Question

For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is :

• A. 4
• B. \(\frac{1}{2}\)
• C. \(\frac{1}{\sqrt 2}\)
• D. \(\frac{1}{4}\)

Answer 1

The Correct Option is B

To solve this problem, we need to understand the relationship between the kinetic energy and potential energy of a satellite orbiting the Earth. A satellite in orbit around the Earth is under the influence of gravitational force, and its motion can be analyzed using classical mechanics. 

When a satellite orbits the Earth, its total energy is the sum of its kinetic energy (\(K\)) and potential energy (\(U\)). In circular orbits, it is known from physics that:

\(K = \frac{1}{2}mv^2\) and \(U = -\frac{GMm}{r}\)

where:

• \(m\) is the mass of the satellite,
• \(v\) is the velocity of the satellite,
• \(G\) is the gravitational constant,
• \(M\) is the mass of the Earth,
• \(r\) is the radius of the orbit.

For a satellite in a stable orbit, the relationship between the kinetic energy and potential energy is given by the virial theorem, which tells us:

\(K = -\frac{1}{2}U\)

This equation implies that the kinetic energy of the satellite is half the magnitude of its potential energy. Therefore, the ratio of kinetic energy to potential energy (\(K:U\)) is:

\(\frac{K}{U} = \frac{(-\frac{1}{2}U)}{U} = \frac{1}{2}\)

Thus, the correct answer is that the ratio of kinetic energy to potential energy for a satellite moving in an orbit around the Earth is \(\frac{1}{2}\).

Therefore, the correct option is:

\(\frac{1}{2}\)