Physics Question

[thinwater]

Originally Posted by anotherT34C

No, it's because at the center of the Earth, it would be weightless.

The center of the Earth is not actually the bottom of the gravity well. As the object fell down the hole, it would be increasingly pulled on (slowed down) by the fraction of the Earth 'behind' it, until it ultimately reaches the center. At that point it would float weightless.

Cool, right?

You are joking, right? 18,000 mph. True, acceleration is reduced the deeper you go, but there is a lot of falling.

http://www.physicscentral.com/explore/poster-earth.cfm

[anotherT34C]

Originally Posted by thinwater

You are joking, right? 18,000 mph. True, acceleration is reduced the deeper you go, but there is a lot of falling...

Yep, you're right. Acceleration goes to zero at the center, not velocity. Stupid thinking on my part (I'll blame the rum) . One period of the object making a round trip up/down should be the same as the object completing one orbit.

[sailorchic34]

Ah, but assuming a hole drilled right on the earth axis, and an object free falling in a vacuum, why would the object disintegrate as it was bounced off the walls of the tube. Why would it bounce. Or do you think it would fall straight through??

Assume the hole is a 1000 feet wide and the object is a foot in diameter.

[Adelie]

Originally Posted by anotherT34C

No, it's because at the center of the Earth, it would be weightless.

The center of the Earth is not actually the bottom of the gravity well. As the object fell down the hole, it would be increasingly pulled on (slowed down) by the fraction of the Earth 'behind' it, until it ultimately reaches the center. At that point it would float weightless.

Cool, right?

Wrong. It would gain speed the whole way, albeit at a steadily decreasing acceleration. Right at the center the acceleration would be zero (weightless) but the object would continue thru the center and decelerate at it rose back to the surface at the north pole.

[sailorchic34]

Originally Posted by anotherT34C

Yep, you're right. Acceleration goes to zero at the center, not velocity. Stupid thinking on my part (I'll blame the rum) . One period of the object making a round trip up/down should be the same as the object completing one orbit.

Hum diameter is 7917 miles, the circumference is 24,000 miles. Orbit will take a bit longer.

[Hard Rock Candy]

Ok! then does the weight change, of a screen wire cage with a butterfly in it, when the butterfly takes wing Mac


Or is Schroders Cat still in the box, after all, these years and is he dead or alive?

[Adelie]

Originally Posted by thinwater

You are joking, right? 18,000 mph. True, acceleration is reduced the deeper you go, but there is a lot of falling.

http://www.physicscentral.com/explore/poster-earth.cfm

Actually 18000MPH is the orbital speed at about 110 mile above the earth.

Starting at rest at the surface
A radius of 6,371,000m
A uniform density of 5,510 kg/M^3

An object would be traveling at 11,356kph or 7041mph as it passed thru the center of the earth.

[sailorchic34]

Originally Posted by Adelie

Actually 18000MPH is the orbital speed at about 110 mile above the earth.

Starting at rest at the surface
A radius of 6,371,000m
A uniform density of 5,510 kg/M^3

An object would be traveling at 11,356kph or 7041mph as it passed thru the center of the earth.

Except the center of mass is not the axis of earths rotation about its axis. The center of mass lays 1062 miles below the crust in a line from the center axis of the earth to the center of the moon. Called the barycenter. Tides would affect the falling object and it would not fall straight down, but tumble against the wall. Big ouch.

The Barycenter is also the point that orbits the sun. The earth and moon wobble about it.

[Adelie]

Originally Posted by sailorchic34

Except the center of mass is not the axis of earths rotation about its axis. The center of mass lays 1062 miles below the crust in a line from the center axis of the earth to the center of the moon. Called the barycenter. Tides would affect the falling object and it would not fall straight down, but tumble against the wall. Big ouch.

The Barycenter is also the point that orbits the sun. The earth and moon wobble about it.

The earth's axis of rotation passes thru the earth's center of mass. By having the hole start at the south pole (not underwater like the north pole) you avoid all Coriolis effects associated with having an initial tangential velocity.

The earth-moon system rotates around the barycenter you described and has a synodic period of 29.53 days.

And the barycenter for the earth-moon-sun system is below the surface of the sun and its sidereal period is 365.256 days

The barycenter for the earth-moon-sun-milky way system is somewhere and its period is about 238 Myr.

Let's assume it takes about 1hr (3600s) to fall thru the earth. Then we can ignore the barycenters for the sun and milky way, the kinematic changes for these are small enough to ignore for the period of interest.

