# The radius of the moon's orbit about the Earth is about 3.6 * 10^8 m. The moon's period is 2.3 * 10^6 seconds. What is the centripetal acceleration of the moon?

Dec 27, 2016

I found $0.00268 \frac{m}{s} ^ 2 \approx 0.003 \frac{m}{s} ^ 2$

#### Explanation:

Centripetal acceleration is given by:

${a}_{c} = {v}^{2} / r$

Where $v$ is the linear velocity and $r$ the radius of the circular
orbit.

We assume uniform velocity and a circular orbit; velocity will be:

$v = \text{distance"/"time"="circumference"/"period} = \frac{2 \pi r}{T}$

Our centripetal acceleration will become:

${a}_{c} = \frac{4 {\pi}^{2} {r}^{\cancel{2}}}{{T}^{2} \cancel{r}}$

${a}_{c} = \frac{4 {\pi}^{2} r}{{T}^{2}}$

${a}_{c} = \frac{4 {\pi}^{2} \cdot 3.6 \cdot {10}^{8}}{{\left(2.3 \cdot {10}^{6}\right)}^{2}}$

${a}_{c} = \frac{39.478 \cdot 3.6 \cdot {10}^{8}}{5.29 \cdot {10}^{12}}$

${a}_{c} = \frac{39.478 \cdot 3.6 \cdot {10}^{8} \cdot {10}^{-} 12}{5.29}$

${a}_{c} = \frac{39.478 \cdot 3.6 \cdot {10}^{-} 4}{5.29}$

${a}_{c} = \frac{142.1208 \cdot {10}^{-} 4}{5.29}$

${a}_{c} = 26.86 \ldots . \cdot {10}^{-} 4$

${a}_{c} = 0.00268 \frac{m}{s} ^ 2$

${a}_{c} \approx 0.003 \frac{m}{s} ^ 2$