Question

An artificial satellite of the earth is launched in a circular orbit in the equator plane of the earth, and the satellite is moving from west to east. With respect to a person on the equator, the satellite is completing one round trip in 24 h. Mass of the earth is \[M = 6 \times {10^{24}}kg\]. For this situation, the orbital radius of the satellite is
$
  {\text{A}}{\text{. 2}}{\text{.66}} \times {\text{1}}{{\text{0}}^4}km \\
  {\text{B}}{\text{. 6400km}} \\
  {\text{C}}{\text{. 36000km}} \\
  {\text{D}}{\text{. 29600km}} \\
 $

Answer 1

Hint: We know the formula for orbital velocity of a satellite around earth and also the relation between angular velocity and the linear velocity. First we need to find the angular velocity of the satellite using the given time period, then solve for radius using the expression for orbital velocity.

Formula used:
The angular velocity of an object is given in terms of its time period by the following expression:
$\omega = \dfrac{{2\pi }}{T}$
The orbital velocity of satellite is given as
$v = \sqrt {\dfrac{{GM}}{r}} $
Velocity is related to the angular velocity by the following relation.
$v = r\omega $

Complete step by step answer:
We are given an artificial satellite. With respect to a person on the equator, the satellite is completing one round trip in 24 h.
${T_S} = 24h$
The satellite is moving from west to east so the angular velocity of this satellite with respect to axis of rotation of earth will be equal to the sum of angular velocity satellite as observed by a person on the equator and angular velocity of earth. It is given as
$
  \omega = {\omega _S} + {\omega _E} \\
   = \dfrac{{2\pi }}{{{T_S}}} + \dfrac{{2\pi }}{{{T_E}}} \\
 $
Here ${T_S}$ represents the time period of revolution of the satellite as observed by an observer on earth while ${T_E}$ represents the time period of rotation of earth. We know that ${T_E} = 24h$, we get the angular velocity of the satellite with respect to axis of rotation of earth to be
$\omega = \dfrac{{2\pi }}{{24}} + \dfrac{{2\pi }}{{24}} = \dfrac{{4\pi }}{{24}} = \dfrac{\pi }{6}rad/hr = 1.45 \times {10^{ - 4}}rad/s$
Now we can write the orbital velocity of the satellite by using the following formula.
$
  v = \sqrt {\dfrac{{GM}}{r}} \\
  r\omega = \sqrt {\dfrac{{GM}}{r}} \\
  {r^2}{\omega ^2} = \dfrac{{GM}}{r} \\
  {r^3}{\omega ^2} = GM \\
  r = {\left( {\dfrac{{GM}}{{{\omega ^2}}}} \right)^{\dfrac{1}{3}}} \\
 $
In the second step we have used the relation that \[v = r\omega \].
Now we can insert the known values in the above expression. Doing so, we get
$
  r = {\left( {\dfrac{{6.67 \times {{10}^{ - 11}} \times 6 \times {{10}^{24}}}}{{{{\left( {1.45 \times {{10}^{ - 4}}} \right)}^2}}}} \right)^{\dfrac{1}{3}}} \\
   = {\left( {19.03 \times {{10}^{21}}} \right)^{\dfrac{1}{3}}} \\
   = 2.67 \times {10^7}m \\
   = 2.67 \times {10^4}km \\
 $
Hence, the correct answer is option A.

Note:
It should be noted that the distance r is calculated as the distance of the satellite from the centre of the earth. By subtracting the radius of the earth from this value we can also obtain the distance of the satellite from the surface of earth.