Question

10)Calculate the period of revolution of a polar satellite orbiting close to the surtace of Earth.Given R=6400km, g=9.8m/s2 *

Answer 1

Explanation:

Given: Radius of earth = R = 6400 km = 6.4 x 106 m, Time period = T = 27.3 days = 27.3 x 24 x 60 x 60, g = 9.8 m/s6.

Answer 2

Solution:

let, T be the period of revolution of a polar satellite orbiting close to the surface of Earth.

     R be the radius of the earth and g be the acceleration due to gravity.

here, given in the question that

R = 6400km and g = 9.8 m/s^2

as the polar satellite orbiting close to the surface of Earth. so we can consider the radius of satellite as same as the radius of the earth.

now, we know that

   T =2π√R/g

   T = 2*3.14√6400/9.8

   T = 6.28√653.06

   T = 6.28*25.55

   T = 160.454 sec.