Centripetal acceleration of shuttle 400 km above Earth

[warfreak131]

Homework Statement

A shuttle is in orbit 400km above the Earths surface, and circles it every 90 minutes. Find the centrip. acceleration and find it in terms of g. (He says for sake of ease, assume gravity = 10 m/s^2)

Earth radius = 6400 km

Homework Equations

$$a_{c}=\frac{v^2}{r}, v=\frac{2{\pi}r}{t}$$

The Attempt at a Solution

First I converted everything to meters and seconds:

6400 km = 6400000m, 90 minutes = 5400 seconds

$$v=\frac{2{\pi}r}{t}$$
$$v=\frac{2{\pi}(6400000+400000)}{5400}$$
$$v=7900\frac{m}{s}$$

$$a_{c}=\frac{v^2}{r}$$
$$a_{c}=\frac{7900^2}{6800000}$$
$$a_{c}=\frac{6.3{\cdot}10^7}{6800000}$$
$$a_{c}=9.2\frac{m}{s^2}$$

= .92g's

 

[kuruman]

Looks right, but I am not the one grading it.

 

[warfreak131]

damn, because i re did the question when posting it on this forum, but i remember that my answer on the test was nothign close to that

i hope i can get some extra credit.