# Question #1e981

Mar 28, 2015

The answer is: ${a}_{c} = 2.4 \cdot {10}^{-} 3 \frac{m}{s} ^ 2$.

Since the centripetal acceleration is:

${a}_{c} = {v}_{t}^{2} / R$, where ${v}_{t}$ is the tangential speed, that is unknown.

But $v = \frac{s}{t}$

where, in this case:

$v = {v}_{t}$,

and

$s = 2 \pi R$

(the lenght of the circle, good approximation of the moon elliptic orbit),

$t = T$ period of the moon revolution.

So:

${a}_{c} = {v}_{t}^{2} / R = {\left(\frac{2 \pi R}{T}\right)}^{2} / R = \frac{4 {\pi}^{2} {R}^{2} / {T}^{2}}{R} = 4 {\pi}^{2} \frac{R}{T} ^ 2 =$

$= \frac{4 \cdot {3.14}^{2} \cdot 3.8 \cdot {10}^{8} m}{2.5 \cdot {10}^{6} s} ^ 2 = \frac{4 \cdot {3.14}^{2} \cdot 3.8}{2.5} ^ 2 \cdot {10}^{8} / {10}^{12} \frac{m}{s} ^ 2 \cong$

$\cong 24 \cdot {10}^{-} 4 \frac{m}{s} ^ 2 = 2.4 \cdot {10}^{-} 3 \frac{m}{s} ^ 2$.