{dt})_r + \omega \times [/math]

 

It's a mathematical operator, relating a time derivative in an inertial reference frame, to a time derivative in a reference frame which is spinning in the inertial frame.

 

They pull it right out of thin air, so even if I follow the whole derivation, I will still not understand it, all because I don't know where they came up with that coordinate transformation.

 

Given that time is absolute, it seems to me that the following would be true:

 

[math] (\frac{d}{dt})_i = (\frac{d}{dt})_r [/math]

 

I'm going to go back to my approach:

 

[math] x = r cos \theta[/math]

[math] y = r sin \theta [/math]

 

Differentiate both sides of each equation above with respect to universal time t. Let a dot above a letter correspond to a time derivative of that variable. So for example, the instantaneous change in the theta coordinate is given by:

 

[math] \frac{d\theta}{dt} = \dot \theta [/math]

 

So we have:

 

[math] \dot x = \dot r cos \theta - r\dot\theta sin\theta[/math]

[math] \dot y = \dot r sin \theta + r\dot\theta cos\theta[/math]

 

So we can now express the position vector of this particle, and velocity vector of this particle as follows:

 

[math] r \hat r= x\hat i + y \hat j = r cos \theta \hat i + r sin \theta \hat j [/math]

 

Keeping in mind that the unit vectors i^ and j^ cannot change direction for any reason, that is

 

[math] \frac{d\hat i}{dt} = 0 [/math]

[math] \frac{d\hat j}{dt} = 0 [/math]

 

we have:

 

[math] \vec V = \dot r \hat r + r \frac{d\hat r}{dt} = (\dot r cos \theta - r\dot\theta sin\theta)\hat i +(\dot r sin \theta + r\dot\theta cos\theta)\hat j

 

[/math]

 

And the instantaneous change in theta with respect to time is the angular velocity [math] \omega [/math] so that we can express the instantaneous velocity as:

 

[math] \vec V = (\dot r cos \theta - r\omega sin\theta)\hat i +(\dot r sin \theta + r\omega cos\theta)\hat j

 

[/math]

 

A special case, would be for an object constrained to move in a circle, in which case dr/dt=0, so that for the special case of a particle constrained to move in a circle we have:

 

[math] \vec V = - r\omega sin\theta \hat i + r\omega cos\theta \hat j

 

[/math]

 

Now, notice the following:

 

[math] r \hat r = r cos \theta \hat i + r sin \theta \hat j [/math]

 

Therefore:

 

[math] \hat r = cos \theta \hat i + sin \theta \hat j [/math]

 

You can readily note that r^ is a unit vector, from the Pythagorean relation sin^2 + cos^2 = 1. Now, the dot product of perpendicular vectors is zero, so notice that the following unit vector is perpendicular to r^:

 

[math] \hat \theta = -sin \theta \hat i + cos \theta \hat j

 

 

[/math]

 

Thus, for circular motion, the velocity of the particle can be written as:

 

[math] \vec V = r\omega\ \hat \theta

 

[/math]

 

I'm gonna go try to find that book, and read about centrifugal force. Be back later. Thanks Martin

[swansont]

the way I remember the date for Hipparchus is that it was right around 137 BC (when he found the distance to the moon) and 137 is one of the few numbers I am confident of always remembering.

 

That makes you an alpha male, right?

[Martin]

That makes you an alpha male, right?

 

it is certainly consistent with Hipparchus being a topdog scientist

 

cant think of anyone to equal him until Johnny Kepler circa 1600 which is

like 1700 years. both those guys were millennium-class, no?

 

yes. Alpha. bleeve Feynman said every physicist should write 1/137 on a postit and put it up on his wall to remind himself to wonder why it was that and not some other number

[J.C.MacSwell]

I am curious' date=' but what you say ignores local gravitational field effects of all other bodies in the universe, [b']which matters according to Mach, they contribute to the local gravitational potential[/b], which displaces the CM of the earth-moon system from where it would be if there were no other material in the universe, yes? At any rate...

