Question

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

• A. $2R$
• B.
  $8R$
• C.
  $7R$
• D.
  $R$

Answer 1

The correct option is C

$7R$
Given, the relation between the orbital speed and escape velocity of a satellite is, $$v_0=\frac{v_e}{4}$$ $\Rightarrow v_0=\frac{1}{4}\sqrt{\frac{2GM}{R}}$

$\Rightarrow v_0=\sqrt{\frac{GM}{8R}}...\left(1\right)$

The orbital velocity of a satellite in circular orbit is given by $$v_0=\sqrt{\frac{GM}{r}}...\left(2\right)$$ where $r$ is the distance from the center of the earth.

So, from equation $\left(1\right)$ and $\left(2\right)$,

$\sqrt{\frac{GM}{r}}=\sqrt{\frac{GM}{8R}}$

$\Rightarrow r=8R$

Hence, height of the satellite above earth's surface

$=8R-R$

$=7R$

Hence, option (c) is the correct answer.

Key Concept : The escape velocity for a planet is given by $$V_e=\sqrt{\frac{2GM}{R}}$$ and the orbital velocity of a satellite in circular orbit is given by $$v_0=\sqrt{\frac{GM}{r}}$$