# Re: Explorer Power Gain Problem

```
>From Brian,

>>Kelly
>> The formula is right, but the number is only valid when the velocity of
the
>> starship is 5 m/s. The problem with the document is that the meaning of
the
>> number is not told.
>
>Now I remember.  I calculated the power the drive system would need for a
one
>second boost, and assumed it would be constant for the flight.  I.E. if
>it takes X watts to push the ship at a ship G for a secound.  Multiply that
>by the number of seconds of boost to get to desired speed, and ....
>
>Given that the engine should need to output the same power, to accelerate
the
>same mass, at the same rate, at diferent speeds.  It seems like it should
>work.

>Tim
>Ah that makes some sence, you indeed can use this trick to calculate the
>power needed for a self-fueled ship. But keep in mind that the mass of the
>ship may decrease significantly due to the "burning" of the fuel.

Brian.
Don't bother sending me a reply to this e-mail (not because I don't like
you, but because I have to be going off-line for the rest of the school
year).

Over break I looked very closely at Kelly's Explorer paper power
requirements.  I noted one very confusing thing that he did when he
calculated his power requirements.  He said that Watts could be divided by
time (example: W/s).

After studying my physics textbook for a few nights, I now believe that
there is no such thing as a W/s because a Watt is already a Joule per
second.  How can you have Joules per second per second?  I don't believe you
can and I couldn't find anything in two different physics textbooks to
suggest otherwise.

I thought that I was missing something because I seemed to have remembered
hearing that the Earth received 1400 W/m^2 per second.  That is wrong.  The
texts available to me say that we get 1400 W/m^2 period.

Anyway, if you follow my reasoning, you can see why this error explains the
low numbers on the Explorer paper.  Otherwise the math seemed good to me (I
think).

```