The earth (mass M = 6 times 10^{24} kg) revo | vedclass

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(D) Given: Mass of the earth $m = 6 	imes 10^{24} \ kg$,angular velocity $\omega = 2 	imes 10^{-7} \ rad/s$,and radius of orbit $R = 1.5 	imes 10^8

Vedclass · Accepted Answer

$36 	imes 10^{21}$

Vedclass · Answer

(D) Given: Mass of the earth $m = 6 	imes 10^{24} \ kg$,angular velocity $\omega = 2 	imes 10^{-7} \ rad/s$,and radius of orbit $R = 1.5 	imes 10^8 \ km = 1.5 	imes 10^{11} \ m$.The centripetal force required for the circular motion of the earth is provided by the gravitational force exerted by the sun.The formula for centripetal force is $F = m \omega^2 R$.Substituting the given values into the formula:$F = (6 	imes 10^{24}) 	imes (2 	imes 10^{-7})^2 	imes (1.5 	imes 10^{11})$$F = (6 	imes 10^{24}) 	imes (4 	imes 10^{-14}) 	imes (1.5 	imes 10^{11})$$F = (6 	imes 4 	imes 1.5) 	imes (10^{24} 	imes 10^{-14} 	imes 10^{11})$$F = 36 	imes 10^{21} \ N$.

The earth (mass $M = 6 \times 10^{24} \ kg$) revolves around the sun with an angular velocity $\omega = 2 \times 10^{-7} \ rad/s$ in a circular orbit of radius $R = 1.5 \times 10^8 \ km$. The force exerted by the sun on the earth in newtons is:

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