If M is the mass of earth and m the mass of satellite the gravitational force at the satellite altitude H is

`F = G*m*M/R^2` , assuming R is measured from the center of the Earth

By writing also `F =m*g(R)` we observe that the gravitational acceleration at altitude R is

`g(R) = G*M/R^2`

which need to be equal to the satellite centripetal acceleration.

`a_("satelite") =G*M/R^2`

For the aircraft that is moving in a horizontal circle of radius r just above the earth, the centripetal acceleration is

`a_("plane") = V^2/r`

Both accelerations are equal means that

`G*M/R^2= V^2/r`

`V^2 =G*M*r/R^2`

The numerical values are

`G =6.67*10^-11 m^3/(kg*s^2)`

`M =5.97*10^24 kg`

`R =6.7*10^6 m`

Therefore

`V=sqrt(6.67*10^-11*5.97*10^24*15/(6.7*10^6)^2)=sqrt(133)=11.54 m/s`

The speed of the plane need to be 11.54 m/s to have the same centripetal acceleration as the satellite has.