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Express the constant k of Eq. (8.38) in ...

Express the constant k of Eq. (8.38) in days and kilometres. Given k = `10^(–13) s^(2) m^(–3)` . The moon is at a distance of `3.84 x× 105` km from the earth. Obtain its time-period of revolution in days.

Text Solution

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`implies` From Kepler.s third law ,
`T^(2) = kr^(3) " "...(1)`
but `k = 10^(-13) (s^(2))/(m^3) " "...(2)`
and `24xx3600 s = 1` day
then , ` 1 s = ? `
`1s =1/(24xx3600)days`
and `1m = 1/(10^3)km`
`:. k = 10^(-13)(1/(24xx3600))^2xx1/((1/10^3)^3)`
( `:.` Substituting in equation (2)),
`= 10^(-13) xx (1.157 xx10^(-5))^(2) xx 10^(9)`
`=10^(-4) xx 1.339 xx 10^(-10)`
`= 1.339 xx 10^(-14) (("day")^(2))/(("km")^3)`
Now, from equations (1),
`T^(2) = kr^(3)`
` = 1.339 xx 10^(-14) xx (3.84 xx10^(5)^(3)`
`[ :. r = 3.84 xx 10^(5) km]`
= 758.18
`:. T = 27.53` days
`:. T = 27.5 ` days
Note : If `k= 1.33 xx 10^(-14) ("day")^2/((km)^3)` them
`T = 27.4` days , but answer from textbook is not 27.3
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