Question

Suppose the space shuttle is in orbit 300 km from the Earth's surface, and circles the earth about once every 90

Answer 1

The centripetal acceleration of the space shuttle is $9.0 m/s^2$ (0.9g)

Explanation:

The period of revolution of the space shuttle is

$T=90 min \cdot 60 = 5400 s$

From the period, we can calculate the angular speed of the space shuttle, which is given by

$\omega = \frac{2\pi}{T}=\frac{2\pi}{5400}=1.16\cdot 10^{-3} rad/s$

Now we can calculate the centripetal acceleration of the space shuttle, given by

$a=\omega^2 r$

where

$\omega=1.16\cdot 10^{-3} rad/s$ is the angular speed

$r=6.40 \cdot 10^6 m +300\cdot 10^3 m = 6.7\cdot 10^3 m$ is the distance of the shuttle from the Earth's centre (it is the  sum of the Earth's radius, 6400 km, and of the altitude of the space shuttle, 300 km)

Substituting and solving, we find

$a=(1.16\cdot 10^{-3})^2(6.7\cdot 10^6)=9.0 m/s^2$

And since $g=9.8 m/s^2$, this can be rewritten as

$a=\frac{9.0}{9.8}=0.9g$

Learn more about centripetal acceleration:

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