Question

A remote-sensing satellite of earth revolves in a circular orbit at a height of $25\times {10}^4$$\text{m}$ above the surface of earth. If earth's radius is $6.4\times {10}^6\text{m}$ and $g=10$${\text{ms}}^{-2}$, then the orbital speed of the satellite is

• A. $4\text{km/s}$
• B. $6\text{km/s}$
• C. $9.6\text{km/s}$
• D. $8\text{km/s}$

Answer 1

The correct option is D $8\text{km/s}$

The orbital velocity of a satellite at a height $h$ from the surface of a planet having mass $M$ and radius $R$ is given by, $$v=\sqrt{\frac{GM}{(R+h)}}=\sqrt{\frac{g{R}^2}{(R+h)}}$$ Given, $R=6.4\times {10}^6\text{m}$,
$h=2.5\times {10}^4\text{m}$

Since, $h<<R,$

$\Rightarrow v=\sqrt{gR}$

$\Rightarrow v=\sqrt{10\times 6.4\times {10}^6}$

$\Rightarrow v=8\times {10}^3\text{m/s}$

$\Rightarrow v=\frac{8\times {10}^3}{1000}\text{km/s}$

$\Rightarrow v=8\text{km/s}$

Hence, option (d) is correct answer.