Centripetal Acceleration Space Shuttle Problem

AI Thread Summary
The discussion focuses on calculating the centripetal acceleration of the space shuttle in orbit at 400 km altitude, with a period of 90 minutes. The user initially calculates the velocity using the formula v = 2πr/t, resulting in approximately 465.2 m/s. They then attempt to find centripetal acceleration using a = v^2/t, yielding 0.084 m/s², which they convert to 0.844 g. However, they are confused as the textbook states the answer should be 0.9 g, prompting questions about potential errors in their calculations. A participant points out a misunderstanding regarding the use of the formula for acceleration, highlighting a unit mismatch.
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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
 
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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
 
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