μ = √(1-81N/f2 )
0.75 = √(1-81N/(10*106 )
Nmax =5.4*1011
fc = 9√(Nmax )
=6.614*106 Hz
and
Dskip = 2h √((fmuf /fc)2 -1)
=2*350√((12*106 /(6.614*106 ))2-1)
Dskip = 631.64 km
2070 magh (Q). The reflection takes place at a height of 350km and the maximum density in the ionosphere corresponds to a 0.75 refractive index at 10MHz. What will be the range (assume the earth is flat) for which MUF is 12MHz.
by 