 

.

 

I think that, if this is the case, then they "contribute" in such a way as to define

an inertia frame. In other words they all cancel (balance would be a better term because if Mach was right we might all have less inertia) and can be ignored.

[Johnny5]

I think that' date=' if this is the case, then they "contribute" in such a way as to define

an inertia frame. In other words they all cancel (balance would be a better term because if Mach was right we might all have less inertia) and can be ignored.[/quote']

 

Actually, I am not convinced Mach was wrong, so if you know how to prove he was wrong go for it. Not too long ago, I was reading about an experiment designed to prove he was wrong, but there was a null result.

 

Scientifically speaking, without a mathematical analysis, I cannot just say that what he said was wrong. If Newton's hypothesis was true, as Mach reasoned, then the gravitational field here, would have contributions from all other bodies in the universe. Although the gravitational potential here and now locally, would depend on the positions of those bodies in the past, due to the fact that gravitational effects travel at exactly the speed of light relative to the source. The whole issue is incredibly mind boggling.

[J.C.MacSwell]

Actually, I am not convinced Mach was wrong, so if you know how to prove he was wrong go for it. .

 

I'm not convinced Mach was wrong either. I have a butt cheek on each side of that fence. I don't think he had a space-time view of space and time however.

[Johnny5]

Yesterday, I was asked to figure out how my weight would be reduced, if I were standing upon the equator of the earth, rather than at one of its poles.

 

(My first thought was that since you are farther away from the center of earth's gravity when you are at the equator than when you are at its poles there is nothing to think about. Just use Newton's gravity formula, and figure out the answer. I am no longer sure this is correct.)

 

There was discussion on the centrifugal force, as being responsible for a decrease in my weight. I am trying to decide whether or not this is true.

 

In classical mechanics, Newton's third law states that for every action, there is an equal an opposite reaction. One way to interpret this statement, is that in any inertial reference frame, you will never find a non-rocket which causes itself to self-accelerate.

 

Things don't just change speeds on their own, an external force must be applied (F=ma) which is Newton's second law.

 

I have been reading things about centripetal and centrifugal force a bit, in order to answer the question.

 

One of things that I read, was that the centrifugal (center fleeing) force is the action/reaction pair force, for the centripetal (center seeking) force.

 

Mathematical analysis of circular motion reveals that an object in circular motion is accelerating, even though it's speed is constant. The reason the object is accelerating, is because acceleration isn't change in speed, it is change in velocity, and velocity is speed times direction.

 

In the case of circular motion, an object moving in a circle is changing direction in some frame. Since its direction of travel in the frame is changing, its velocity in the frame must be changing, hence the object is accelerating in the frame.

 

Upon analysis, you will discover that the direction of the acceleration always points at the center of the circular path of the object, and has the following magnitude:

 

[math] \frac{v^2}{R} [/math]

 

Where v is the speed of the object in the reference frame, and R is the orbital radius of this object.

 

The quantity above, has units of acceleration, and using polar coordinates we can write this "centripetal acceleration" as:

 

[math] \vec a_c= -\frac{v^2 \hat r}{r} [/math]

 

in a coordinate system whose origin is the center of the circular path. The symbol r^ is a unit vector which always points from the origin to the particle. So the minus sign indicates that the direction of this acceleration, is opposite to the direction of r^, and therefore this acceleration points to the point the particle is orbiting.

 

If we now multiply the centripetal acceleration by the inertial mass m of the object, we will have the quantity known as the centripetal force.

 

[math] \vec F_c = -\frac{mv^2 \hat r}{r} [/math]

 

In the case of gravitation, one thinks of gravity as pulling us towards the center of the earth, and Newton's gravity formula is:

 

[math] \vec F_G = -\frac{GMm\hat r}{r^2} [/math]

 

Where m is the inertial mass of a body sitting on the surface of the earth, and M is the inertial mass of the earth, and r is the distance from the center of inertia of earth to the center of inertia of the body sitting on the surface of earth.

