Express the constant k of T^{2}=k(R_{E}+h)^{3 | vedclass

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(C) Given $k = 10^{-13} \; s^{2} \; m^{-3}$.To convert $k$ into $d^{2} \; km^{-3}$:$1 \; s = \frac{1}{24 	imes 60 	imes 60} \; d = \frac{1}{86400} \

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$27.3$

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(C) Given $k = 10^{-13} \; s^{2} \; m^{-3}$.To convert $k$ into $d^{2} \; km^{-3}$:$1 \; s = \frac{1}{24 	imes 60 	imes 60} \; d = \frac{1}{86400} \; d$$1 \; m = 10^{-3} \; km$Substituting these values into $k$:$k = 10^{-13} 	imes (\frac{1}{86400})^{2} \; d^{2} 	imes (10^{-3})^{-3} \; km^{-3}$$k = 10^{-13} 	imes \frac{1}{7.465 	imes 10^{9}} 	imes 10^{9} \; d^{2} \; km^{-3}$$k \approx 1.33 	imes 10^{-14} \; d^{2} \; km^{-3}$Now,using $T^{2} = k(R_{E}+h)^{3}$ where $R_{E}+h = 3.84 	imes 10^{5} \; km$:$T^{2} = (1.33 	imes 10^{-14}) 	imes (3.84 	imes 10^{5})^{3}$$T^{2} = (1.33 	imes 10^{-14}) 	imes (56.623 	imes 10^{15})$$T^{2} \approx 753.08$$T = \sqrt{753.08} \approx 27.4 \; d$.The closest option is $27.3 \; d$.

Express the constant $k$ of $T^{2}=k(R_{E}+h)^{3}$ in days and kilometres. Given $k = 10^{-13} \; s^{2} \; m^{-3}$. The moon is at a distance of $3.84 \times 10^{5} \; km$ from the earth. Obtain its time period of revolution in days.

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