Question

If the escape velocity from the surface of the earth 11.2km/s find the critical velocity of the body orbiting close to the surface from the earth

Answer 1

Hello dear,

◆ Answer -

vo = 7.92 km/s

● Explanation -

For satellite orbiting very close to the earth,

vo = √(gR)

Rearranging,

vo = √(2gR) / √2

As we know, ve = √(2gR), so now

vo = ve / √2

vo = 11.2 / √2

vo = 7.92 km/s

Therefore, the critical velocity of the body orbiting close to the surface from the earth is 7.92 km/s.

Thanks dear...