1.

If the orbital velocity of the moon is 1020 m s^(-1), find the time taken by the moon to complete one revolution around the Earth. Explain why this period is different from the period that is observed from the Earth, which is 29.5 days. (Take the distance of the moon from the Earth as 3.4 xx 10^(8) m, "and" (1)/(86400) = 1.157 xx 10^(-5))

Answer»

Solution :(i) Here the force of attraction between the moon and the Earth can be obtained by using the formula `F = (GM_(e)m_(m))/(r^(2)) = (m_(m)V^(2))/(r)`
Find the VALUE of v and r from given data But,`""v = (2 pi r)/(T)`
Obtain the value of 'T'.
Since 1 day has 86400 s.
Find the NUMBER of DAYS the moon takes for one revolution around the Earth as observed from earth.
Divide 'T' by 86400 s to get the value of 'T' in terms of the number of days.
What is a sidereal month and a synodic month?
(ii) 29.5 days


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