The work done in one complete revolution of the moon around the earth is equal to:

Question

Prepp Team · Accepted Answer

Understanding Work Done in Orbital Motion

The question asks about the work done during one complete revolution of the moon around the earth. Work done by a force on an object is defined as the product of the force, the displacement of the object, and the cosine of the angle between the force and the displacement vector.

Mathematically, work (\(W\)) is given by:

\(W = \vec{F} \cdot \vec{d} = Fd \cos\theta\)

where:

\(F\) is the magnitude of the force
\(d\) is the magnitude of the displacement
\(\theta\) is the angle between the force vector \(\vec{F}\) and the displacement vector \(\vec{d}\)

Analyzing the Forces and Motion of the Moon

The moon orbits the earth primarily due to the gravitational force exerted by the Earth. This gravitational force acts as the centripetal force, keeping the moon in its nearly circular orbit.

The gravitational force vector is always directed towards the center of the Earth.
The moon moves along the orbital path, which is approximately a circle.
The instantaneous displacement of the moon at any point in its orbit is tangential to the orbit, and thus perpendicular to the radius vector connecting the moon to the Earth's center.

Calculating Work Done by Gravitational Force

At any instant, the gravitational force acting on the moon is directed radially inwards towards the Earth, while the moon's instantaneous displacement is tangential to the orbit. This means the angle (\(\theta\)) between the gravitational force vector and the instantaneous displacement vector is \(90^{\circ}\).

Using the work formula \(W = Fd \cos\theta\), the work done by the gravitational force over a very small displacement \(dd\) is:

\(dW = F \cdot dd \cdot \cos(90^{\circ})\)

Since \(\cos(90^{\circ}) = 0\), the work done over this small displacement is:

\(dW = F \cdot dd \cdot 0 = 0\)

For one complete revolution, the total work done is the sum of the work done over all the small displacements along the path. Since the work done over each small displacement is zero, the total work done over the entire revolution is also zero.

This result is consistent with the property of conservative forces. The gravitational force is a conservative force, and the work done by a conservative force over a closed path (like a complete orbit) is always zero.

Evaluating the Options

Let's look at the given options based on our analysis:

Option 1: Zero - This aligns with our calculation that the work done by the gravitational force during one complete revolution is zero because the force is always perpendicular to the displacement.
Option 2: Gravitational force × diameter of the orbit of the moon - This is incorrect. Work involves the component of force in the direction of displacement. Simply multiplying force by diameter (or any distance) without considering the angle is wrong for this scenario.
Option 3: Centripetal force × radius of the orbit of the moon - The centripetal force here is the gravitational force. Again, work is not simply force times radius. The force is perpendicular to the instantaneous displacement, resulting in zero work.
Option 4: Gravitational force × circumference of the orbit of the moon - Similar to option 2, this product \(F \times d\) would represent work only if the force was constant and acted along the circumference, which is not the case here. The force is radial, not tangential.

Therefore, the work done in one complete revolution of the moon around the earth by the gravitational force is zero.

Concept	Description
Work Done	\(W = Fd \cos\theta\)
Gravitational Force Direction	Towards Earth's center (Radial)
Moon's Displacement Direction	Tangent to orbit (Tangential)
Angle (\(\theta\)) between Force and Displacement	\(90^{\circ}\)
\(\cos(90^{\circ})\)	0
Work Done Over Small Displacement	0
Total Work Done Over One Revolution	0
Nature of Gravitational Force	Conservative
Work by Conservative Force over Closed Path	Zero

Revision Table: Key Physics Concepts for Orbital Work

Term	Definition/Role	Relevance to Moon's Orbit
Work Done	Energy transferred by force over distance. \(W = Fd \cos\theta\)	Calculated based on force, displacement, and angle.
Gravitational Force	Attractive force between masses (Earth and Moon).	Provides the centripetal force for the orbit.
Centripetal Force	Force directed towards the center of a circular path.	In this case, it's gravity. Always perpendicular to velocity/displacement.
Displacement	Change in position. Instantaneous displacement is along the tangent.	Direction is crucial for calculating work via the dot product.
Angle (\(\theta\))	Angle between force and displacement vectors.	\(90^{\circ}\) for a central force in circular motion.
Conservative Force	Force for which work done is path-independent; work over a closed path is zero.	Gravitational force is a conservative force.
Closed Path	A path that starts and ends at the same point.	One complete revolution of the moon is a closed path.

Additional Information: Conservative vs. Non-Conservative Forces

Understanding conservative and non-conservative forces helps explain why work done over a closed path can be zero or non-zero.

Conservative Forces:
- Work done depends only on the initial and final positions, not the path taken.
- Work done over any closed path is zero.
- Examples: Gravitational force, electrostatic force, elastic spring force.
- Associated with potential energy (e.g., gravitational potential energy, elastic potential energy).
Non-Conservative Forces:
- Work done depends on the path taken.
- Work done over a closed path is generally non-zero.
- Examples: Friction, air resistance, viscous drag.
- These forces typically dissipate energy (e.g., as heat).

Since gravity is a conservative force and the moon's orbit is a closed path, the total work done by gravity over one orbit is zero. This does not mean there are no forces acting or no energy changes (kinetic energy might fluctuate slightly in an elliptical orbit, but potential energy changes balance it such that total mechanical energy is conserved if only gravity acts). However, the *work done by the gravitational force* itself is zero for a complete loop.