1. When you have different masses for each sphere, how does the force that the larger mass sphere exerts on the smaller mass sphere exerts on the larger mass sphere?
2. What happens to the force between the two spheres of the same mass when the distance between the mass sphere is halved?
3. The force of attraction between 2 objects is called gravity. What do you think about an object, like your notebook, pull towards you? Explain your reasoning.

In this part of the problem, we want to compare the gravitational force exerted by the larger mass sphere on the smaller mass sphere to the force exerted by the smaller mass sphere to the larger mass sphere.

We can do this by using Newton's third law of motion, which states that:

"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"

In this problem, we can identify the larger mass sphere as object A and the smaller mass sphere as object B. This law tells us that the two forces are equal in magnitude and opposite in direction: therefore, the gravitational force exerted by the larger mass sphere on the smaller mass sphere is equal to the force exerted by the smaller mass sphere to the larger mass sphere.

2)

The magnitude of the gravitational force between the two spheres is given by

[tex]F=G\frac{m_1 m_2}{r^2}[/tex]

where

G is the gravitational constant

[tex]m_1, m_2[/tex] are the masses of the two spheres

r is the separation between the two spheres

In this problem, we are asked to find what happens when the distance between the spheres is halved, therefore when the new distance is

[tex]r'=\frac{r}{2}[/tex]

Substituting into the equation, we find

[tex]F'=G\frac{m_1 m_2}{r'^2}=G\frac{m_1 m_2}{(r/2)^2}=4(\frac{Gm_1 m_2}{r^2})=4F[/tex]

So, the force between the two spheres will quadruple.

3)

We can give an estimate for the gravitational force exerted by your notebook on you.

As we said, the magnitude of the gravitational force is

[tex]F=G\frac{m_1 m_2}{r^2}[/tex]

Where:

[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}[/tex] is the gravitational constant

Let's estimate the following:

[tex]m_1 = 60 kg[/tex] is your mass

[tex]m_2 = 2 kg[/tex] is the mass of the notebook

[tex]r=1 m[/tex], assuming the notebook is at 1 metre from you

Substituting,

[tex]F=(6.67\cdot 10^{-11})\frac{(60)(2)}{1^2}=8.0\cdot 10^{-9} N[/tex]

We see that this force has an extremely small value: therefore, it is almost negligible in daily life, where other much stronger forces act on you.

Learn more about gravity:

brainly.com/question/1724648

brainly.com/question/12785992

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