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Radius of orbit of satellite of earth is $R$. Its kinetic energy is proportional to
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Verified Answer
The correct answer is:
$\frac{1}{R}$
Kinetic energy of the satellite
$\mathrm{KE}=\frac{1}{2} m v_0^2$ ...(1)
where $v_0=\sqrt{\frac{G M}{R}}$
Now putting the value of $v_0$ is eq. (1), we get
$\mathrm{KE}=\frac{1}{2} m\left(\sqrt{\frac{G M}{R}}\right)^2=\frac{1}{2} \frac{m G M}{R}$
Hence, $\mathrm{KE} \propto \frac{1}{R}$
$\mathrm{KE}=\frac{1}{2} m v_0^2$ ...(1)
where $v_0=\sqrt{\frac{G M}{R}}$
Now putting the value of $v_0$ is eq. (1), we get
$\mathrm{KE}=\frac{1}{2} m\left(\sqrt{\frac{G M}{R}}\right)^2=\frac{1}{2} \frac{m G M}{R}$
Hence, $\mathrm{KE} \propto \frac{1}{R}$
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