An earth satelites S has orbit radius which $4$ times that of communication satellite C. The period of revolution of S will be.

Question

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Answer 1

Larger the distance of planet from the sun, larger will be its period of revolution around the sun.
Kepler's third low planetary motion the square of the period of revolution of any planet around the sun is directly proportional to the cube of the semi-major axis of its elliptical orbit.
$∴ T^{2} \propto R^{3}$
$\frac{T _{s}}{T _{c}} = (\frac{R _{s}}{R _{c}})^{3 / 2}$
given $R_{s} = 4 R_{c}$
$\frac{T _{s}}{T _{c}} (\frac{4 R _{c}}{R _{c}})^{3 / 2} = 8$
for $T_{c} = 1$ day
$T_{s} = 8$ days.

Answer 2

Larger the distance of planet from the sun, larger will be its period of revolution around the sun.
Kepler's third low planetary motion the square of the period of revolution of any planet around the sun is directly proportional to the cube of the semi-major axis of its elliptical orbit.

∴ T^{2} \propto R^{3}

\frac{T _{s}}{T _{c}} = (\frac{R _{s}}{R _{c}})^{3 / 2}

given

R_{s} = 4 R_{c}

\frac{T _{s}}{T _{c}} (\frac{4 R _{c}}{R _{c}})^{3 / 2} = 8

for

T_{c} = 1

day

T_{s} = 8

days.

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