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A planet is moving at constant speed in a circular orbit around a star. In one complete orbit, what is the net amount of work done on the planet by the star’s gravitational force: positive, negative, or zero? What if the planet’s orbit is an ellipse, so that the speed is not constant? Explain your answers.

Short Answer

Expert verified

The net work done on the planet by the star’s gravitational force in the circular orbit is zero.

the net work done on the planet by the star’s gravitational force in the elliptical orbit is zero.

Step by step solution

01

Significance of the gravitational force

The attractive force between two objects by virtue of their mass is called gravitational force.

The concept of the gravitational force gives the net amount of work done.

02

Determination of the net amount of work done if the orbit is circular

As the gravitational force is regarded as a conservative force, then the work done is the difference in the potential energy of the two points.

Let the planet has moved in a conservative field of force from the point A to the point B in the path C, then the equation of the work done is expressed as:

W=CF.dr=UB-UA

Here, W is the work done, F is the force exerted, dr is the distance travelled, U(B) is the potential energy at point B and U(A) is the potential energy at point A.

As the orbit has been completed and as it is circular, then the planet will start and end both at the point A. Hence, the work done will be-

W=U(A)-U(A)=0

Thus, the net work done on the planet by the star’s gravitational force in the circular orbit is .

03

Determination of the work done if the orbit is ellipse

In case of elliptical orbit, as the planet is moving closer to the star, it increases its speed and hence the work done is positive. In the same way, as the planet moves away from the star, it reduces its speed and hence does equal amount of negative work. So net work done in one complete revolution is zero.

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