To solve this problem it is necessary to apply the concepts presented in Kepler's third law in which the period is described, as well as the theorems developed for acceleration based on gravity.
Acceleration in gravitational terms can be expressed as
[tex]a = \frac{GM}{r^2}[/tex]
Where,
G = Gravitational Universal Constant
M = Mass of Earth
r = Distance
At the same time the Period by Kepler's law the Period is described as
[tex]T^2 = \frac{4\pi^2r^3}{GM}[/tex]
That is equal to
[tex]T^2 = \frac{4\pi^2r}{\frac{GM}{r^2}}[/tex]
Using the equation of acceleration,
[tex]T^2 = \frac{4\pi^2r}{a}[/tex]
Re-arrange to find a,
[tex]a = \frac{4\pi^2r}{T^2}[/tex]
Our values are given as
[tex]r = 1.5*10^{11}m[/tex]
[tex]T = 365days(\frac{86400s}{1days}) = 31536000s[/tex]
Replacing we have,
[tex]a = \frac{4\pi^2r}{T^2}[/tex]
[tex]a = \frac{4\pi^2(1.5*10^{11})}{(31536000)^2}[/tex]
[tex]a = 0.005954m/s^2[/tex]
Therefore the magnitude of the acceleration of the earth is [tex]0.005954m/s^2[/tex]
