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Question:
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Match the following columns.
Codes
$\begin{array}{llll}A & B & C & D\end{array}$
Options:
Codes
$\begin{array}{llll}A & B & C & D\end{array}$
Solution:
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Verified Answer
The correct answer is:
None of these
The potential energy of satellite at distance $r$ from earth is
$$
U=\frac{-G M m}{r}
$$
So, it is always negative as it is due to attractive gravitation force of earth's field. The kinetic energy of satellite is
$$
K=\frac{G M m}{2 r}
$$
It is always positive.
The total energy of satellite is
$$
\begin{aligned}
\mathrm{TE} & =K+U=\frac{G M m}{2 r}-\frac{G M m}{r} \\
& =-\frac{G M m}{2 r}
\end{aligned}
$$
which is negative.
The gravitation potential energy of satellite at infinity is
$$
U=-\frac{G M m}{r}=-\frac{G M m}{\infty}=0
$$
So, the correct match is $\mathrm{A} \rightarrow \mathrm{II}, \mathrm{B} \rightarrow \mathrm{II}, \mathrm{C} \rightarrow \mathrm{I}, \mathrm{D} \rightarrow$ III.
No option is correct.
$$
U=\frac{-G M m}{r}
$$
So, it is always negative as it is due to attractive gravitation force of earth's field. The kinetic energy of satellite is
$$
K=\frac{G M m}{2 r}
$$
It is always positive.
The total energy of satellite is
$$
\begin{aligned}
\mathrm{TE} & =K+U=\frac{G M m}{2 r}-\frac{G M m}{r} \\
& =-\frac{G M m}{2 r}
\end{aligned}
$$
which is negative.
The gravitation potential energy of satellite at infinity is
$$
U=-\frac{G M m}{r}=-\frac{G M m}{\infty}=0
$$
So, the correct match is $\mathrm{A} \rightarrow \mathrm{II}, \mathrm{B} \rightarrow \mathrm{II}, \mathrm{C} \rightarrow \mathrm{I}, \mathrm{D} \rightarrow$ III.
No option is correct.
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