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The time period of revolution of a satellite ( T ) around the earth depends on the radius of the circular…
The time period of revolution of a satellite ( T ) around the earth depends on the radius of the circular orbit ( R ), mass of the earth (M) and universal gravitational constant (G). The expression for T , using dimensional analysis is ( K is constant of proportionality)
$\mathrm{K} \sqrt{\frac{\mathrm{R}^2}{\mathrm{GM}}}$ $K \sqrt{\frac{R}{G M}}$ $\mathrm{K} \sqrt{\frac{\mathrm{R}^3}{\mathrm{GM}}}$ $\mathrm{K} \sqrt{\frac{\mathrm{R}^3}{\mathrm{GM}^2}}$ Solution
$\begin{aligned} & \text { } \mathrm{T} \propto \mathrm{R}^{\mathrm{a}} \mathrm{M}^{\mathrm{b}} \mathrm{G}^{\mathrm{c}} \Rightarrow \mathrm{T}=\mathrm{k} \mathrm{R}^{\mathrm{a}} \mathrm{M}^{\mathrm{b}} \mathrm{G}^{\mathrm{c}} \\ & \Rightarrow[\mathrm{T}]=[\mathrm{L}]^{\mathrm{a}}[\mathrm{M}]^{\mathrm{b}}\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^{\mathrm{c}} \\ & \therefore \mathrm{a}+3 \mathrm{c}=0 \\ & \mathrm{~b}-\mathrm{c}=0 \Rightarrow \mathrm{~b}=\mathrm{c} \\ & -2 \mathrm{c}=1 \Rightarrow \mathrm{c}=-\frac{1}{2} \\ & \therefore \quad \mathrm{a}+3\left(-\frac{1}{2}\right)=0 \Rightarrow \mathrm{a}=\frac{3}{2} \\ & \therefore \quad \mathrm{~T}=\mathrm{KR}^{\frac{3}{2}} \mathrm{M}^{-\frac{1}{2}} \mathrm{G}^{-\frac{1}{2}}=\mathrm{K} \sqrt{\frac{\mathrm{R}^3}{\mathrm{GM}}}\end{aligned}$
Asked in: AP EAMCET 2024 (21 May Shift 1)
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