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  • 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?

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 the acceleration of object on the surface of the earth towards its centre? What is it at latitude \theta? How does these accelerations compare with g=9.8\; m/s^{2}?

b) Earth also moves in circular orbit around sun once every year with an orbital radius of 1.5\times 10^{11}\; m. What is the acceleration of earth towards the centre of the sun? How does this acceleration compare with g=9.8\; m/s^{2}?

\left ( Hint: acceleration\frac{V^{2}}{R}=\frac{4\pi ^{2}R}{T^{2}} \right )

Answers (1)

The angular acceleration will have its direction towards the centre and the value will be a=w^{2}R

Now, w=\frac{2\pi }{T}

R=6.4\times 10^{6}m\; and T=24\times 3600=86400s

Acceleration of the person on the surface of the earth,

a=w^{2}R=\left ( \frac{2\pi }{T} \right )^{2}\times R

=\frac{4\times22\times22\times6.4\times10^{6}}{7\times 7\times 24\times 24\times 3600\times 3600}=\frac{1210}{81\times49\times9}=0.034\; m/s^{2}

Now, At equator latitude = 0

Hence \frac{a}{g}=\frac{0.034}{9.8}=\frac{1}{288}=0.0034 which is much smaller than the previous case.

b) acceleration of Earth revolving around the sun

R=1.5\times10^{11}m and w=\frac{2\pi }{T}

T=365\times24\times3600=3.15\times10^{7}seconds

Now, a=w^{2}R=(\frac{2\pi }{T})^{2}\times R

=\frac{4\times 3.14\times3.14\times1.5\times10^{11}}{3.15\times3.15\times10^{7}\times10^{7}}=6\times10^{-3}m/s^{2}

Also, \frac{a}{g}=\frac{0.0006}{9.8}

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