Question

The escape velocity of a body at a height h above the surface of the earth is 36% of that from the surface of earth. if v0 be the orbital velocity of this body near the surface of the earth, then what will be its orbital velocity at a height h?

Answer 1

it's orbital velocity at height h will be ...

Answer 2

Answer:

The orbital velocity of the body at height h will be 2.85 Km/s.

Explanation:

escape velocity(Ve) at the surface of the earth will be=$\sqrt{2GM/R}$  (1)

orbital velocity(V0) at the surface of the earth will be=$\sqrt{GM/R}$        (2)

G=universal gravitational constant

R=radius of the earth

M=mass of the earth

When a body is placed at a certain height above the surface of the earth then-new escape velocity and orbital velocity,

V'e=$\sqrt{2GM/(R+H)}$        (3)

V'o=$\sqrt{GM/(R+H)}$          (4)

relation between Ve and Vo:

Ve=$\sqrt{2}$Vo                          (5)

Ve=11.2  Km/s    (it is constant)    (6)

as per question V'e will be 0.36Ve

so from equations 1 and 3, we get:

H=6.716R

by putting the value of H in equation (4) we get:

V'o=0.36$\sqrt{GM/R}$=0.36Vo

by using equations( 5) and (6) we get:

V'o=2.85 Km/s