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’s surface. Total energy is always constant at any orbit. The height of the satellite above the surface of the earth is?
A- 6400 km
B- 3200 km
C- 9600 km
D- 4800 km

Answer 1

satellite moves in the orbit due to the influence of gravitational force. It is a text-book example of a body in a uniform circular motion although the orbits are not spherical in reality, they are elliptical but the motion can be considered as a circular motion.

Step by step answer:
When the satellite is moving in the orbit the motion is dragged by centripetal force and in such a motion satellite has a constant velocity that can be found at any place in the orbit by drawing the tangent at that time.
Now radius of the earth, R= 6400 km
Orbital speed of satellite, ${{V}_{0}}=\dfrac{{{v}_{e}}}{2}$
We know the relationship between orbital speed and escape speed is given as $\sqrt{2}{{v}_{0}}={{v}_{e}}$
But ${{v}_{0}}=\dfrac{\sqrt{G{{M}_{e}}}}{\sqrt{R+h}}$ and ${{v}_{e}}=\dfrac{\sqrt{2G{{M}_{e}}}}{\sqrt{R}}$
Putting the values, we get,

$$\dfrac{{{v}_{e}}}{2}=\dfrac{\sqrt{G{{M}_{e}}}}{\sqrt{R+h}} \\\ \Rightarrow \dfrac{\sqrt{2G{{M}_{e}}}}{2\sqrt{R}}=\dfrac{\sqrt{G{{M}_{e}}}}{\sqrt{R+h}} \\\ \Rightarrow \dfrac{2G{{M}_{e}}}{4R}=\dfrac{G{{M}_{e}}}{(R+h)} \\\ \Rightarrow (R+h)=2R \\\ \therefore R=h \\\$$

Thus at height equal to radius (6400 km) the condition satisfies, hence, the correct option is (A).

Note: All heavenly bodies move under the influence of mutual gravitational forces which are an example of conservative forces. The analogy of this problem is a stone tied to a circle and whirled in a vertical circle and when the thread breaks the stone fly away tangentially.