Question #245770

(Acceleration due to rotation and orbit of the Earth) A body is at rest at a location on the Earth's equator. Find its acceleration due to the Earth's rotation.[Take the Earth's radius at the equator to be 6400 km.] Find also the acceleration of the Earth in its orbit around the sun.[Take the Sun to be fixed and regard the Earth as a particle following a circular path with centre the Sun and radius 15*1010m.]


1
Expert's answer
2021-10-04T16:52:35-0400
a=ω2R=(2πT)2Ra=\omega^2R=(\dfrac{2\pi}{T})^2R

a)


R=6400 km=6.4×106 mR=6400\ km=6.4\times10^6\ m


T=24 h=24×3600 s=8.64×104 sT=24\ h=24\times3600\ s=8.64\times10^4\ s

a=(2π8.64×104 s)2(6.4×106 m)a=(\dfrac{2\pi}{8.64\times10^4\ s})^2(6.4\times10^6\ m)

=0.033846 m/s2=0.033846\ m/s^2

a=0.033846 m/s2a=0.033846\ m/s^2 directed towards the centre of the Earth.



b)


R=15×1010 mR=15\times10^{10}\ m




T=365×24 h=365×24×3600 s=31536000 sT=365\times24\ h=365\times24\times3600\ s=31536000\ s

a=(2π31536000 s)2(15×1010 m)a=(\dfrac{2\pi}{31536000\ s})^2(15\times10^{10}\ m)




=0.005954 m/s2=0.005954\ m/s^2

a=0.005954 m/s2a=0.005954\ m/s^2 directed towards the centre of the Sun.


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Comments

Mohammed
05.10.21, 18:59

Thank you

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