Centripetal Acceleration Space Shuttle Problem

[C^2]

Homework Statement

Suppose the 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. Express your answer in terms of g, the gravitational accleration at the Earth's surface.

Known Variables:

time(t)= 5400seconds
centripetal acceleration(a)= ?
radius(r)=400,000 meters

Homework Equations

v=2pir/t
a=v^2/t
g=9.8 m/s^2

The Attempt at a Solution

So, what I thought I was supposed to do was use the first equation and plug in the numbers:

v= 2(3.14)(400,000)/5400s

=465.2 m/s

then I plugged the result into the second equation:

a= (465.2)^2/5400

=.084 m/s^2

and since they ask to express the answer in terms of g,

.084 m/s^2 * 9.8 m/s^2

=.844 m/s^2

but the book says the answer should be .9g's. I don't know if I'm just splitting hairs here, but .844 does not round up to .9.
Can someone point out where I went wrong?
Thanks in advance

 

[einstein_a_go_go]

Hi,

where did you get a = v^2 / t ?

v is a velocity in metres per second
t is a time in seconds
a is an acceleration in metres per second squared

the dimensions (units) do not add up!

Hope that helps cheers

 

[baba89]

Your calculations are correct. The book may have rounded up to .9 for simplicity, but your answer of .844 m/s^2 is more accurate. Sometimes in scientific calculations, we have to round off the numbers for simplicity, but it is important to keep track of the significant figures. In this case, your answer has three significant figures, while .9 only has one. So, your answer of .844 m/s^2 is more precise. Keep up the good work in your calculations!