Question

If an artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth, the height of the satellite above the surface of the earth is:

2R
R/2
R
R/4

Answer 1

v= 1/2 (escape velocity)
v^2 = (1/2)^2 (escape velocity)^2
GM/(R+h)=1/4[2GM/R]
1/(R+h)=1/2R
2R=R+h
h= R

Answer 2

Answer:

R

Explanation:

The orbital speed = √ GM / D

G = the universal gravitational constant,

M = Mass of Earth

D = Distance of Object from center of Earth

Escape Velocity = √ 2GM / R

The orbital speed = (1/2) Escape Velocity

=> √ GM / D  =  (1/2) √ 2GM / R

Squaring both sides

=> GM/D  = (1/4) * 2GM/R

=> D = 2R

height of the satellite above the surface of the earth is = D - R

= 2R - R

= R