Question

An artificial satellite is revolving in a circular orbit at height of $1200\text{km}$ above the surface of planet. If the radius of the planet is $6400\text{km}$ and mass is $9\times {10}^{23}\text{kg}$, the orbital speed is ($G=6.7\times {10}^{-11}$${\text{Nm}}^2/{\text{kg}}^2$)

• A. $2.1{\text{kms}}^{-1}$
• B. $2.8{\text{kms}}^{-1}$
• C. $9{\text{kms}}^{-1}$
• D. $1{\text{kms}}^{-1}$

Answer 1

The correct option is B $2.8{\text{kms}}^{-1}$

The orbital velocity of a satellite in circular orbit of radius $R$ around a planet having mass $M$ is given by $$v=\sqrt{\frac{GM}{R}}$$Where, $R=\text{radius of the planet+height above the planet surafce}$

$\therefore R=6400+1200=7600\text{km}=76\times {10}^5\text{m}$

Substituting the values,

$v=\sqrt{\frac{6.7\times {10}^{-11}\times 9\times {10}^{23}}{76\times {10}^5}}$

$\therefore v=2817\text{m/sec}\approx 2.8\text{km/sec}$

Hence, option (b) is the correct answer.

Key Concept: The orbital speed of a satellite is given by $$v=\sqrt{\frac{GM}{r}}$$