To solve this problem we will apply the concept related to the kinematic equations of angular motion, as well as the definition of the period as a function of angular velocity and Newton's second law. For this purpose we will have to:
[tex]a = \omega^2r[/tex]
Where,
a = Acceleration
[tex]\omega[/tex]= Angular velocity
r = Radius
At the same time the angular frequency defined in terms of period:
[tex]\omega = \frac{2*\pi}{T}[/tex]
Replacing at the first equation we have that
[tex]a = \frac{4\pi^2r}{T^2}[/tex]
Applying this concept of acceleration in Newton's second law we have to
[tex]F= ma[/tex]
[tex]F = m(\frac{4\pi^2r}{T^2})[/tex]
Our values are given as,
[tex]T = 365.25days(\frac{24hours}{1 day})(\frac{3600s}{1hour})[/tex]
[tex]T = 31557600 s[/tex]
[tex]R = 1.5*10^8km = 1.5*10^{11}m[/tex]
[tex]m = 6*10^{24}kg[/tex]
Replacing we have,
[tex]F = m(\frac{4\pi^2r}{T^2})[/tex]
[tex]F = (6*10^{24})(\frac{4\pi^2(1.5*10^{11})}{(31557600)^2})[/tex]
[tex]F = 3.5677*10^{22}N[/tex]
Therefore the correct answer is 7.
The centripetal force exerted on the Earth by the Sun is [tex]3.56 \times 10^{22} \ N[/tex].
The given parameters;
The angular velocity of the Earth around the sun is calculated as follows;
[tex]\omega = \frac{2\pi }{T} \\\\\omega = \frac{2\pi }{31,557,600} \\\\\omega = 1.99 \times 10^{-7} \ rad/s[/tex]
The centripetal force exerted on the Earth by the Sun is calculated as;
[tex]F = ma_c\\\\F= m\times (\omega )^2 r\\\\F = (6\times 10^{24}) \times (1.99\times 10^{-7})^2 \times 1.5 \times 10^{11}\\\\F = 3.56 \times 10^{22} \ N[/tex]
Thus, the centripetal force exerted on the Earth by the Sun is [tex]3.56 \times 10^{22} \ N[/tex].
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