### Unit 3 Breakdown

You are on Lesson 1 of 4

**Unit 3.1 | Understanding circular motion and centripetal forces****Unit 3.2 | Solving circular motion problems using FBDs****Unit 3.3 | The gravitational force**(Current Lesson)**Unit 3.4 | Combining circular motion and gravitation (satellites, orbits, and more)**

#### In this lesson:

- We will learn about gravitational force
- Where the gravitational force applies in common scenarios
- The relationship between centripetal and gravitational forces and solving problems that involve both

### The force between any two object

It turns out there’s an *attractive* force that exists between *any *two or more masses.

For example, you and your computer’s screen; The desk and your notebook; You and the earth; The Earth and the Sun.

Any two masses will share an equally attractive force between them.

This is the **gravitational force**.

### Gravitational force equation

[katex] F_g = \frac{Gm_1m_2}{r^2} [/katex]

In this equation:

- [katex] G [/katex] is the gravitational constant [katex] 6.69 \times 10^{-11} [/katex]
**NOT**to be confused with the gravitational field [katex] 9.81 \, m/s^2[/katex]. - [katex] m_1 [/katex] and [katex] m_2 [/katex] are the masses of each object in no particular order
- [katex] r [/katex] is the distance between the
**centers**of each mass

### A force you already know

Consider a person and the earth. These two masses are pulled together by the gravitational force.

And we already have a name for this attraction: the force due to gravity ([katex] F_g [/katex]. We also call it weight ([katex] mg [/katex]).

So why do we have two names for the same force?

It’s because they are not used in the same cases. You can only use [katex] mg [/katex] if you know the value of the [katex]g[/katex] the gravitational field.

On the surface of earth, [katex] g [/katex] is [katex] 9.81 \, m/s^2[/katex].

But what if you are **not** on the surface? What happens to the value of gravity, [katex]g[/katex] if you’re in the international space station (ISS)?

#### Diminishing Gravitational Fields

As you move further away from earth, the magnitude of the gravitational feild decreases.

A satellite in orbit for example, experiences a much lesser gravitational field compared to when at the surface ([katex] 9.81 \, m/s^2[/katex].

This means that the satellite weighs more on the surface than it does in orbit!

To prove this we can calculate weight of the satellite using [katex]mg[/katex]. But **only** if we know the value of [katex]g[/katex].

So instead we use the [katex]F_g[/katex] (gravitation) formula. Plug the mass of the satellite, the mass of the earth, and the distance between the two to find the weight of the person.

Bonus Question: What happens to your weight if you move extremely far away from earth?

Answer: As you move far away from earth, the gravitational field acting on you would decrease to 0. This means you would weight nothing as [katex] g = 0 [/katex].

### Calculating gravitational fields

To find the gravitational forces we simply use the gravitation formula, but can we also calculate the value for [katex]g[/katex] as well?

It turns out we can. Here’s a quick derivation:

- As discussed, weight is equal to the gravitational force. Thus, we can set the two equal to each other. Let [katex] m_p[/katex] be the mass of the person and [katex] m_E[/katex] be the mass of the earth (or any other planet)
- [katex] m_pg = G\frac{m_pm_E}{r^2} [/katex]

- In this step we can cancel out [katex] m_p[/katex] from both sides to get our final equation for [katex]g[/katex].
- [katex] g = G\frac{m_E}{r^2} [/katex]

You can use this equation to find the magnitude of the gravitational field any distance anywhere from the center of the earth.

Question: What value for g would you get if you plugged in the radius of earth?

Answer: [katex] 9.81 \, m/s^2[/katex]!

To recap, you can find the gravitational field any distance away from any mass (not just the earth) using this equation:

[katex] g = G\frac{m_E}{r^2} [/katex]

### PS: Finding the equilibrium distance between two masses

In this video we’ll solve the following question:

*D*away from Earth, such that the spacecraft experiences zero net force. The distance between the Moon and Earth is ~3.844 x 10

^{8}m. NOTE: You may need the mass of the earth and moon. You can find this in the formula table.

Locked Video Content

### You Try – 4 practice questions

Find the net gravitational force on a 2.0 kg sphere midway between a 4.0 kg sphere and a 7.0 kg sphere that are 1.2 m apart.

The exoplanet HD 69830b has a mass 10 times that of the Earth and a radius 5 times that of the Earth. The value of *g* on HD 69830b is most nearly

2 m/s

^{2}4 m/s

^{2}8 m/s

^{2}10 m/s

^{2}15 m/s

^{2}

The magnitude of the gravitational field on the surface of a new planet is 20 N/kg. The planet's mass half the mass of Earth. The radius of Earth is 6400 km. What is the radius of the new planet?

^{6}m from the center of a larger object whose mass is 7.4 x 10

^{22}kg.

### Preview Lesson 3.4

The last lesson of this Unit will make the connection between gravitational and centripetal forces. After this you’ll be able to solve every type of circular motion problem.