Centripetal Acceleration Space Shuttle Problem

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SUMMARY

The discussion focuses on calculating the centripetal acceleration of the Space Shuttle in orbit at 400 km above the Earth's surface, completing a revolution every 90 minutes. The correct formula for centripetal acceleration is derived from the velocity equation, where velocity (v) is calculated as v = 2πr/t, resulting in a velocity of 465.2 m/s. The centripetal acceleration is then calculated as a = v²/r, yielding an acceleration of 0.084 m/s², which is approximately 0.9g when expressed in terms of gravitational acceleration (g = 9.8 m/s²). The discrepancy in rounding was clarified, emphasizing the importance of unit consistency in calculations.

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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 acceleration 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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