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- Thread starter BLUE PEARL
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In summary, the net force on each star of equal mass M, moving with constant speed V in a circular orbital of radius R about their common center of mass, is given by the formula net force=GM^2/4R^2. This net force is not equal to zero because, despite being in translational equilibrium, the stars are still subject to a centripetal force that keeps them in their circular orbit. This is due to Newton's first and second laws of motion.

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BLUE PEARL said:

Any help in this regard would be appreciated...Thank You in anticipation

If something is in a circular orbit, how can the force on it be 0? Newton's first/second laws apply.

Net force is the overall force acting on an object, taking into account the magnitude and direction of all individual forces.

When net force is equal to 0, it means that all individual forces acting on an object are balanced and there is no acceleration in any direction.

Net force is calculated by adding together all the individual forces acting on an object, taking into account their direction and magnitude.

This equation shows the relationship between net force and the gravitational force between two objects, where G is the universal gravitational constant, M is the mass of one object, and R is the distance between the two objects.

Yes, the net force can be greater than GM^2/4R^2 if there are other forces acting on the object in addition to the gravitational force, such as friction or air resistance.

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