A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the centre of earth in new orbit is 2 times that in the earlier orbit. The time period in the second orbit is (1) 4.8 hours (2) 48root2 hours (3) 24 hours (4) 24root2 hours

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Solution

$T i m e p e r i o d o f g e o s t a t i n a r y s a t a l l i t e = T = 24 h o u r s D i s \tan c e = r n e w d i s \tan c e = 2 r n e w t i m e p e r i o d i s T' {(\frac{T'}{T})}^{2} = {(\frac{2 r}{r})}^{3} T' = T {(2)}^{3 / 2} T' = 24 \times 2 \sqrt{2} = 48 \sqrt{2} h o u r s$

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A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the centre of earth in new orbit is 2 times that in the earlier orbit. The time period in the second orbit is (1) 4.8 hours (2) 48root2 hours (3) 24 hours (4) 24root2 hours

Time period of geostatinary satallite = T= 24 hoursDistance = rnew distance = 2rnew time period is T'T'T2=2rr3T' = T23/2T' = 24×22=482 hours

$T i m e p e r i o d o f g e o s t a t i n a r y s a t a l l i t e = T = 24 h o u r s D i s \tan c e = r n e w d i s \tan c e = 2 r n e w t i m e p e r i o d i s T' {(\frac{T'}{T})}^{2} = {(\frac{2 r}{r})}^{3} T' = T {(2)}^{3 / 2} T' = 24 \times 2 \sqrt{2} = 48 \sqrt{2} h o u r s$