The Moon and Earth rotate about their common center of mass, which is located ab
ID: 1631615 • Letter: T
Question
The Moon and Earth rotate about their common center of mass, which is located about 4700 km from the center of Earth. (This is 1690 km below the surface.) (a) Calculate the magnitude of the acceleration due to the Moon's gravity at that point. (b) Calculate the magnitude of the centripetal acceleration of the center of Earth as it rotates about that point once each lunar month (about 27.3 d) and compare it with the acceleration found in part (a). Comment on whether or not they are equal and why they should or should not be.Explanation / Answer
The formula to find the acceleration due to gravity is
g = G*M/R^2
Where
G = gravitational constant
M = mass of planet
R = disatace of the point
(a)
Convert the distance of the point form kilometer to meter
d = (4700 km)*(1.0*10^3 m/1 km)
= 4.7*10^6 m
R = (3.84*10^8)(4.70*10^6)
= 3.79*10^8 m
g = (6.673*10^-11 N- ^2/kg^2)*(7.35*10^22kg)/(3.79810^8 m)^2
= 3.42*10^-5 m/s^2
(b)
Formula to be find the centripetal is
ac = (omega)^2*r
Where
ac = centripetal the acceleration
omega = angular velocity
r = point distance
Convert time period form day to sec.
T = (27.3 day)(8.64*10^3 s/1 day)
= 2.35*10^6 s
Since
omega = (2*pi/T)
Here T = time period
omega = (2*pi/2.35*10^6 s)
= 2.66*10^-6 rad/s
ac = (2.66*10^-6 rad/s)^2(4.7*10^6 m)
= 3.34*10^-5 m/s^2
Hence centripetal acceleration is 3.34*10^-5 m/s^2.
Centripetal acceleration to the acceleration due to gravity of moon in part (a) is almost equal