 

So can we equate these two forces?

 

I just read this in wikipedia:

 

A centripetal force is a force pulling an object toward the center of a circular path as the object goes around the circle. An object can travel in a circle only if there is a centripetal force on it.

 

In the case of an orbiting satellite the centripetal force is its weight and acts towards the satellite's primary; in the case of an object at the end of a rope' date=' the centripetal force is the tension of the rope and acts towards whatever the rope is anchored to.

 

Centripetal force must not be confused with centrifugal force. In an inertial reference frame (not rotating or accelerating), the centripetal force accelerates a particle in such a way that it moves along a circular path. In a corotating reference frame, a particle in circular motion has zero velocity. In this case, the centripetal force appears to be exactly cancelled by a pseudo-force, the centrifugal force. Centripetal forces are true forces, appearing in inertial reference frames; centrifugal forces appear only in rotating frames.

 

[/quote']

 

Part of this seems very clear. In particular, the part which says, "in the case of an orbiting satellite, the centripetal force is its weight."

 

Ok...

 

We are at the surface of the earth, and we have a weight which can be measured by a scale. A scale operationally defines weight. Now, since the earth is spinning, a reference frame attached to an object at the surface of the earth is a non-inertial frame.

 

But the center of mass of the earth isn't at rest in an inertial reference frame, nor is it moving in a straight line at a constant speed in an inertial frame, so that a system of coordinates whose origin is the center of mass of earth isn't an inertial frame either.

 

Mathematical analysis of acceleration

 

 

[math] x = r cos \theta[/math]

[math] y = r sin \theta [/math]

 

[math] \dot x = \dot r cos \theta - r\dot\theta sin\theta[/math]

[math] \dot y = \dot r sin \theta + r\dot\theta cos\theta[/math]

 

So we can now express the position vector of this particle, and velocity vector of this particle as follows:

 

[math] r \hat r= x\hat i + y \hat j = r cos \theta \hat i + r sin \theta \hat j [/math]

 

 

[math] \vec V = \dot r \hat r + r \frac{d\hat r}{dt} = (\dot r cos \theta - r\dot\theta sin\theta)\hat i +(\dot r sin \theta + r\dot\theta cos\theta)\hat j

 

[/math]

 

And the instantaneous change in theta with respect to time is the angular velocity [math] \omega [/math] so that we can express the instantaneous velocity as:

 

[math] \vec V = (\dot r cos \theta - r\omega sin\theta)\hat i +(\dot r sin \theta + r\omega cos\theta)\hat j

 

[/math]

 

A special case, would be for an object constrained to move in a circle, in which case dr/dt=0, so that for the special case of a particle constrained to move in a circle we have:

 

[math] \vec V = - r\omega sin\theta \hat i + r\omega cos\theta \hat j

 

[/math]

 

Now, notice the following:

 

[math] r \hat r = r cos \theta \hat i + r sin \theta \hat j [/math]

 

Therefore:

 

[math] \hat r = cos \theta \hat i + sin \theta \hat j [/math]

 

You can readily note that r^ is a unit vector, from the Pythagorean relation sin^2 + cos^2 = 1. Now, the dot product of perpendicular vectors is zero, so notice that the following unit vector is perpendicular to r^:

 

[math] \hat \theta = -sin \theta \hat i + cos \theta \hat j

 

 

[/math]

 

Thus, for circular motion, the velocity of the particle can be written as:

 

[math] \vec V = r\omega\ \hat \theta

 

[/math]

 

Now, for acceleration in general, we need to compute dV/dt.

 

Recall:

 

[math] \vec V = (\dot r cos \theta - r\omega sin\theta)\hat i +(\dot r sin \theta + r\omega cos\theta)\hat j

 

[/math]

[Martin]

Hi J5,

would you like guidance or suggestions? If you say yes then you might get some help from experts like Tom M. or Swansont. Or I might put in a word or two.

