Question

The minimum energy required to launch a satellite of mass \(m\) from the surface of a planet of mass \(M\) and radius \(R\) in a circular orbit at an altitude of \(3R\) is \(\{G=\text{gravitational constant}\}\)

• A. \(\frac{3GMm}{8R}\)
• B. \(\frac{5GMm}{4R}\)
• C. \(\frac{3GMm}{4R}\)
• D. \(\frac{7GMm}{8R}\)

Hint:

The required energy is the total energy in the orbit minus the total energy at rest on the surface.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The minimum energy needed equals the final mechanical energy in orbit minus the initial mechanical energy on the surface.

Step 2: Initial energy
At rest on the surface, \(E_{i}=-\frac{GMm}{R}\).

Step 3: Final energy
The orbital radius is \(r=R+3R=4R\). In a circular orbit \(E_{f}=-\frac{GMm}{2r}=-\frac{GMm}{8R}\).

Step 4: Calculation
\(\Delta E=E_{f}-E_{i}=-\frac{GMm}{8R}+\frac{GMm}{R}=\frac{7GMm}{8R}\), which is option (D).

Final Answer:
The minimum energy required is \(\frac{7GMm}{8R}\). \[ \boxed{\text{(D) }\dfrac{7GMm}{8R}} \]