Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

# What is a period of revolution of the earth satellite? Ignore the height of satellite above the surface of the earth. Given, the value of gravitational acceleration g = 10 ms -2, radius of the earth Re = 6400 km. (take, π = 3.14)

1) 85 min

2) 156 min

3) 83.73 min

4) 90 min

Solution:

Radius of earth RE = 6400 km = 6.4 x 106 m

Time period of revolution

$$\begin{array}{l}T = 2\pi \sqrt{\frac{R_{E}}{g}}\end{array}$$

Therefore,

$$\begin{array}{l}T = 2\times 3.14\times \sqrt{\frac{6.4\times 10^{6}}{10}}\end{array}$$

T = 5024 s

T = 5024/60 = 83.73 minutes