Please solve this question...
Dear student Let’s start with Kepler’s Third Law: It states, “The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.” Putting it simply, if the time taken for a planet to rotate once in its orbit is T and the radius of the orbit is R (assuming a circular orbit, as must do in such questions), then, T22 is proportional to R33. =>(T1/T2)22 = (R1/R2)33 Here, let T1, R1 be the time period and radius of Earth’s current orbit. Let T2, R2 be the same quantities for the planet’s orbit. Given, R2 = 4*R1 => R1/R2 = 1/4 Substituting in the above equation, => (T1/T2)22 = (1/4)33 = 1/64 => T1/T2 = (1/64)0.50.5 = 1/8 => T2 = 8*T1 Since T1 is one Earth year or 365 Earth days, The time period of revolution of the planet is 8 Earth year or 2920 Earth days. Option B P.S. ‘=>’ means ‘which implies' Regards