Using the data given below,find the height at | vedclass

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(A) Given: $G=6.67 	imes 10^{-11} 	ext{ N-m}^2 	ext{ kg}^{-2}$,$M=5.98 	imes 10^{24} 	ext{ kg}$,$R=6.4 	imes 10^6 	ext{ m}$.For a communication

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$35850$

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(A) Given: $G=6.67 	imes 10^{-11} 	ext{ N-m}^2 	ext{ kg}^{-2}$,$M=5.98 	imes 10^{24} 	ext{ kg}$,$R=6.4 	imes 10^6 	ext{ m}$.For a communication satellite,the time period $T = 24 	ext{ h} = 24 	imes 3600 	ext{ s} = 8.64 	imes 10^4 	ext{ s}$.The orbital radius $r = R+h$ is related to the time period by Kepler's Third Law: $T^2 = \frac{4 \pi^2 r^3}{GM}$.Rearranging for $r$: $r = \left( \frac{T^2 GM}{4 \pi^2} ight)^{1/3}$.Substituting the values:$r = \left( \frac{(8.64 	imes 10^4)^2 	imes 6.67 	imes 10^{-11} 	imes 5.98 	imes 10^{24}}{4 	imes (3.14)^2} ight)^{1/3}$.$r \approx 42.25 	imes 10^6 	ext{ m}$.Since $h = r - R$,we have $h = 42.25 	imes 10^6 	ext{ m} - 6.4 	imes 10^6 	ext{ m} = 35.85 	imes 10^6 	ext{ m}$.Converting to kilometers: $h = 35850 	ext{ km}$.

Using the data given below,find the height at which a communication satellite can reside. $(G=6.67 \times 10^{-11} \text{ N-m}^2 \text{ kg}^{-2}, M=5.98 \times 10^{24} \text{ kg}, R=6.4 \times 10^6 \text{ m})$ (in $\text{ km}$)

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