Determining the gravity acceleration at distant point from earth

A space shuttle is in orbit 400 km from the Earth's surface, and circles the Earth about once every 90 minutes.Find the centripetal acceleration of the space shuttle in its orbit (Earth radius = 6.38 x 10^3 km)
I got the answer 5350 m/s^2 (forgot to factor in the radius of the earth). Then I got the answer 155 m/s^2 (remembered to factor in the radius of the earth)
Apparently the real answer is 0.9g's = 0.9 x 9.8 = 8.82 m/s^2

Now I'm following the steps in my book to the decimal point and I still don't seem to know how they got that answer.
Could someone show me the proper equation?

this is the solution:
to make it easy, we have to calculate at the same system unit.

radius of the space shuttle: 400 km + 6380 km = 6780 km
it circles the earth every 90 minutes (5400 second).
The distance in once circle is = 2*pi*Radius=2*3.14*6780=42578.4 km = 42578400 m
from this we can obtain the velocity of shuttle: distance/time = 42578400/5400= 7884.8 m/s

The centripetal acceleration is = velocity^2/Radius= (7884.8^2)/6780000= 9.16 m/s^2
as we know the g = 9.8 m/s^2, 
so the answer is 9.16/9.8 = 0.93 g