Calculating Work Done by Gravity on a Satellite

The question asks us to calculate the work done by the force of gravity on a satellite moving in an orbit around the Earth. This is a common concept in physics, particularly in the study of mechanics and orbital motion.

Work done by a force is defined as the product of the force, the displacement, and the cosine of the angle between the force and displacement vectors. Mathematically, it is given by:

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

where:

• \( W \) is the work done.
• \( \vec{F} \) is the force vector.
• \( \vec{d} \) is the displacement vector.
• \( F \) is the magnitude of the force.
• \( d \) is the magnitude of the displacement.
• \( \theta \) is the angle between the force vector and the displacement vector.

Gravity Force and Satellite Motion in Orbit

For a satellite orbiting the Earth, the force of gravity always acts towards the center of the Earth. This force is the centripetal force that keeps the satellite in orbit.

In a circular orbit, the satellite moves with a constant speed along a circular path. At any point in its orbit, the instantaneous velocity of the satellite is tangential to the circular path. The direction of the instantaneous displacement is the same as the direction of the instantaneous velocity.

Therefore, for a satellite in a circular orbit:

• The force of gravity is directed radially inward (towards the center of the Earth).
• The instantaneous displacement is tangential to the orbit.

The angle between a radial line and a tangent line to a circle at the same point is always 90 degrees.

So, the angle \( \theta \) between the force of gravity and the instantaneous displacement of the satellite is \( 90^\circ \).

Applying the Work Formula

Now, we can substitute this angle into the work done formula:

\( W = Fd \cos(\theta) \)

\( W = Fd \cos(90^\circ) \)

Since \( \cos(90^\circ) = 0 \), the formula becomes:

\( W = Fd \times 0 \)

\( W = 0 \)

Thus, the work done by the force of gravity on a satellite moving in a circular orbit is zero. The radius of the orbit (40,000 km in this case) is given, but it does not affect the fact that the work done is zero in a circular orbit.

This is a fundamental concept: a force that is always perpendicular to the displacement does no work.

Summary of Calculation

Let's summarise the key points:

• Force of gravity direction: Towards the Earth's center (radial).
• Satellite displacement direction (instantaneous): Tangential to the orbit.
• Angle between force and displacement: 90 degrees.
• Work done \( W = Fd \cos(90^\circ) = Fd \times 0 = 0 \).

Therefore, the work done by the force of gravity when a satellite moves in an orbit of radius 40,000 km around the earth is 0 J.

Revision Table: Work Done by Gravity

Additional Information: Gravity and Energy in Orbit

Gravity is a conservative force. This means that the work done by gravity depends only on the initial and final positions, not on the path taken. However, the calculation here is for instantaneous work or work over a complete cycle in a steady orbit.

For a satellite in a circular orbit, because gravity does no work, the mechanical energy (sum of kinetic and potential energy) of the satellite remains constant. The speed of the satellite is constant in a circular orbit, so its kinetic energy (\( KE = \frac{1}{2}mv^2 \)) is constant. The potential energy (\( PE = -\frac{GMm}{r} \)) is also constant because the radius \( r \) is constant.

If a satellite were in an elliptical orbit, the force of gravity would not always be perpendicular to the displacement. Gravity would do negative work as the satellite moves away from Earth (slowing it down) and positive work as it moves towards Earth (speeding it up). However, over one complete elliptical orbit, the total work done by gravity would still be zero because gravity is a conservative force and the satellite returns to its starting position.

Understanding the relationship between forces, displacement, and work is crucial for solving problems involving motion and energy.