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Can I Get Into Outer Space Using AA Batteries?

August 30th, 2012 | Posted by colin in Random

AA Battery in Space

A long time ago I was quite fascinated by the idea of electro-gravitics, the notion that one might be able to manipulate gravity or otherwise generate thrust using nothing but the careful modulation and configuration of electric fields.

While this was enjoyable folly at the time, on reflection the other night the thought occurred that even if this were possible, and reasonably efficient, how could it be powered, and more specifically, could I get into outer space using nothing but the humble AA battery?

As with most problems in physics the question boils down to energy, and it’s conversion from one form to another. Provided we accept the law of conservation of energy, many a problem or postulation can be coarsely tested simply by evaluating the best case scenario of the energy available and the energy required. If there is insufficient energy to meet requirements then there is no point going any further, or into the details of how efficient the energy may be converted in between.

In the case of our AA battery the first question we should ask is, ignoring the electro-gravitic conversion efficiency, does a AA battery have enough energy to lift itself into outer space?

For the sake of this argument, we’ll use the following parameters;

  • Outer Space starts at 100km above sea level
  • Our AA battery is pretty standard Alkaline type, weighing 23 grams, and has a nominal voltage of 1.5v and a current capacity of 2000 mAh

We can determine the amount of energy available by noting that a Joule is equivalent to a WattSecond, the product of power and time, and power is simply the product of Voltage and Current. So to determine the energy capacity of the AA battery we simply multiply the battery voltage by the current capacity normalised to AmpSeconds instead of mAh (ie. divide by 1000 to get current in Amps, and multiple by 3600 to convert hours to seconds).

P_e=1.5*(2000/1000)*3600 J=10.8 KJ

The amount of energy required to lift the AA battery to a height of 100 km is approximated by the equation for the gravitational potential energy for an object of mass m, at a height h.

P_g=m*g*h = 23*10^-3 * 9.8 * 100*10^3 =  22.54 KJ

Much to my disappointment, what we see immediately is that a AA battery needs twice as much energy just to lift itself into outer space. So we can conclude that, no matter how efficient our electro-gravitic thruster might be, no amount of standard AA batteries will every lift themselves, let alone a payload of my personage, into outer space.
Which just leaves me wondering what the minimum energy density of a battery needs to be to be able to lift itself and a percentage of it’s own weight into space?
TBC.

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