Question

An artificial satellite of earth is launched in circular orbit in equatorial plane of the earth and satellite is moving from west to east. With respect to a person on the equator, the satellite is completing one round trip in $24\text{hours}$. Mass of earth is, $M=6\times {10}^{24}\text{kg}$. For this situation, orbital radius of the satellite is (Take $G=6.67\times$${10}^{-11}{\text{Nkg}}^{-2}{\text{m}}^2$)

• A. $2.66\times {10}^4\text{km}$
• B. $6.4\times {10}^4\text{km}$
• C. $3.6\times {10}^4\text{km}$
• D. $2.9\times {10}^4\text{km}$

Answer 1

The correct option is A $2.66\times {10}^4\text{km}$

Here the time period of satellite w.r.t observer on equator is $24\text{hours}$ and the satellite is moving from west to east, so angular velocity of satellite w.r.t earth's axis of rotation (considered as fixed) is calculated by using,

${\omega }_{se}={\omega }_s-{\omega }_e$

Where, ${\omega }_{se}$ angular velocity of satellite w.r.t earth,
${\omega }_s$ angular velocity of satellite and,
${\omega }_e$ angular velocity of earth.

$\Rightarrow {\omega }_s={\omega }_{se}+{\omega }_e$

$\therefore {\omega }_s=\frac{2\pi }{T_{se}}+\frac{2\pi }{T_e}$

Where, given
$T_s=24\text{hours}$ and $T_e=24\text{hours}$ are time periods of satellite and earth, respectively.

$\Rightarrow {\omega }_s=\frac{2\pi }{24}+\frac{2\pi }{24}$

$\Rightarrow {\omega }_s=\frac{\pi }{6}{\text{hr}}^{-1}$

$\Rightarrow {\omega }_s=\frac{\pi }{6\times 3600}{\text{sec}}^{-1}$

$\Rightarrow w_s=1.45\times {10}^{-4}\text{rad/s}.$

From, $v=\sqrt{\frac{GM}{r}}$

$\Rightarrow r\omega =\sqrt{\frac{GM}{r}}$

$\Rightarrow$ $r^{3/2}=\frac{\sqrt{GM}}{\omega }$

Substituting the values, we get

$\Rightarrow$ $r^{3/2}=\frac{\sqrt{6.67\times {10}^{-11}\times 6\times {10}^{24}}}{1.45\times {10}^{-4}}$

$\Rightarrow r=2.66\times {10}^7\text{m}$

$\Rightarrow r=2.66\times {10}^4\text{km}.$

Hence, option (a) is the correct answer.