Also if you say yes, you dont necessarily have to TAKE any advice or hints anyone offers.

the thing that most impresses me is that you seem prepared to either teach yourself or calculate stuff on your own, but you should be aware that you always have the OPTION of asking for suggestions.

 

 

BTW I didnt realize I had said "centrifugal force" in any post. I try never to say that, but sometimes I may forget.

What I try to say, instead, is centrifugal effect which is the tendency of things to whirl off of a turning merrygoround just because they want to continue in a straight line. what I mean by the centrifugal effect is how things scatter away from the center if they are not holding on or glued in place, or constantly falling inwards because of gravity.

 

I hope i didnt say "centrifugal force", but if I did accidentally say that in some post here then please excuse

 

You seem to be doing fine! just remember there is a kind of safety net.

[Johnny5]

Hi J5' date='

would you like guidance or suggestions? [/quote']

 

Yes, I am not looking to re-invent the wheel, but only a tiny suggestion. Additionally, there is a sort of 'muddled' thing going on in the mathematics, which isn't my fault. It was there long before I got here. Centrifugal effect isn't a technical term, say what you mean please.

 

Regards

[Johnny5]

 

What I try to say' date=' instead, is centrifugal effect which is the tendency of things to whirl off of a turning merrygoround just because they want to continue in a straight line. what I mean by the centrifugal effect is how things scatter away from the center if they are not holding on or glued in place, or constantly falling inwards because of gravity.

 

[/quote']

 

The way you have it stated, inertia is responsible for "centrifugal effect" i think.

 

As Galileo concieved of it, an object carries its inertia with it wherever it goes. So in absense of an externally applied force, an object will obey Galileo's law of inertia, which is that it will either remain at rest OR travel in a straight line at a constant speed.

 

So then for something on a Merry go round, the thing is constrained to orbit a central point. It's inertia wants it to move in a straight line, but the object is 'glued' good word to the merry go round.

 

But the moment the glue breaks or string snaps, off in a straight line goes the object, but that straight line is straight only in a reference frame in which the center of orbit is at rest right?

[Martin]

The way you have it stated, inertia is responsible for "centrifugal effect" i think.

 

Yes!

things appear to whirl away from center simply because (in absence of forces) they 'want' to continue in a straight line

[Martin]

Yes, I am not looking to re-invent the wheel, but only a tiny suggestion....

 

the guy is standing at the equator

calculate his speed V that is simply due to earth rotation

 

you have already derived the formula that his acceleration towards earth center (as he stands there) is V^2/R

but you have not yet plugged in a value for V and a value for R into the formula

[Johnny5]

the guy is standing at the equator

calculate his speed V that is simply due to earth rotation

 

 

Ok, first I have to choose a reference frame to define speed in. The reference frame I am going to choose is one in which the center of inertia of earth is at rest.

 

Now there are all kinds of wildly spinning frames to choose from, but I would like to choose one which is inertial if possible.

 

There are 24 hours in a day.

 

The question now, is how to interpret this.

 

Suppose that there is a reference frame in which the sun is at rest, and the earth is at rest.

 

Focus on a plane which contains the line joining their centers of inertia.

Now let the earth be spinning, and let the axis of rotation be perpendicular to the plane.

 

That axis of rotation is to be the z axis.

[Martin]

Ok' date=' first I have to choose a reference frame to define speed in. The reference frame I am going to choose is one in which the center of inertia of earth is at rest.

 

Now there are all kinds of wildly spinning frames to choose from, but I would like to choose one which is inertial if possible.

 

There are 24 hours in a day.[/quote']

 

keep going (24 hours is approximately the time it takes for the earth to make one rotation relative to the stars, the approximation is good enough)

what is the guy's speed?

[Johnny5]

what is the guy's speed?

