Answer (2)
Sol. $T^{2} \propto r^{3}$
$\mathbf{T}^{2} \propto\left(\mathbf{R}_{\mathrm{E}}+\mathbf{h}\right)^{3}$
$\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{\left(\mathrm{R}_{\mathrm{E}}+6 \mathrm{R}_{\mathrm{e}}\right)^{3}}{\left(\mathrm{R}_{\mathrm{E}}+2.5 \mathrm{R}_{\mathrm{E}}\right)^{3}}$
$\frac{\mathrm{T}_{1}^{2}}{\mathrm{~T}_{2}^{2}}=\frac{7^{3}}{\left(\frac{7}{2}\right)^{3}}$
$\mathrm{T}_{2}=\frac{\mathrm{T}_{1}}{2 \sqrt{2}}$
$T_{2}=\frac{24}{2 \sqrt{2}}$
$T_{2}=6 \sqrt{2} \mathrm{~h}$
The time period of a geostationary satellite is $24 \mathrm{~h}$, at a height $6 R_{E}\left(R_{E}\right.$ is radius of earth) from surface of earth. The time period of another satellite whose height is $2.5 \mathbf{R}_{E}$ from surface will be, (1) $\frac{12}{2.5} h$ (2) $6 \sqrt{2} \mathrm{~h}$ (3) $12 \sqrt{2} \mathrm{~h}$ (4) $\frac{24}{2.5} \mathrm{~h}$
The time period of a geostationary satellite is $24 \mathrm{~h}$, at a height $6 R_{E}\left(R_{E}\right.$ is radius of earth) from surface of earth. The time period of another satellite whose height is $2.5 \mathbf{R}_{E}$ from surface will be, (1) $\frac{12}{2.5} h$ (2) $6 \sqrt{2} \mathrm{~h}$ (3) $12 \sqrt{2} \mathrm{~h}$ (4) $\frac{24}{2.5} \mathrm{~h}$