Question

A satellite revolves around the earth of radius \(R\) in a circular orbit of radius \(3R\). The percentage increase in energy required to lift it to an orbit of radius \(5R\) is

• A. 10%
• B. 20%
• C. 30%
• D. 40%
• E. 67%

Hint:

Satellite energy varies as \(-1/r\); always compare energies using magnitude of initial energy.

Answer 1

The Correct Option is D

Concept: The total mechanical energy of a satellite moving in a circular orbit is given by: \[ E = -\frac{GMm}{2r} \] This shows that energy is negative (bound system) and inversely proportional to orbital radius. When the satellite is lifted to a higher orbit, its total energy increases (becomes less negative), and the difference in energy must be supplied externally.

Step 1: Write energy in initial orbit.
Initial radius \(r_1 = 3R\) \[ E_1 = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R} \]

Step 2: Write energy in final orbit.
Final radius \(r_2 = 5R\) \[ E_2 = -\frac{GMm}{2(5R)} = -\frac{GMm}{10R} \]

Step 3: Calculate energy required.
Energy required is the increase in total energy: \[ \Delta E = E_2 - E_1 \] \[ = -\frac{GMm}{10R} - \left(-\frac{GMm}{6R}\right) = \frac{GMm}{6R} - \frac{GMm}{10R} \]

Step 4: Take LCM and simplify. \[ \Delta E = GMm \left(\frac{5 - 3}{30R}\right) = \frac{2GMm}{30R} = \frac{GMm}{15R} \]

Step 5: Find percentage increase relative to initial energy. \[ %\text{ increase} = \frac{\Delta E}{|E_1|} \times 100 = \frac{\frac{GMm}{15R}}{\frac{GMm}{6R}} \times 100 \]

Step 6: Simplify ratio. \[ = \frac{6}{15} \times 100 = 40% \]

Step 7: Conclusion. \[ \boxed{40%} \]