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Question

(C) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $r$,i.e.,$

Vedclass · Accepted Answer

$664$

Vedclass · Answer

(C) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $r$,i.e.,$T^2 \propto r^3$ or $T \propto r^{3/2}$.For the first satellite,the distance from the center of the Earth is $r_1 = R$,where $R$ is the radius of the Earth. The time period is $T_1 = 83 	ext{ minutes}$.For the second satellite,the orbit is at a distance of $3R$ from the surface,so the distance from the center of the Earth is $r_2 = R + 3R = 4R$.Using the ratio formula: $\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} ight)^{3/2}$.Substituting the values: $\frac{T_2}{83} = \left( \frac{4R}{R} ight)^{3/2} = (4)^{3/2}$.Calculating the value: $(4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$.Therefore,$T_2 = 83 	imes 8 = 664 	ext{ minutes}$.

The rotation period of an earth satellite close to the surface of the earth is $83$ minutes. The time period of another earth satellite in an orbit at a distance of three earth radii from its surface will be .......... $\min$

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