**The moon revolves around the earth in almost a circle of radius 382400 Km in 27.3 days. What is the centripetal acceleration?**

**Solution**

**Strategy:** Centripetal acceleration is given by

So we need to find the velocity first. Then use this formula to find the centripetal acceleration. Similarly time for one revolution is 27.3 days and one day is 24 hours. So we have to convert this time in seconds.

**Given **Radius of the circular path = 382400 Km = 382400 * 10^{3} m

Time for one revolution = t = 27.3 days

= 27.3 x 24 x 60 x 60 s = 2358720 s

**Required **Centripetal acceleration = a_{c} = ?

Velocity= v =?

Now the circumference of the circle is

This is in fact the distance moon covers in 27.3 days. So apply the formula (v = s/t) to find the velocity of the moon.

Therefore,

Put this value of velocity to find the acceleration.

Pingback:numerical-problem-1-rotatory-and-circular-motion-physics-11 – msa

Pingback:Numerical problem on converting linear acceleration to angular acceleration – msa

Pingback:Numerical Problems on Rotatory and Circular Motion, Physics Grade 11 – msa

Pingback:numerical-problems-on-circular-and-rotatory-motion-chapter-4-physics-11