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Question

(B) 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 $(r)$: $

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) 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 $(r)$: $T^2 \propto r^3$,which implies $T \propto r^{3/2}$.Given the initial time period $T_1 = 7 \, hours$ and the initial radius $r_1 = r$.The new radius is $r_2 = 3r$.Using the ratio: $\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}ight)^{3/2}$.Substituting the values: $\frac{T_2}{7} = \left(\frac{3r}{r}ight)^{3/2} = 3^{3/2} = 3\sqrt{3}$.$T_2 = 7 	imes 3\sqrt{3} = 21\sqrt{3} \, hours$.Since $\sqrt{3} \approx 1.732$,$T_2 \approx 21 	imes 1.732 = 36.372 \, hours$.Therefore,the approximate new time period is $36 \, hours$.

The time period of a satellite revolving around earth in a given orbit is $7 \, hours$. If the radius of orbit is increased to three times its previous value,then the approximate new time period of the satellite will be ...... $hours$.

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