The time period of a satellite of Earth is 5 hrs. If the separation between the Earth and the satellite is increased to 4 times the previous value, the new time period will become,

Question

TestHub · Accepted Answer

According to Kepler's third law, the square of the time period of a satellite is directly proportional to the cube of its orbital radius.

$T^{2} \propto R^{3}$

$(\frac{T _{2}}{T _{1}})^{2} = (\frac{R _{2}}{R _{1}})^{3}$

Given $T_{1} = 5$ hrs and $R_{2} = 4 R_{1}$ .

$(\frac{T _{2}}{5})^{2} = (4)^{3} = 64$

$\frac{T _{2}}{5} = 8$

$T_{2} = 40$ hrs.

Answer: $40$ hrs.

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