Question

How much time will a satellite in an orbit at height 35780 km above earths surface moving with tangential velocity of 5km|s take to complete one revolution around earth if radius of earth is 6400 km

Answer 1

Explanation:

Given G = gravitational constant = 6.67 x 10-11 N m2/kg2

M mass of earth = 4 x 6 x 1024 kg

R = 6400 km

 

h = height of satellite above earth’s surface = 35780 km

R+h = 6400 + 35780 = 42180 km = 42180 x 103 m

 

v = ?

we know

v = √GM/R+H

= $\sqrt{6.67*10^{-11} *4*6*10^{24}/42180*10^{3} }$

v = 6.16 km/s

We know v = distance/time

= circumference/time

= 2πr/T

T =  $2\pi (R+h)/v$

T =  $2*3.14*42180/6.616$

T = 43001.68 sec

T = 11.94 hr or approximately 12 hours.

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Answer 2

Explanation:

Given G = gravitational constant = 6.67 x 10-11 N m2/kg2

M mass of earth = 4 x 6 x 1024 kg

R = 6400 km

 

h = height of satellite above earth’s surface = 35780 km

R+h = 6400 + 35780 = 42180 km = 42180 x 103 m

 

v = ?

we know

v = √GM/R+H

= \sqrt{6.67*10^{-11} *4*6*10^{24}/42180*10^{3} }6.67∗10−11∗4∗6∗1024/42180∗103

v = 6.16 km/s

We know v = distance/time

= circumference/time

= 2πr/T

T =  2\pi (R+h)/v2π(R+h)/v

T =  2*3.14*42180/6.6162∗3.14∗42180/6.616

T = 43001.68 sec

T = 11.94 hr or approximately 12 hours.

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