 

 

Speed=distance/time = distance/24 hours

 

distance = 2 pi R

 

where R is the radius of the earth (assuming earth is spherical )

 

Speed = 2 pi R/24 hours

 

R = 6' date='400,000 meters (approximately)

 

Time in seconds = 24h(60m/h)(60s/m) = 24*3600 seconds

 

[math'] speed/2pi = \frac{6,400,000}{24*3600} = \frac{64,000}{24*36}= \frac{2,000}{3*9}= 74 meters /second [/math]

 

Now, rounding 2pi off to 6 gives...

 

[math] speed = 444 m/s [/math]

 

Now I am thinking about whether or not this makes sense.

[Martin]

Speed=distance/time = distance/24 hours

 

distance = 2 pi R

 

where R is the radius of the earth (assuming earth to be spherical)

 

Speed = 2 pi R/24 hours

 

R = 6' date='400,000 meters (approximately)

 

Time in seconds = 24h(60m/h)(60s/m) = 24*3600 seconds

 

[math'] speed/2pi = \frac{6,400,000}{24*3600} = \frac{64,000}{24*36}= \frac{2,000}{3*9}=74 [/math]

 

 

let's see what I get (i didnt do it yet )

I think the circumf of earth is 40,000 km so divide by 24 hours and get

1667 km/h and change that to meters per second and get

463 meters per second

 

that is just an approximation (actually one complete rotation of earth takes a bit less than 24 hours and the circumf is not exactly 40,000 km, but lets just get some rough answer and worry about accuracy later if we want)

 

what did you get for his speed? either of us could have made a mistake so its best to compare

 

oh, I see your answer----you got essentially the same speed your answer amounts to around 464 meters per second.

 

so what is the guy's acceleration towards the center of the earth, as he stands peacefully there on the equator (we know he is accelerating because otherwise he would fly off on a straightline tangent and not get back in time to take his wife to the movies)

[Johnny5]

what did you get for his speed? either of us could have made a mistake so its best to compare

 

 

I got 444 meters per second, but I rounded 2 pi to 6, and I rounded the radius of earth up from 6,378,100 meters to 6,400,000 meters. All my work is there. I didn't use a calculator, so I could easily have made a mistake, but our answers agree.

[Johnny5]

so what is the guy's acceleration towards the center of the earth' date=' as he stands peacefully there on the equator (we know he is accelerating because otherwise he would fly off on a straightline tangent and not get back in time to take his wife to the movies)[/quote']

 

centripetal acceleration is:

 

[math] \vec a_c = \frac{v^2}{R} [/math]

 

[math] v^2 = 464*464 = 215,296 [/math]

 

[math] \frac{v^2}{R} = \frac{215,296 }{6,400,000}[/math]

 

Units are meters per second squared. Rounding off 215,296 to 200,000 gives:

 

[math] \frac{v^2}{R} = \frac{200,000}{6,400,000} =\frac{2}{64} = \frac{1}{32} = .03125

 

[/math]

 

I rounded down the numerator a bit, so I will approximate the answer as:

 

[math] \vec a_c = .04 \frac{m}{s^2} [/math]

[Martin]

centripetal acceleration is:

...

 

Units are meters per second squared. Rounding off 215' date='296 to 200,000 gives:

 

[math'] \frac{v^2}{R} = \frac{200,000}{6,400,000} =\frac{2}{64} = \frac{1}{32} = .03125

 

[/math]

 

I rounded down the numerator a bit, so I will approximate the answer as:

 

[math] \vec a_c = .04 \frac{m}{s^2} [/math]

 

Yes! congratulations. I will see what my handbook says. It will be quite close to this I think (even tho you did a lot of rounding off)

 

Yes, I checked in the book and it says that the centrip. accel at the equator is 0.033915 meter per second squared

[Johnny5]

Yes' date=' I checked in the book and it says that the centrip. accel at the equator is 0.033915 meter per second squared

 

[/quote']

 

That is a whole lot less than the acceleration of earth's gravity g = 9.8 m/s^2. What does the fact that it is negligible in relation to g mean exactly? What are the important inferences that can be drawn from this?

