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A satellite is orbiting the earth. It's orbital radius is reduced to half of its initial value then the change in its total energy is
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Verified Answer
The correct answer is:
$100 \%$
Total energy of satellite revolving around the earth,
$$
\begin{aligned}
& E=\frac{-G M_t m}{r} \\
& \text { i.e } \quad E \propto \frac{1}{r} \\
& \Rightarrow \quad \frac{E_2}{E_1}=\frac{r_1}{r_2} \\
& =\frac{r_1}{\frac{r_1}{2}} \quad\left(\therefore \text { Given } r_2=\frac{r_1}{2}\right) \\
& =2 \\
& E_2=2 E_1 \\
&
\end{aligned}
$$
$\therefore \%$ change in total energy of satellite
$$
\begin{aligned}
& =\frac{E_2-E_1}{E_1} \times 100 \\
& =\frac{2 E_1-E_1}{E_1} \times 100=100 \%
\end{aligned}
$$
$$
\begin{aligned}
& E=\frac{-G M_t m}{r} \\
& \text { i.e } \quad E \propto \frac{1}{r} \\
& \Rightarrow \quad \frac{E_2}{E_1}=\frac{r_1}{r_2} \\
& =\frac{r_1}{\frac{r_1}{2}} \quad\left(\therefore \text { Given } r_2=\frac{r_1}{2}\right) \\
& =2 \\
& E_2=2 E_1 \\
&
\end{aligned}
$$
$\therefore \%$ change in total energy of satellite
$$
\begin{aligned}
& =\frac{E_2-E_1}{E_1} \times 100 \\
& =\frac{2 E_1-E_1}{E_1} \times 100=100 \%
\end{aligned}
$$
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