Physics Question

[onestepcsy37]

Lakesuperior, your explanation is superior. I think I am now on my way to becoming a theoretical physicist.

[TeddyDiver]

Originally Posted by Adelie

For the specific situation you described there would be no net gravitational attraction between you and the space craft wherever you are within the sphere. Attraction to opposing sides cancels out.

There is a discussion and maybe a proof in the 2nd quarter of the Halliday&Resnick college physics book for charged spheres and the net electrical field within. Since the attraction force equations are the same the effect is the same for gravitational fields and effects.

I once used this to calculate the potential energy between the surface and center of the earth falling in a tube under vacuum conditions.

I take that as no..

[Adelie]

Originally Posted by TeddyDiver

I take that as no..

Gravity Force Inside a Spherical Shell

If you need help working your way thru the math let me know.

Now what's the potential energy between the surface and center of the earth?

[onestepcsy37]

Originally Posted by Adelie

Now what's the potential energy between the surface and center of the earth?

Are you referring to oil wells?

[thinwater]

General relativity = all non-accelerating frames of reference are equivalent.

While there are corrections others have pointed out (angular acceleration of the earth and wind), the practical result is plainly obvious and not worth discussion after 11th grade. Pretty obvious by 6th grade, I would think.

[sailorchic34]

Originally Posted by onestepcsy37

1. I am in a rocket way out in space. The rocket is neither accelerating or decelerating.
2. I am in a rocket way out in space. The rocket is accelerating.

Question. Assuming there is no gravitational force present, does one of these situations require more effort for me to move about inside the rocket?

For 1 you are weightless. For 2 the constant acceleration, you would not be weightless as the rocket is accelerating and you are being supported on a aft bulkhead. You could drop a ball and it would float/fall to the engines.

Now for a constant velocity, you would be weightless, But under acceleration you are not.

[sailorchic34]

Originally Posted by DefinitelyMe

Yes, it is. See post #36, which took me forever to write but it seems was too long for people to read :-( The feeling of despair is overwhelming!

Oh Sorry, your post #36 was not quite correct. The man is using the the same amount of energy walking up an escalator as stairs. That is for each step he or she takes, the step takes the same energy.

Now your are correct in that over period time or journey, the man on the escalator will use less energy overall on a moving escalator then on a equal flight of stairs as the escalator is moving and he will arrive sooner at the top. But it takes the same amount of calories (roughly 0.17 calories per step) to take a step on a escalator as on fixed step of equal height.

The escalator will use 47 times (Ok maybe only 25 times) the energy to transport the man to the top, as the man would use alone walking a fixed flight of steps.

[anotherT34C]

Hi Sailorchic,

Your answer is only approximately correct. You will use approximately the same amount of energy climbing say 20 steps on an escalator as on a stationary staircase, but in fact you'll use less on the escalator because the escalator will raise you farther out of the Earth's gravity well, making each subsequent step that much easier.

Cheers.

[sailorchic34]

Originally Posted by anotherT34C

Hi Sailorchic,

Your answer is only approximately correct. You will use approximately the same amount of energy climbing say 20 steps on an escalator as on a stationary staircase, but in fact you'll use less on the escalator because the escalator will raise you farther out of the Earth's gravity well, making each subsequent step that much easier.

Cheers.

You are of course correct, but the net reduction in weight per foot is roughly 0.000014 pounds ( I calculated it for a 200# man). so if the stairs were say 33000 feet high, the net reduction in weight at 33,000 feet would be .56 pounds for a net energy savings of 0.0005 calories per step. Not really something you can really measure, or feel.

Of course you have to factor in that at 33,000 feet there is less air pressure so you will weight slightly more (just like an anchor weighs more in air then water), so the net difference at 33,000 feet would be about 0.495 pounds.

Then there is a very tiny negative or is it positive force, that effect's satellites during a gravity assist flyby to achieve a higher velocity. Something on the order of 0.12m/s greater acceleration then should exist. No one know why either.

[onestepcsy37]

I want to thank you all for the valuable information I have gleaned from this thread. Not only I have I been introduced to a whole new way (for me) of looking at gravity and acceleration and relativity, but you have convinced me not to make a rather expensive upgrade to my yacht.

I will NOT be replacing my companionway steps with an escalator.

[anotherT34C]

Gravity assist on satellites works by transferring some of the larger body's angular momentum into kinetic energy for the smaller body. This is how the Moon is slowly leaving the Earth. The Earth's rotation is slowing... transferring kinetic energy to the moon, which then moves to a higher orbit, thus increasing its potential energy.

I suppose the configuration of the planets will have some measurable effect on your staircase answer as well, but I wasn't thinking to delve down that rabbit hole.

[Adelie]

Originally Posted by onestepcsy37

Are you referring to oil wells?

No.

If you bored a hole thru the earth from the South Pole and evacuated the air what would the speed be of an object dropped down the hole when it reached the center?

This question has an analogy in the escape velocity of the earth. If you dropped a rock from infinity towards the surface of the earth what would its velocity be when it hit ignoring air resistance? The corollary is how fast would a bullet fired straight up for in never to fall straight back to earth? This is not just an academic question, it is something the NASA folks use as a starting point when calculating what it take to launch satellites.


Sent from my iPhone using Cruisers Sailing Forum

[anotherT34C]

Originally Posted by Adelie

No.

If you bored a hole thru the earth from the South Pole and evacuated the air what would the speed be of an object dropped down the hole when it reached the center?

I know you're not asking me, but the answer is zero. Know why?

[Adelie]

Originally Posted by anotherT34C

I know you're not asking me, but the answer is zero. Know why?

Because I forgot to say ignore the fact that the heat in the hole would cause it to melt and evaporate.

[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?