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?

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Answer 1 · 2016-12-27T11:56:08

I found #0.00268m/s^2~~0.003m/s^2#

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="distance"/"time"="circumference"/"period"=(2pir)/T#

Our centripetal acceleration will become:

#a_c=(4pi^2r^cancel(2))/(T^2cancel(r))#

#a_c=(4pi^2r)/(T^2)#

Using your data for the radius and period:

#a_c=(4pi^2*3.6*10^8)/((2.3*10^6)^2)#

#a_c=(39.478*3.6*10^8)/(5.29*10^12)#

#a_c=(39.478*3.6*10^8*10^-12)/(5.29)#

#a_c=(39.478*3.6*10^-4)/(5.29)#

#a_c=(142.1208*10^-4)/5.29#

#a_c=26.86.... *10^-4#

#a_c=0.00268m/s^2#

#a_c~~0.003m/s^2#