Question:
A satellite is orbiting the earth. If its orbital radius is reduced to half of its initial value, then the change in its total energy is
A satellite is orbiting the earth. If its orbital radius is reduced to half of its initial value, then the change in its total energy is
Show Hint
The total energy of a satellite in circular orbit is inversely proportional to orbital radius:
\[
E\propto -\frac{1}{r}
\]
So, halving the radius doubles the magnitude of total energy.
Updated On: Jun 26, 2026
- \(100\%\)
- \(75\%\)
- \(50\%\)
- \(25\%\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Recall the formula for total energy of a satellite.
The total mechanical energy of a satellite in circular orbit is \[ E=-\frac{GMm}{2r} \] where \[ G=\text{gravitational constant}, \] \[ M=\text{mass of earth}, \] \[ m=\text{mass of satellite}, \] and \[ r=\text{orbital radius} \]
Step 2: Find the initial total energy.
Let the initial orbital radius be \[ r \] Then initial total energy is \[ E_i=-\frac{GMm}{2r} \]
Step 3: Find the final total energy.
The orbital radius is reduced to half: \[ r_f=\frac{r}{2} \] So, \[ E_f=-\frac{GMm}{2(r/2)} \] \[ E_f=-\frac{GMm}{r} \] Now compare with \[ E_i=-\frac{GMm}{2r} \] Thus, \[ E_f=2E_i \] Hence, the magnitude of total energy becomes double.
Step 4: Calculate percentage change.
Initial magnitude of energy: \[ |E_i| \] Final magnitude of energy: \[ |E_f|=2|E_i| \] Increase in magnitude: \[ 2|E_i|-|E_i|=|E_i| \] Therefore, \[ \text{Percentage change} = \frac{|E_i|}{|E_i|}\times 100 \] \[ =100\% \]
Step 5: Final conclusion.
Hence, the change in total energy is \[ \boxed{100\%} \]
The total mechanical energy of a satellite in circular orbit is \[ E=-\frac{GMm}{2r} \] where \[ G=\text{gravitational constant}, \] \[ M=\text{mass of earth}, \] \[ m=\text{mass of satellite}, \] and \[ r=\text{orbital radius} \]
Step 2: Find the initial total energy.
Let the initial orbital radius be \[ r \] Then initial total energy is \[ E_i=-\frac{GMm}{2r} \]
Step 3: Find the final total energy.
The orbital radius is reduced to half: \[ r_f=\frac{r}{2} \] So, \[ E_f=-\frac{GMm}{2(r/2)} \] \[ E_f=-\frac{GMm}{r} \] Now compare with \[ E_i=-\frac{GMm}{2r} \] Thus, \[ E_f=2E_i \] Hence, the magnitude of total energy becomes double.
Step 4: Calculate percentage change.
Initial magnitude of energy: \[ |E_i| \] Final magnitude of energy: \[ |E_f|=2|E_i| \] Increase in magnitude: \[ 2|E_i|-|E_i|=|E_i| \] Therefore, \[ \text{Percentage change} = \frac{|E_i|}{|E_i|}\times 100 \] \[ =100\% \]
Step 5: Final conclusion.
Hence, the change in total energy is \[ \boxed{100\%} \]
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