The time period of a satellite of Earth is 5, | vedclass

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(C) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is proportional to the cube of the semi-major axis $(r)$,i

Vedclass · Accepted Answer

$40$

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 semi-major axis $(r)$,i.e.,$T^2 \propto r^3$ or $T \propto r^{3/2}$.Given the initial time period $T_1 = 5\, h$ and the initial separation $r_1 = r$.The new separation is $r_2 = 4r$.Using the ratio formula: $\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}ight)^{3/2}$.Substituting the values: $\frac{T_2}{5} = \left(\frac{4r}{r}ight)^{3/2}$.$\frac{T_2}{5} = (4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$.Therefore,$T_2 = 5 	imes 8 = 40\, h$.

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

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