a) Earth can be thought of as a sphere of radius 6400 km. Any object is performing circular motion around the axis of earth due to earth’s rotation. What is acceleration of object on the surface of the earth towards its centre? What is it at latitude θ? How does these accelerations compare with g = 9.8 m/s2? b) Earth also moves in circular orbit around sun once every year with an orbital radius of 1.5 × 1011m. What is the acceleration of earth towards the centre of the sun? How does this acceleration compare with g = 9.8 m/s2?
a) Earth can be thought of as a sphere of radius 6400 km. Any object is performing circular motion around the axis of earth due to earth’s rotation. What is acceleration of object on the surface of the earth towards its centre? What is it at latitude θ? How does these accelerations compare with g = 9.8 m/s2? b) Earth also moves in circular orbit around sun once every year with an orbital radius of 1.5 × 1011m. What is the acceleration of earth towards the centre of the sun? How does this acceleration compare with g = 9.8 m/s2?

Answer:

(a) According to the question, we have been given that,
Radius of the earth (R)=6400 km =6.4 \times 10^{6} m.
Time period of the motion (T)=1 day =24 \times 60 \times 60 s =86400 s
As we know that Centripetal acceleration,

    \[\begin{aligned} \left(a_{c}\right)=\omega^{2} R=\frac{4 \pi^{2} R}{T} &=\frac{4 \times(22 / 7)^{2} \times 6.4 \times 10^{6}}{(24 \times 60 \times 60)^{2}} \\ &=\frac{4 \times 484 \times 64 \times 10^{6}}{49 \times(24 \times 3600)^{2}}=0.034 m / s ^{2} \end{aligned}\]

At equator, latitude makes an angle \theta=0^{\circ}

  acg=0.0349.8=1288\therefore \quad \frac{a_{c}}{g}=\frac{0.034}{9.8}=\frac{1}{288}

 

b) The earth’s orbital radius around the sun is measured in meters. (R)=1.5 \times 10^{11} m

Time period =1 yr =365 days =365 \times 24 \times 60 \times 60 s =3.15 \times 10^{7} s

Centripetal acceleration

    \[\begin{aligned} &\left(a_{c}\right)=R \omega^{2}=\frac{4 \pi^{2} R}{T^{2}}=\frac{4 \times(22 / 7)^{2} \times 1.5 \times 10^{11}}{\left(3.15 \times 10^{7}\right)^{2}}=5.97 \times 10^{-3} m / s ^{2} \\ &\frac{a_{c}}{g}=\frac{5.97 \times 10^{-3}}{9.8}=\frac{1}{1642} \end{aligned}\]