Question

Consider a 0.300 -kg apple (a) attached to a tree and (b) falling. Does the apple exert a gravitational force on the Earth? If so, what is the magnitude of this force?

Short answer

Yes, the 0.300-kg apple exerts a gravitational force on the Earth. By applying Newton's Third Law of Motion and the gravitational force formula, the magnitude of this force is approximately 2.94 N. This force is the same whether the apple is attached to the tree or falling.

Step-by-step solution

State the known quantities and their units.

We know that the apple has a mass (m) of 0.300 kg and the Earth's mass (M) is approximately 5.97 x 10^24 kg. Also, we know the gravitational constant (G) is 6.674 x 10^-11 N m²/kg².

Apply Newton's Third Law of Motion.

Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. This means that if the apple is exerting a gravitational force on the Earth, the Earth must also be exerting the same force on the apple but in the opposite direction.

Apply the gravitational force formula.

To find the magnitude of the gravitational force exerted by the apple on Earth, we can use the following formula: F = G * (m * M) / r² You'll notice that we're missing the value for the distance between the apple and Earth (r). However, in this case, since the apple's mass is so small and its distance from the Earth is so insignificant, the force exerted would be the same whether the apple was attached to the tree or falling.

Calculate the gravitational force.

Now plug the known values into the formula and solve for F: F = (6.674 x 10^-11 N m²/kg²) * (0.300 kg * 5.97 x 10^24 kg) / (6,371,000 m)² F ≈ 2.94 N

Write the conclusion.

Yes, the 0.300-kg apple exerts a gravitational force on the Earth, and the magnitude of this force is approximately 2.94 N. Note that this force is the same whether the apple is attached to the tree or falling due to Newton's Third Law of Motion.