Question

calculated the height of the satellite from the surface of the earth if it's orbiting with a velocity of 4 km/sec g= 6.67× 10-11 Nm2 mass 6×1024 kg Radius 6400​

Answer 1

Answer:

The distance of the satellite from the surface of the Earth is  2.56128 ×10¹⁴ km

Explanation:

We know that $V_{orbital}$ = $\sqrt{\frac{Gm}{R} }$

where $V_{orbital}$ = orbital velocity

                  G = Gravitational constant

                  m = Mass of the Earth

                  R = Distance of the satellite from the center of the Earth

Let the distance of the satellite from the surface of the Earth be H

On substituting the values,

=> 4 = √(6.67 × 10¹¹ × 6×1024)/(6400+H)

Squaring both sides,

=> 16 = (6.67 × 10¹¹ × 6×1024)/(6400+H)

=> 16 (6400+H) = 6.67 × 10¹¹ × 6×1024

=> 6400+H = 6.67 × 10¹¹ × 6×64

=> H = 40.02×64× 10¹¹ - 6400

=> H = 64( 40.02×10¹¹ - 100)

=> H = 64 (4.002×10¹² )

=> H = 2.56128 ×10¹⁴ km

Therefore, the distance of the satellite from the surface of the Earth is  2.56128 ×10¹⁴ km