The barycenter for the earth-moon system is 6380 km from the center of the earth. In a 1 hour period the center of the earth will have moved 1/708th of a circle around the barycenter or about 1/2 of a degree. The falling body will have started with the same tangential velocity as the earth's axis. Assuming the body continued in a straight line while the axis of the earth follows its orbit around the barycenter, the displacement between the 2 will be about 251m (6380km -6380km * cos(0.5 derees)). So if the hole is 252m across we won't bounce on the side.

But the body does not travel in a straight line the moon attracts it so let's look at what the acceleration due to the moon's gravity is. Again all the other attractive effects are small enough to be ignored for the time period of interest.

Given the moon's mass of 7.348 × 10^22 kg
and average distance of 384,400,000m

then the acceleration due to the moon's gravity is 3.319 x 10^-5 m/s^2

Xfinal = Xinitial + Vinitial x t + .5 x a x t^2.

Assuming the Xinitial and Vinitial are both 0 then

Xfinal - .5 x 3.319 x 10^-5 x (3600)^2 = 215m towards the barycenter. That means that a 36m diameter hole is all that's required.

But what if you give yourself a slight lateral nudge just as you start to fall so you are drifting towards the barycenter. By pushing off slightly as the fall starts the body drifts across the hole almost skims the other side as it passes the center and drifts back to the original side as it gets back to the surface.

Since it is now past midnight and I don't want to work thru this math I'm just going to say that the diameter of the hole is now down to the 10-18m range.

[sailorchic34]

Originally Posted by Adelie

The barycenter for the earth-moon system is 6380 km from the center of the earth. In a 1 hour period the center of the earth will have moved 1/708th of a circle around the barycenter or about 1/2 of a degree. The falling body will have started with the same tangential velocity as the earth's axis. Assuming the body continued in a straight line while the axis of the earth follows its orbit around the barycenter, the displacement between the 2 will be about 251m (6380km -6380km * cos(0.5 derees)). So if the hole is 252m across we won't bounce on the side.

Impressive. But I think you forgot about pi.

The center of the earth rotates around a baryceter. By my calc's the barycenter is 2896 miles or 4660 km from the center of the earth. The orbit of the center of the earth around the Barycenter would then be approximately 18,1999 miles or 29288 km (2r*pi) and the movement in one hour would be ~25 miles or 40 km, or roughly 2200 feet per second.... Splat.

[Adelie]

Originally Posted by sailorchic34

Impressive. But I think you forgot about pi.

The center of the earth rotates around a baryceter. By my calc's the barycenter is 2896 miles or 4660 km from the center of the earth. The orbit of the center of the earth around the Barycenter would then be approximately 18,1999 miles or 29288 km (2r*pi) and the movement in one hour would be ~25 miles or 40 km, or roughly 2200 feet per second.... Splat.

Standing at the pole you have the same tangential (horizontal) velocity as the axis around the barycenter.

If you continued to move with the same velocity in a straight line and the axis moved with the same velocity in an arc of a circle, how far would they diverge in one hour? For very small angles near 0 degrees cosine is very small too. That's how I figured the 251m hole. I ignored the speed and looked at the difference between a straight line and an arc.

However the moon is providing a gravitational attraction that pulls you toward it so you don't continue on in a straight line, you travel in an arc too. Because the hole is thru the axis and that is always on the opposite side of the barycenter from the moon attraction of the moon is less on you than it would be at the barycenter so the curve you follow will be outside of the curve the axis follows. The attraction of the moon gave me the 215m difference from the straight line, that's 36m different than the previous calculation, hence to 36m hole.

So if you drop into the hole with no velocity across the hole you need to do so on the moon side of the hole and when you get to the north pole you will be on the side of the hole away from the moon.

If you enter with a nudge you need to start on the side away from the moon, you will cross the hole and be on the moon side as you pass the center and then drift back to the away side when you get to the north pole. That's how you get the hole diameter down even further.

I don't think the 29,288kph number is correct, the radius from the axis to the barycenter is 6380km with a period of 29.53d (708.72hr). The circumference is 40,087k, so speed of the axis around the barycenter is 56.56kph. If the speed of the axis around the barycenter were any significant fraction of the speed needed to orbit the earth, you would be able to feel the difference in your weight thru the day depending on where you were relative to the moon. This would also require the moon to be a LOT closer.

Oh yeah, I got 6380km for the earth-moon barycenter from here , its the R1 value.

