A geostationary satellite is orbiting the earth at a height of 4 R above the surface of the earth, where R is the radius of the earth. Another satellite is orbiting the earth at a height 1.5 R from th

Question

A geostationary satellite is orbiting the earth at a height of 4 R above the surface of the earth, where R is the radius of the earth. Another satellite is orbiting the earth at a height 1.5 R from the surface of the earth with periodic time 'T' in hour. The value of 'T' in hour is

TestHub · Accepted Answer

The orbital radius of the geostationary satellite is $r_{1} = R + 4 R = 5 R$ .

The time period of a geostationary satellite is $T_{1} = 24$ hours.

The orbital radius of the second satellite is $r_{2} = R + 1.5 R = 2.5 R$ .

Let the time period of the second satellite be $T_{2} = T$ .

According to Kepler's third law, the square of the time period is proportional to the cube of the orbital radius:

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

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

$(\frac{T}{24})^{2} = (\frac{2.5 R}{5 R})^{3}$

$(\frac{T}{24})^{2} = (\frac{1}{2})^{3} = \frac{1}{8}$

$\frac{T}{24} = \frac{1}{2 2}$

$T = \frac{24}{2 2} = 62$ hours

Answer: $62$

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