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(D) The time period of a satellite is given by $T = 2\pi \sqrt{\frac{r^3}{GM}}$.From this,we have $T \propto \sqrt{\frac{r^3}{M}}$,which implies $\fra

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$1.05 	imes 10^4 	ext{ km}$

Vedclass · Answer

(D) The time period of a satellite is given by $T = 2\pi \sqrt{\frac{r^3}{GM}}$.From this,we have $T \propto \sqrt{\frac{r^3}{M}}$,which implies $\frac{T_1}{T_2} = \left(\frac{r_1}{r_2}ight)^{3/2} \left(\frac{M_2}{M_1}ight)^{1/2}$.Here,$T_1 = 6 	ext{ hours}$,$T_2 = 24 	ext{ hours}$ (for Earth's geostationary orbit).$M_1 = \frac{M_e}{4}$ and $M_2 = M_e$.$r_2 = 4.2 	imes 10^4 	ext{ km}$.Substituting the values: $\frac{6}{24} = \left(\frac{r_1}{4.2 	imes 10^4}ight)^{3/2} \left(\frac{M_e}{M_e/4}ight)^{1/2}$.$\frac{1}{4} = \left(\frac{r_1}{4.2 	imes 10^4}ight)^{3/2} 	imes (4)^{1/2}$.$\frac{1}{4} = \left(\frac{r_1}{4.2 	imes 10^4}ight)^{3/2} 	imes 2$.$\frac{1}{8} = \left(\frac{r_1}{4.2 	imes 10^4}ight)^{3/2}$.Taking the power of $2/3$ on both sides: $\left(\frac{1}{8}ight)^{2/3} = \frac{r_1}{4.2 	imes 10^4}$.$\frac{1}{4} = \frac{r_1}{4.2 	imes 10^4}$.$r_1 = \frac{4.2 	imes 10^4}{4} = 1.05 	imes 10^4 	ext{ km}$.

$A$ satellite revolving around a planet in a stationary orbit has a time period of $6 \text{ hours}$. The mass of the planet is one-fourth the mass of the Earth. What is the radius of the orbit of the satellite? (Given: Radius of the geostationary orbit for Earth is $4.2 \times 10^4 \text{ km}$)

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