Question

The orbital radius of moon around the earth is $3.8\times {10}^8$ meter and its time period is 27.3 days. The centripetal acceleration of moon will be

• A. $-2.4\times {10}^{-3}m/s^2$
• B. $11.2m/s^2$
• C. $2.7\times {10}^{-3}m/s^2$
• D. $9.8m/s^2$

Answer 1

Answer: (C)

$2.7\times {10}^{-3}m/s^2$

The centripetal acceleration = Net centripetal force / mass
Net centripetal force = $\frac{GMm}{R^2}-m(\omega {)}^{2}R$
$\omega =\frac{2\pi }{27.3\times 24\times 3600},R=3.8\times {10}^8,M=5.92\times {10}^{2}4$
centripetal acceleration = $\frac{\frac{GMm}{R^2}-m(\omega {)}^{2}R}{m}=\frac{6.67\times {10}^{-11}\times 5.92\times {10}^{2}4}{(3.8\times {10}^8{)}^2}-(\frac{2\pi }{27.3\times 24\times 3600}{)}^2\times 3.8\times {10}^8=2.7\times {10}^{-3}m.s^{-2}$