You can
put this solution on YOUR website!You don't know the diameter of the planet, so let's call that
![D[p]](/cgi-bin/plot-formula.mpl?expression=D%5Bp%5D&x=0003)
miles. The first orbit is one mile above the planet, so we know that the diameter of the orbit is
![D[p]+2](/cgi-bin/plot-formula.mpl?expression=D%5Bp%5D%2B2&x=0003)
. We had to add 2 because 1 mile of altitude adds 1 mile to the radius, and the diameter is twice that. Now, if we move out to a point 2 miles above the surface, the diameter of the new orbit is
![D[p]+4](/cgi-bin/plot-formula.mpl?expression=D%5Bp%5D%2B4&x=0003)
.
We know that the distance travelled in the first orbit is the circumference of the circle with diameter
![D[p]+2](/cgi-bin/plot-formula.mpl?expression=D%5Bp%5D%2B2&x=0003)
, or
![C[1]=(D[p]+2)*pi](/cgi-bin/plot-formula.mpl?expression=C%5B1%5D=%28D%5Bp%5D%2B2%29%2Api&x=0003)
, and the distance travelled in the second orbit is
![C[2]=(D[p]+4)*pi](/cgi-bin/plot-formula.mpl?expression=C%5B2%5D=%28D%5Bp%5D%2B4%29%2Api&x=0003)
.
Now we can subtract:
![C[2]-C[1]](/cgi-bin/plot-formula.mpl?expression=C%5B2%5D-C%5B1%5D&x=0003)
=>
![((D[p]+4)pi)-((D[p]+2)pi)](/cgi-bin/plot-formula.mpl?expression=%28%28D%5Bp%5D%2B4%29pi%29-%28%28D%5Bp%5D%2B2%29pi%29&x=0003)
=>
![(D[p]pi+4pi)-(D[p]pi+2pi)](/cgi-bin/plot-formula.mpl?expression=%28D%5Bp%5Dpi%2B4pi%29-%28D%5Bp%5Dpi%2B2pi%29&x=0003)
=>

miles. Not too far, really, just a little over six and a quarter miles.
Hope this helps,
John