[Martin]

Now we have to compare that with the usual gee

which is the acceleration toward the center the guy would have if you dropped him down an elevator shaft and had people at each floor with stop watches

 

It happens that at the equator, at sea level, this is 9.78 meter per second square.

 

this is the actual observed acceleration measured by dropping people down elevator shafts and duely recorded written down in my book.

 

this is what the other little accel has to be compared with

[Johnny5]

Now we have to compare that with the usual gee

which is the acceleration toward the center the guy would have if you dropped him down an elevator shaft and had people at each floor with stop watches

 

 

Assuming that he is in free-fall... As he falls' date=' the earth rotates a little so that he wont hit the surface of the earth right above where you drop him at. The earth will rotate slightly underneath him as he falls. Suppose you drop him at a height of 9.78 meters above the surface of the earth. He will strike the ground 1 second later, but in this time, the point on the surface of the earth which he was dropped above, will move slightly. It will accelerate at .0339 m/s^2 underneath him.

 

We can now figure out how much the ground moved.

 

[math'] S = r \theta [/math]

 

I will treat the surface of the earth as flat, in order to get a rough approximation of the distance.

 

Assume acceleration is constant.

 

[math] D = v_0 t + 1/2 a t^2 [/math]

 

Suppose that he is holding onto a pole which is embedded in the surface of the earth, before he lets go and starts falling. Then he lets go, and is decoupled from the earth's rotation, when prior to letting go of the pole he was coupled to the earth, and spun with it.

 

Ignore the atmosphere.

 

So he is falling through vacuum, towards a spinning earth.

 

The acceleration of the earth is only .033 m/s^2, it is slight but it is nonzero.

 

Assume this acceleration is constant, and that the earth is flat.

 

We are assuming that he is at a height of 9.78 meters above the earth. Once he lets go, he will hit the ground one second later.

 

The problem is to find how much the base of the pole moves (presuming incorrectly that its path is linear)

 

We are carrying out the analysis in the rest frame of the man who is falling towards the earth.

 

In that mans frame, the base of the pole moves say to his right, while he is falling.

 

Initially, the base of the pole is at rest in the man's rest frame, so v₀ = 0

 

So we have to solve:

 

[math] D = 1/2 a t^2 [/math]

 

We know the time for him to fall 9.78 meters, is exactly 1 second so t=1, and t^2=1, hence the distance is given by:

 

[math] D = 1/2 a [/math]

 

Where a = .033 m/s^2.

 

So we have the following:

 

[math] D = \frac{.033}{2} =.0165 [/math]

 

So once the man lets go of the pole, the base of the pole will move to his right a distance of .0165 meters, 100cm=1 meter, so the distance the base of the pole moves relative to the falling man is:

 

1.65 centimeters

 

This seems false. It certainly is testable, but perhaps I made a math error.

[Martin]

That is a whole lot less than the acceleration of earth's gravity g = 9.8 m/s^2. What does the fact that it is negligible in relation to g mean exactly? What are the important inferences that can be drawn from this?

 

 

You started it J5

you said find the mass of the moon by doing a handstand (i.e. feeling the gravity) and you asked did it make any difference WHERE you do the handstand

 

and what we are exploring is how much the gravity of the earth changes as you go around to different places

 

and frankly, it does nt change very much does it?

 

A kind of average value which you learn when you learn metric units is the "standard" gee which is 9.80665 meter per second square.

 

And at the equator it is 9.78 meter per second square

 

and at the north pole it would be a fraction percent MORE than 9.80665

 

 

So a kilogram mass is going to weigh different amounts as you move it around----sometimes it weighs 9.80 newtons of force

sometimes it weighs 9.78 newtons of force

 

I cant say that is important, just a little thing to be ware of

[Martin]

I think what is important is V^2/R

 

(the fact that V^2/R makes things slightly lighter at the equator is not so important)

 

 

PROBLEM

what speed does a satellite have to go if it is going in a circle orbit around the earth with radius 7000 km?