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In Kepler’s law of periods T^2 = kr^3, t...

In Kepler’s law of periods `T^2 = kr^3`, the constant `k = 10^(–13) s^2m^(–3)`. Express the constant k in days and kilometers. The moon is at a distance of `3.84 xx 10^5` km from the earth. Obtain its time period of revolution in days.

Text Solution

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We are given that `lambda= 7xx 10^(-7)m, a= 10^(-3)m, D= 4xx 10^(8)m`
Clearly, angular spread, `theta= (lambda/a)= (7xx 10^(-7)m)/(10^(-3)m)= 7xx 10^(-4)rad`
Linear spread, `x= D theta`
`=(4xx10^(-8)m)(7xx 10^(-4))= 28xx 10^(4)m`
Areal spread `x^(2)= (28xx 10^(4)m)^(2)= 7.84xx 10^(10) m^(2)`.
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