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Standard XII

Physics

Satellites

What is a per...

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

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156 minutes

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83.73 minutes

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Solution

The correct option is B 83.73 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

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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.

The correct option is B 83.73 minutesRadius 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 $π = 3.14$ .

The correct option is B 83.73 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