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$(A)$ Given: Altitude $h = 1000 \,km = 10^6 \,m$, Radius of Earth $R = 6.4 	imes 10^6 \,m$, Mass of Earth $M = 6 	imes 10^{24} \,kg$, Gravitational

Vedclass · Accepted Answer

$105$

Vedclass · Answer

$(A)$ Given: Altitude $h = 1000 \,km = 10^6 \,m$, Radius of Earth $R = 6.4 	imes 10^6 \,m$, Mass of Earth $M = 6 	imes 10^{24} \,kg$, Gravitational constant $G = 6.67 	imes 10^{-11} \,Nm^2 \,kg^{-2}$.The orbital radius is $r = R + h = 6.4 	imes 10^6 + 1.0 	imes 10^6 = 7.4 	imes 10^6 \,m$.The time period $T$ of a satellite is given by $T = 2 \pi \sqrt{\frac{r^3}{GM}}$.Substituting the values:$T = 2 	imes 3.14 	imes \sqrt{\frac{(7.4 	imes 10^6)^3}{6.67 	imes 10^{-11} 	imes 6 	imes 10^{24}}}$$T = 6.28 	imes \sqrt{\frac{405.224 	imes 10^{18}}{40.02 	imes 10^{13}}}$$T = 6.28 	imes \sqrt{10.125 	imes 10^5} = 6.28 	imes \sqrt{1012500} \approx 6.28 	imes 1006.23 \approx 6319 \,s$.Converting to minutes: $T = \frac{6319}{60} \approx 105.3 \,min$.Thus, the time period is approximately $105 \,min$.

$A$ satellite is placed in a circular orbit around the Earth at an altitude of $1000 \,km$. The time period of the satellite in minutes is approximately (mass of the Earth $= 6 \times 10^{24} \,kg$, radius of the Earth $= 6.4 \times 10^6 \,m$, $G = 6.67 \times 10^{-11} \,Nm^2 \,kg^{-2}$).

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