[sailorchic34]

Originally Posted by Adelie

I don't think the 29,288kph number is correct, the radius from the axis to the barycenter is 6380km with a period of 29.53d (708.72hr). The circumference is 40,087k, so speed of the axis around the barycenter is 56.56kph. If the speed of the axis around the barycenter were any significant fraction of the speed needed to orbit the earth, you would be able to feel the difference in your weight thru the day depending on where you were relative to the moon. This would also require the moon to be a LOT closer.

Oh yeah, I got 6380km for the earth-moon barycenter from here , its the R1 value.

AH, the R1 value is the radius of the earth and not the radius to the barycenter. r1 is the radius to the barycenter, that's 4670 km. I had 4460, but I had calculated it. I suspect 4670 may be closer to correct at the equator and 4460 is polar. Of course it varies as the earth pivits on its axis 23 degrees ish.

We don't feel the velocity as we are part of the reference frame and there is no acceleration. No more then we feel the 1000 mph as the earth spins on it's axis or the 60,000 mph as we rotate around the sun or the 245,000 mph as we rotate around the milkway. So a 25 MPH rotation around the barycenter, does not register.

Standing on the north or south pole, there is no motion felt. In free fall however, we are no longer constrained to the same reference frame as the earth. I expect there will be a bit of drift.

[Adelie]

Originally Posted by sailorchic34

AH, the R1 value is the radius of the earth and not the radius to the barycenter. r1 is the radius to the barycenter, that's 4670 km. I had 4460, but I had calculated it. I suspect 4670 may be closer to correct at the equator and 4460 is polar. Of course it varies as the earth pivits on its axis 23 degrees ish.

We don't feel the velocity as we are part of the reference frame and there is no acceleration. No more then we feel the 1000 mph as the earth spins on it's axis or the 60,000 mph as we rotate around the sun or the 245,000 mph as we rotate around the milkway. So a 25 MPH rotation around the barycenter, does not register.

Standing on the north or south pole, there is no motion felt. In free fall however, we are no longer constrained to the same reference frame as the earth. I expect there will be a bit of drift.

You are correct about the axis-barycenter radius, I misread the table. Recalculating for this the first round hole would need to be 178m across, the second would be 37m and with a nudge about 10-15m across.

Actually we are standing in several different frames of reference, but because the effects of most of them are small compared to the gravity of the earth we don't notice them. If the moon were a lot closer (25,000 mi instead of 250,000mi it would be enough to feel, or if the earth's gravity were a lot weaker the others might start to be noticeable.

[sailorchic34]

Originally Posted by Adelie

You are correct about the axis-barycenter radius, I misread the table. Recalculating for this the first round hole would need to be 178m across, the second would be 37m and with a nudge about 10-15m across.

Actually we are standing in several different frames of reference, but because the effects of most of them are small compared to the gravity of the earth we don't notice them. If the moon were a lot closer (25,000 mi instead of 250,000mi it would be enough to feel, or if the earth's gravity were a lot weaker the others might start to be noticeable.

I created a to scale drawing of the diameter of the barycenter to earth axis (I'm an engineer I can do that). My simple calc's say in one hour the pole will move 0.503 degrees around the barycenter orbit. During that time the pole has moved 40.756 km. But the distance between the two points is now (earth axis to tangent object falling in the hole) 0.8424km or 2700 feet.

Thoughts?

[Adelie]

Originally Posted by sailorchic34

I created a to scale drawing of the diameter of the barycenter to earth axis (I'm an engineer I can do that). My simple calc's say in one hour the pole will move 0.503 degrees around the barycenter orbit. During that time the pole has moved 40.756 km. But the distance between the two points is now (earth axis to tangent object falling in the hole) 0.8424km or 2700 feet.

Thoughts?

I make the divergence between the arc and a straight line tangents to be 180m or 363m depending on whether I approach from radius and cosine or travel distance and sin. See attached sketch.

Obviously one of them is incorrect. What I get for trying to do this while cooking crepes for breakfast and helping the 7yr old with her first crossword puzzle. She did pretty well with it, sharp as a tack, takes after her mother.

Then you factor in the moon's gravity and an initial cross-hole velocity and the hole diameter can get much smaller.

I figured your were an engineer or physicist, you were the only interested person finding my mistakes and extending the problem. What flavor of engineer? I'm civil.

Going back an extra post, the max declination of the moon is 18d17m or so, the earth's tilt to the sun is 23deg. Either way, drop when the moon is above the equator and no correction needed.