Dot Physics

Note: This is an old post from the time before my blog was in wordpress. I noticed there was some incoming link for this, and I never moved it over. Here it is in it’s unaltered (except for this part) format.

I don’t know why I even suggest a new energy source. Fusion power is only a few years away in the future (just like it as always been). This will replace any other sources of energy that we could come up with. But, I can’t help myself, I need to share my idea and save the world. It’s what I do. (call me a superhero is you want).

We can get all of our energy from the rotation of the Earth.

The basic idea is to use the rotation energy of the Earth to power our Nintendos and computers and stuff. How much energy could we get out of this and what would we lose? First – what we lose. If we use the rotational energy of the Earth, it would spin at a slower rate. I will start off assuming we make the day 1 second longer.

Currently, the Earth takes 23.9345 hours to rotate. It take the 24 hours for the Sun to be back in the same location. This is the difference between sidereal day and synodic day. I am concerned with the rotational rate of the Earth on its axis, so I need to use the sidereal day. Check the wikipedia link for a great diagram that shows the difference between the two days.

This leads to an angular speed of:

i-9659ab52ca1a1f73a727dbb6ddb030ee-dotphys_2rwsw.jpg

Now, suppose I want to increase the length of the sidereal day by 1 second, this would give a new angular speed of:

i-de0ee12265c9240a30caa06092109d10-dotphys_2rwsw_1.jpg

How much energy would this produce? (assuming it could all be turned into useful energy).
The energy of motion for an object rotating is:

i-62b93c70272c1b4f365273910949161a-dotphys_2rwsw_2.jpg

Where (1/2) is (1/2), I is the moment of inertia about the axis you are rotating and ? is the angular speed. The moment of inertia is sort of like the “rotational mass”. An object with a higher moment of inertia is more difficult to change its rotational motion. In this case, we are dealing with a spherically-shaped object (the Earth is mostly spherical). The moment of inertia for this can be approximated by assuming it is a uniform sphere, but it isn’t. The density in the center of the Earth is much greater than on the surface. So, for this calculation, I will use someone else’s determination of the moment of inertia of the Earth – here is Wolfram’s Research value

i-3c72db5bea6c891518e7e62693e4ffa7-dotphys_2rwsw_3.jpg

From this, I can calculate the change in energy by slowing the Earth down.

i-7444cd29a83ea4429c085a235cc9f521-dotphys_2rwsw_4.jpg

and putting in the values from above:

i-c4e38fd672ff31a2376857b844a81c57-dotphys_2rwsw_5.jpg

This is how much energy the Earth loses, so we could use this for other stuff. Is this enough energy?

Energy Usage:

Let me just look at the energy usage by the U.S.A. because they would be the ones to harness the rotational energy of the Earth (but really, because I found the data for US energy usage first). http://tonto.eia.doe.gov/ask/electricity_faqs.asp#home_consumption – this site has the data that I started with. It has a spreadsheet with average monthly usage for residential, commercial and industrial:

  • Residential: There were 122,471,071 consumers that used an average of 920 kilowatt hours per month.
  • Commercial: There were 17,172,499 users that used an average of 6,307 kilowatt hours per month.
  • Industrial: There were 759,604 users that used an average of 110,946 kilowatt hours per month.
  • NOTE: by “users” I mean companies or places or whatever the spreadsheet meant.

A kilowatt is a unit of power – or the rate at which energy is used. A kilowatt-hour is a unit of energy. Since 1 watt is a Joule per second, 1 kilowatt-hour would be:

i-e6d7f6f454f2f2b1fc1ae6f6536aea69-dotphys_2rwsw_6.jpg

and the monthly usage in the US would be:

i-34ee65771b1d238f6d6d03f8604c8f34-dotphys_2rwsw_7.jpg

that is per month.
So how many months (and years) would this energy last for the US assuming a steady energy usage? Actually, I will also assume that only 50% of the rotational energy of the Earth goes to useful stuff (like nintendos) and the rest is wasted. This would mean the amount of useful energy would be 2.5484 x 1024 Joules. So

i-9cb29cfe282adad3e030b1df84394172-dotphys_2rwsw_8.jpg

This is a long time. So the length of the sidereal day would only increase by 1 second over this time period (that way wouldn’t have to store all this energy, but rather generate it as we use it).

Now for the details:

How exactly do I propose that this rotational energy be harnessed? I will leave that as an exercise for the reader – but I will give a hint: Magnets and wire (I have already said too much).

Also, this method produces no greenhouse gases.

Comments

  1. #1 Tom
    June 23, 2009

    And we are already slowing down by some fraction of a second per year (depending on the year) due to tidal friction with the moon and other factors. Nobody will notice that this extra increase happened.

  2. #2 Nick
    June 23, 2009

    Any harness I can think of for this kind of energy requires the first step to be “build space elevator” (or space tether, at least). And, as we’ve seen, that has some of its own issues.

  3. #3 weathercast forecaster
    March 26, 2011

    So how many months (and years) would this energy last for the US assuming a steady energy usage? Actually, I will also assume that only 50% of the rotational energy of the Earth goes to useful stuff (like nintendos) and the rest is wasted. This would mean the amount of useful energy would be 2.5484 x 1024 Joules.

  4. #4 john wolff
    September 2, 2011

    surely the earth has some reluctance so we just need brushes!?

    John

  5. #5 bahram kasmai
    March 4, 2012

    i have a new idea about using energy of earth rotation