Question #1e981

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Answer 1 · 2015-03-28T10:57:43

The answer is: #a_c=2.4*10^-3m/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=s/t#

where, in this case:

#v=v_t#,

and

#s=2piR#

(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=((2piR)/T)^2/R=(4pi^2R^2/T^2)/R=4pi^2R/T^2=#

#=(4*3.14^2*3.8*10^8m)/(2.5*10^6s)^2=(4*3.14^2*3.8)/2.5^2*10^8/10^12m/s^2~=#

#~=24*10^-4m/s^2=2.4*10^-3m/s^2#.