Question

There are two satellites A and B. Their periods are T and TB respectively. If 8TA = TB and Height of A equal to 300 km, RE = 6400 km. Find Height of B.​

Answer 1

Answer:  Calculate the speed and period of revolution of a satellite orbiting at a height ... Given: height of satellite above the surface of earth = h = 700 km, ... To Find: speed of satellite = vc = ?, Period = T = ? ... Given: radius of earth = R = 6400 km = 6.4 x 108 m, radius of orbit of ... If yes what is the nature of the orbit?

Explanation:

Answer 2

Hey there!

we know that ,

$T^{2} \ is \ proportional \ to\ R^{3}$

∴ hence ,  

•     $\frac{T_{A} ^{2}}{T_{B} ^{2}} = \frac{R_{A} ^{3}}{R_{B} ^{3}}$

 

•      $\frac{T_{A} ^{2}}{64T_{A} ^{2}} = \frac{(300+ 6400_{}) ^{3}}{(H_{B}+6400) ^{3}}$

•      $(\frac{1}{64} )^\frac{1}{3} = \frac{(300+ 6400_{}) { }}{(H_{B}+6400) {}}$

•       $(\frac{1}{4^3} )^\frac{1}{3} =\frac{300+ 6400}{H_B +6400}$

•          $\frac{1}{4} = \frac{6700}{H_B+ 6400}$  

•       $H_B + 6400 = 26800$  

•       $H_B = 26800 -6400$  

•       $H_B = 20400Km$  

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