What is a period of revolution of earth satellite? Ignore the height of satellite above the surface of earth.
Given : (1) The value of gravitational acceleration $g = 10 m s^{- 2}$ .
(2) Radius of earth $R_{E} = 6400$ km. Take $π = 3.14$ .
85 minutes
83.73 minutes
90 minutes
156 minutes

Question

What is a period of revolution of earth satellite? Ignore the height of satellite above the surface of earth.
Given : (1) The value of gravitational acceleration $g = 10 m s^{- 2}$ .
(2) Radius of earth $R_{E} = 6400$ km. Take $π = 3.14$ .
85 minutes
83.73 minutes
90 minutes
156 minutes

Toppr Admin · Accepted Answer

Radius of earth -xA0- RE-6400 km -6-4-xD7-106 mTime period of revolution -xA0- -xA0- T-2-x3C0-x221A-REg-x2234- -xA0-xA0-T-2-xD7-3-14-xD7-x221A-6-4-xD7-10610 -xA0- -xA0- -xA0- -xA0- -xA0-xA0-xA0- -x27F9-T-5024 s-x27F9- -xA0- T-502460-83-73 minutes

What is a period of revolution of earth satellite? Ignore the height of satellite above the surface of earth.Given : (1) The value of gravitational acceleration g=10ms−2. (2) Radius of earth RE=6400 km. Take π=3.14.85 minutes83.73 minutes90 minutes156 minutes

Radius of earth RE=6400 km =6.4×106 mTime period of revolution T=2π√REg∴ T=2×3.14×√6.4×10610 ⟹T=5024 s⟹ T=502460=83.73 minutes

What is a period of revolution of earth satellite? Ignore the height of satellite above the surface of earth.
Given : (1) The value of gravitational acceleration $g = 10 m s^{- 2}$ .
(2) Radius of earth $R_{E} = 6400$ km. Take $π = 3.14$ .
85 minutes
83.73 minutes
90 minutes
156 minutes

Radius of earth $R_{E} = 6400$ km $= 6.4 \times 10^{6}$ m
Time period of revolution $T = 2 π \sqrt{\frac{R_{E}}{g}}$
$∴$ $T = 2 \times 3.14 \times \sqrt{\frac{6.4 \times 10^{6}}{10}}$ $⟹ T = 5024$ s
$⟹$ $T = \frac{5024}{60} = 83.73$ minutes