If the particle shown in the figure is projected with orbital speed $v_{0} = \frac{GM}{2 R}$ , it returns after time T. If it is projected such that time period of revolution is 8T then the maximum distance of the particle from the centre of the earth is XR, the value of X is :-

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Launch radius r = R + h = 2 R . Vis-viva: v 0 2 ​ = GM ( r 2 ​ − a 1 ​ 1 ​ ) ⇒ a 1 ​ = 2 R . New period 8T: T 2 ∝ a 3 ⇒ a 2 ​ = 8 2/3 a 1 ​ = 4 × 2 R = 8 R . Launch point is the perigee, so r min ​ = 2 R . r min ​ + r ma x ​ = 2 a 2 ​ = 16 R ⇒ r ma x ​ = 14 R . X = 14. Answer: 14.

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