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### Course: High school physics - NGSS > Unit 2

Lesson 1: Newton's law of universal gravitation# Newton’s law of universal gravitation

Review your understanding of Newton's law of universal gravitation in this free article aligned to NGSS standards.

## Key terms

Term | Meaning |
---|---|

Gravitational force ( | Attractive force between two objects with mass. |

Gravitational field ( | A model explaining the influence an object extends to produce a force on other objects. |

Gravitational mass ( | The property of matter that causes it to experience a force in a gravitational field. Two objects that balance each other on a scale have the same gravitational mass. Gravitational mass is experimentally equivalent to inertial mass, and has SI units of |

## Equations

Equation | Symbol breakdown | Meaning |
---|---|---|

The gravitational force between point-like mass | ||

The gravitational field strength is directly proportional to mass creating the field and inversely proportional to the square of the distance. |

## Newton’s law of universal gravitation

Gravitational force ${F}_{g}$ is always attractive, and it depends only on the masses involved and the distance between them. Every object in the universe attracts every other object with a force along an imaginary line between them.

The equation for Newton’s law of gravitation is:

where:

The force is

*directly*proportional to the product of the masses. It is also*inversely*proportional to the square of the distance between the centers of mass. This is sometimes called an inverse-square law.For example, if we double the distance between Earth and the Moon, the attractive force between them would decrease (because it has an inverse relationship to distance), and it would go down by a factor of $4$ instead of $2$ (because $r$ is squared).

Newton’s law of universal gravitation describes objects falling down as well as objects in a circular orbit, such as a satellite orbiting Earth.

## How to find gravitational field strength

All objects attract other objects by producing a gravitational field $g$ , which is defined as gravitational force per unit mass. We can find the strength of the gravitational field of mass ${m}_{1}$ on any object with mass ${m}_{2}$ by dividing the above equation by ${m}_{2}$ , and simplifying.

## What else should I know about Newton’s law of universal gravitation?

**Gravity causes attraction between all objects.**Every mass attracts every other mass. That means you are gravitationally attracted to your friend, your pet, and even your pizza.**The variable**We measure the distance between the objects from their centers, not their surfaces. is the distance between the centers of mass.$r$

## Want to join the conversation?

- Why does gravitational potential energy increase as the masses get farther apart?(22 votes)
- Because you are doing work against gravity. Moving the object up parallel in the direction the gravity is acting.

Work = Force * displacement parallel to the force

Work done against gravity = weight * Δheight

= (mass * gravity) * Δheight

Ug = Gravitational potential energy

ΔEnergy = work done

ΔUg = work done against gravity = (mass * gravity) * Δheight

Ug = (mass * gravity) * height

In this case energy is proportional to height. As height increases energy increases too.

//Don't know how to get rid of the delta sign but should make sense intuitively if you think of the initial height as 0 and final height as h(2 votes)

- I'm confused about calculating all numbers like this "6.67 x 10^-11 m^3/kg*s^2". I know that is gravitational constant but how to solve that kind of number?(10 votes)
- Remember that 10² is 100. 10³ is 1,000. So 10^10 is a one followed by ten zeros, or ten billion. 10^-10 is going backwards: a decimal point followed by nine zeros and then a one. Whatever this is then multiplies the number in front, and at the end are just a bunch of units. For example, 2.345*10^-6 m²/kgs is the same thing as 0.000001*2.345 meters squared / kilogram seconds, which is 0.000002345 of the same thing.(9 votes)

- Wait, so, I'm kind of confused. If g = G*(m1/r^2), then does that mean that the mass of m2 does not affect the gravitational effect that m1 has on it?(2 votes)
- It does, but it is mostly not obvious. Everything has a gravitational pull, but it is not noticeable except for some exceptions; your dog, for instance, won't fly up to you the moment you get 2 feet close to it, but the Moon circles the Earth because of the gravitational pull. To put it simply, the mass of the objects still have a pull on each other, but many times it is small, so it can be ignored, except if you want to be super specific or have huge objects, like Jupiter's moons.

Hope this helps!(4 votes)

- Is the "little g" just a gravitational field? And is it specifically Earth's gravitational field?(2 votes)
- Little g is the numerical value of Earth's gravitational field's pull at the Earth's surface.

Another planet would have a different value for its own gravitational pull.(4 votes)

- How do I calculate after I input the numbers in the formula? What does N stand for?(2 votes)
- N stands for Newtons, which is a unit of force. Could you clarify what you mean about inputting the numbers in the formula?(1 vote)

- How would you calculate the distance between an object and an object with a gravitational pull?(2 votes)
- Will
`G`

always be 6.67 x 10^-11 m^3/kg*s^2?(1 vote)- Yes. By definition, G is a constant.(3 votes)

- what are the SI units for 'r'?(1 vote)
- The SI units for the radius are metres.(1 vote)

- The gravitational force between a 105kg person and the 5.97*10^24 Earth is 1030N.

Calculate the distance from the person to the center of Earth.

Will someone please help me?(1 vote)- okay, so...

to answer your question there are 3 equations.

1) for getting the a mass: M2 = (Fg * r^2)/(G * M1)

"note, there is a more complicated one if all the masses is unknown, that ill give you if you want it, just ask!"

2) for getting the Force: Fg = G * (M1 * M2)/(r)^2

3) for getting the distance: r^2 = G * (M1 * M2)/(Fg)

you will be using "3)" for calculating the distance

................(M1 * M2)

r^2 = G -------------

.....................(Fg)

Now you can just fill in the information you have...

....................................(105)(5.97*10^24)

r^2 = 6.67*10^-11* -----------------------

.............................................(1030)

"note, you don't have to put in the (N, kg, m) and all that units, you can leave that"

then you can start doing the problem...

.........................................6.2685*10^26

= r^2 = 6.67*10^-11* -------------------

................................................1030

= r^2 = 6.67*10^-11* 6.08592233*10^23

= r^2 = 4.059310194*10^13

(to get the "^2" away from the left hand side of the equation to make it only "r =" you put the right hand side in a square root"

.............*_____________________*

= r = \|4.059310194*10^13

= r = 6371271.611 "you just have to round your answer on how they want you to round it!"

and there you have your answer!!

hopes this helps.!(1 vote)

- 6.67

\times 10^{-11} \dfrac{\text{m}^3}{\text{kg} \cdot \text{s}^2}

how do you plug this into calculators if

A: mass, kilograms, and seconds are not a function

B: this isn't really a number(1 vote)- it's too big of a number to be answered with a calculator.(1 vote)