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Question: A geostationary satellite is orbiting the earth at a height \(6\) R above the earth’s surface, where...

A geostationary satellite is orbiting the earth at a height 66 R above the earth’s surface, where R is the radius of the earth. The time period of another satellite at height 2.52.5 R from the earth’s surface would be
a) 2424 h
b) 62.5\dfrac{6}{2.5} h
c) 256\dfrac{25}{6} h
d) 626\sqrt{2} h

Explanation

Solution

Hint : The time period of a geostationary satellite is 24 hours. Use the time period of revolution in the height of a satellite equation. Then compare it to any other satellites height to find its respective time period of revolution.

Complete solution:
Before we begin with the question let us know the conditions for a satellite to be geostationary,

  1. The satellite should revolve in the orbit concentric and coplanar to the equatorial plane of the earth.
  2. The satellite should have the same angular speed as that of the rotation of the earth i.e. from west to east. Hence the relative angular velocity is zero and the period of revolution is the same as the rotation of the earth. Therefore for an observer on earth it would appear the satellite to be stationary.
    The height of any satellite is given by,
    h=[T2R2g4π2]13−Rh={{\left[ \dfrac{{{T}^{2}}{{R}^{2}}g}{4{{\pi }^{2}}} \right]}^{\dfrac{1}{3}}}-R
    In the above expression ,
    H is the height of the satellite from the surface of the earth,
    T is the period of revolution,
    R is the radius of the earth.
    g is the acceleration due to gravity.
    Now let us substitute the height of geostationary satellite in the above equation,
    h=[T2R2g4π2]13−Rh={{\left[ \dfrac{{{T}^{2}}{{R}^{2}}g}{4{{\pi }^{2}}} \right]}^{\dfrac{1}{3}}}-R
    Since the time period of geostationary satellite is 24hrs24hrs
    6R=[242R2g4π2]13−R6R={{\left[ \dfrac{{{24}^{2}}{{R}^{2}}g}{4{{\pi }^{2}}} \right]}^{\dfrac{1}{3}}}-R
    Taking R on the other side and taking cube of the equation,
    (6R+R)3=[242R2g4π2]33{{(6R+R)}^{3}}={{\left[ \dfrac{{{24}^{2}}{{R}^{2}}g}{4{{\pi }^{2}}} \right]}^{\dfrac{3}{3}}}
    (7R)3=[242R2g4π2]{{(7R)}^{3}}=\left[ \dfrac{{{24}^{2}}{{R}^{2}}g}{4{{\pi }^{2}}} \right]
    Let us call this as equation 1
    To find the period of revolution other satellite let us substitute the height of the other satellite,
    h=[T2R2g4π2]13−Rh={{\left[ \dfrac{{{T}^{2}}{{R}^{2}}g}{4{{\pi }^{2}}} \right]}^{\dfrac{1}{3}}}-R
    2.5R=[T2R2g4π2]13−R2.5R={{\left[ \dfrac{{{T}^{2}}{{R}^{2}}g}{4{{\pi }^{2}}} \right]}^{\dfrac{1}{3}}}-R
    Taking R on the other side and taking cube of the equation,
    (2.5R+R)=[T2R2g4π2]33(2.5R+R)={{\left[ \dfrac{{{T}^{2}}{{R}^{2}}g}{4{{\pi }^{2}}} \right]}^{\dfrac{3}{3}}}
    (3.5R)3=[T2R2g4π2]{{(3.5R)}^{3}}=\left[ \dfrac{{{T}^{2}}{{R}^{2}}g}{4{{\pi }^{2}}} \right]
    Let us call this as equation 2
    Dividing equation 2 by 1
    [(3.5R)3(7R)3]=[T2R2g4π2242R2g4π2]\left[ \dfrac{{{(3.5R)}^{3}}}{{{(7R)}^{3}}} \right]=\left[ \dfrac{\dfrac{{{T}^{2}}{{R}^{2}}g}{4{{\pi }^{2}}}}{\dfrac{{{24}^{2}}{{R}^{2}}g}{4{{\pi }^{2}}}} \right]
    Dividing the terms and using the law of exponent the above equation becomes,
    (3.57)3=(T24)2{{\left( \dfrac{3.5}{7} \right)}^{3}}={{\left( \dfrac{T}{24} \right)}^{2}}
    (12)3=(T24)2{{\left( \dfrac{1}{2} \right)}^{3}}={{\left( \dfrac{T}{24} \right)}^{2}}
    Taking square root on both the sides,
    122=T24\dfrac{1}{2\sqrt{2}}=\dfrac{T}{24}
    After cross multiplying on both the sides,
    T=2422T=\dfrac{24}{2\sqrt{2}}
    Multiplying and dividing by 2\sqrt{2} in the above equation we get
    T=1222T=\dfrac{12\sqrt{2}}{2}
    T=62hrsT=6\sqrt{2}hrs

Note: The time period of geostationary satellites is 24hrs and not any other satellites. There can be many other satellites which are geostationary but positioned differently though they exist at approximately the same height.