Answer:
The value is [tex]a_r = 3.81 *10^{-3} m/s^2[/tex]
Explanation:
Generally the moon's radial acceleration is mathematically represented as
[tex]a_r = r * w^2[/tex]
Here [tex]w[/tex] is the angular velocity which is mathematically represented as
[tex]w =\frac{2 \pi }{ T}[/tex]
substituting [tex]1760 * 10^3 \ seconds[/tex] for T(i.e the period of the moon ) we have
[tex]w =\frac{2 * 3.142 }{ 1760 * 10^3}[/tex]
=> [tex]w = 3.57 *10^{-6} \ rad/s [/tex]
From the question r(which is the radius of the orbit ) is evaluated as
[tex]r = R + H[/tex]
substitute [tex]3.60 * 10^6 m[/tex] for R and [tex]295.0 * 10^6 \ m[/tex] H
[tex]r = 295.0 * 10^6 +3.60 * 10^6[/tex]
=> [tex]r = 2.986 *10^{8} \ m [/tex]
So
[tex]a_r = 2.986 *10^{8} * ( 3.57 *10^{-6} )^2[/tex]
[tex]a_r = 3.81 *10^{-3} m/s^2[/tex]
