Unit 2.C | The Gravitational Force

Earth rising over the cratered horizon of the Moon, seen from lunar orbit
  • 16 minutes
  • 3 formulas
  • 11 questions
  • 1 videos

This lessons will cover gravitation — another linear force responsible for pulling two masses together.

Unit 2.A–2.D Breakdown

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 objects

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] 6.67 \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
Each mass pulls on the other with the same size force \(F_g\), in opposite directions. \(r\) runs from center to center.

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 field 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!

Earth's gravitational field vs. distance

Double the distance, a quarter of the field: \(g\) falls off as \(1/r^2\).

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 satellite.

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 weigh 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:

  1. 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]
  2. 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

Try the following question:

Open in UBQ
Question 1
A spacecraft somewhere in between the Earth and the Moon experiences zero net force acting on it. This is because the Earth and the Moon pull the…
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You Try – 4 practice questions

Preview Lesson 2.D

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.

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This article was written by Jason