Question

A remote sensing satellite of earth revolves in a height of 250 KM above the earth surface.What is the 1)orbital speed 2) Period of revolution of satellite​

Answer 1

Answer:

v = 39916.532 m/s , T = 0.49 π sec

Explanation:

1) Orbital Speed: v = √G.M / √R

Where G= gravitational constant = 6.673 x 10⁻¹¹ Nm² / Kg²

           M= Mass of Earth= 5.972 x 10²⁴ Kg

           R= Radius of the orbit = 250 Km= 25 x 10⁴ m

           v= √3.983 x 10¹⁴ / √25 x 10 ⁴

             = 39916.532 m/s

2) Period of revolution: T² = 4 π² R³ / G.M

T² = 4 . π² . 24 x 10¹² / 3.983 x 10¹⁴

   = 0.24 π²

T = 0.49 π sec