Two satellites A and B of masses 200 , kg and | vedclass

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(A) The time period of a satellite at a height $h$ from the Earth's surface is given by $T = 2 \pi \sqrt{\frac{(R+h)^{3}}{GM}}$.For satellite $A$: $h_

Vedclass · Accepted Answer

$1.33 	imes 10^{3} \, s$

Vedclass · Answer

(A) The time period of a satellite at a height $h$ from the Earth's surface is given by $T = 2 \pi \sqrt{\frac{(R+h)^{3}}{GM}}$.For satellite $A$: $h_{A} = 600 \, km = 0.6 	imes 10^{6} \, m$,$R = 6.4 	imes 10^{6} \, m$. Orbital radius $r_{A} = R + h_{A} = 7.0 	imes 10^{6} \, m$.$T_{A} = 2 \pi \sqrt{\frac{(7.0 	imes 10^{6})^{3}}{6.67 	imes 10^{-11} 	imes 6 	imes 10^{24}}} \approx 5800 \, s$.For satellite $B$: $h_{B} = 1600 \, km = 1.6 	imes 10^{6} \, m$. Orbital radius $r_{B} = R + h_{B} = 8.0 	imes 10^{6} \, m$.$T_{B} = 2 \pi \sqrt{\frac{(8.0 	imes 10^{6})^{3}}{6.67 	imes 10^{-11} 	imes 6 	imes 10^{24}}} \approx 7133 \, s$.$T_{B} - T_{A} = 7133 - 5800 = 1333 \, s = 1.33 	imes 10^{3} \, s$.

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