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Given that T^(2) = kR^(3), express the c...

Given that `T^(2) = kR^(3)`, express the constant `k` of the above relation in days and kilometres. Given, `k= 10^(-13)s^(2) m^(-3)`. 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

Verified by Experts

By Kepler's law of periods,
` T^(2) = kr^(3)`
where k is a constant.
Data : ` k = 10^(-13) s^(2)// m^(3) , r = 3.85 xx 10^(5) km, 1 d = 8.64 xx 10^(4) s`
` k = 10^(-13) s^(2)/m^(3) = (10^(-13))/((8.64 xx 10^(4))^(2) xx (10^(-3))^(3)) d^(2)/(km^(3))`
` = (10^(-12))/((8.64)^(2)) d^(2)/(km^(3))`
` :. T^(2) = (10^(-12))/((8.64)^(2))(3.85 xx 10^(5))^(3)`
` = ((3.85)^(3) xx 10^(-12) xx 10^(15))/((8.64)^(2))`
` = 0.7647 xx 10^(3) = 764.7 d^(2)`
` :. T = sqrt(764.7) = 27.66 ` d
`{:(log 8.64,,0.9365),(,,ul(" "xx2)),(,,1.8730):}`
`{:(log 3.85,," "0.5855),(,,ul(" "xx3)),(,," "1.7565),(,,ul(-1.8730)),(,,ul(" "bar(1).8835)):}`
AL ` bar(1).8835 = 0.7647`
`{:(log 764.7,,2.8835),(,,ul(" "xx1/2)),(,,ul(1.4418)):}`
` AL 1.4418 = 27.66`
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