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# Gravity on Earth versus on Moon

by Ron Kurtus (15 April 2020)

The force of * gravity on the Earth* is greater than the

*, because of the greater mass of the Earth. However, the Earth has a greater radius than the Moon, so that is a factor in the ratios of the forces of gravity.*

**gravity on the Moon**Calculation of the force of gravity of each starts with Newton's *Universal Gravitation Equation*. Then values of the mass and the radius of each are used to determine the acceleration and force.

Questions you may have include:

- What is the gravitation equation?
- What is the value of gravity at the Earth's surface?
- What is gravity on the Moon's surface?
- How do the values of gravity vary with mass and radius?

This lesson will answer those questions. Useful tool: Units Conversion

## Gravitation equation

The acceleration due to gravity constant comes from Newton's *Universal Gravitation Equation*, which shows the force of attraction between any two objects—typically astronomical objects:

F = GMm/R^{2}

where

**F**is the force of attraction, as measured in newtons (N) or kg-m/s^{2}**G**is the Universal Gravitational Constant: 6.674*10^{−11}m^{3}/s^{2}-kg**M**and**m**are the masses of the objects in kilograms (kg)**R**is the separation of the centers of the objects in meters (m)

(

See Universal Gravitation Equation for more information.)

If a much smaller object is near the surface of the Earth, the separation **R** becomes close to the radius of the Earth. Likewise for the separation on the Moon.

Since force equals mass times acceleration, the gravity equation becomes:

F = mg=GMm/R^{2}

Thus:

g = GM/R^{2}

This is the gravity constant or acceleration due to gravity.

(See Gravity Acceleration is Constant for more information.)

## Acceleration due to gravity on Earth

You can find the value of **g _{E}** by substituting the following items into the equation:

g_{E}= GM_{E}/R_{E}^{2}

where

**G**= 6.674*10^{−11}m^{3}/s^{2}-kg**M**= 5.974*10_{E}^{24}kg (mass of Earth)**R**= 6.371*10_{E}^{6}m (radius of Earth - distance from center to surface)**R**= 40.590*10_{E}^{2}^{12}m^{2}

Substituting in values:

g= (6.674*10_{E}^{−11}m^{3}/s^{2}-kg)(5.974*10^{24}kg)/(6.371*10^{6}m)^{2}

g= (6.674*10_{E}^{−11})(5.974*10^{24})/(40.590*10^{12}) m/s^{2}

∴ g= 9.823 m/s_{E}^{2}This value is close to the official value of

g =9.807 m/s^{2}or 32.174 ft/s^{2}, defined by the internationalGeneral Conference on Weights and Measuresin 1901. Factors such as the rotation of the Earth and the effect of large masses of matter, such as mountains were taken into effect in their definition.

Although, the value of **g** varies from place to place around the world, we use the common values of:

g9.8 m/s_{E}=^{2}or 32 ft/s^{2}

## Acceleration due to gravity on the Moon

You can find the value of **g _{M}** on the Moon by substituting the following items into the equation:

g_{M}= GM_{M}/R_{M}^{2}

where

**G**= 6.674*10^{−11}m^{3}/s^{2}-kg**M**= 7.35*10_{M}^{22}kg (mass of the Moon)**R**= 1.738*10_{M}^{6}m (radius of the Moon)**R**= 3.02*10_{M}^{2}^{12}m^{2}

Substituting in values:

g= (6.674*10_{M}^{−11}m^{3}/s^{2}-kg)(7.35*10^{22}kg)/(1.738*10^{6}m)^{2}

g= (6.674*10_{M}^{−11})(7.35*10^{22})/(3.02*10^{12}) m/s^{2}

∴ g= 1.62 m/s_{M}^{2}

Thus, the acceleration due to gravity on the Moon is about 1.6 m/s^{2}

## Comparison

Compare the mass and radius of the Earth and Moon to see the gravity ratio.

### Mass

Mass of Earth: **M _{E}** = 5.974*10

^{24}kg

Mass of Moon: **M _{M}** = 7.35*10

^{22}kg

M/_{E}M= 5.974*10_{M}^{24}/7.35*10^{22}= 81.3

M= 81.3*_{E}M_{M}

In other words, the mass of the Earth is about 81 times that of the Moon

### Radius

Radius of Earth: **R _{E}** = 6.371*10

^{6}m

Radius of Moon: **R _{M}** = 1738*10

^{3}m

R= 6.371*10_{E}/R_{M}^{6}/1738*10^{3}= 3.66

(R= 13.4_{E}/R_{M})^{2}

R= 13.4*_{E}^{2}R_{M}^{2}

In other words, the square of the Earth's radius is about 13.4 times that of the Moon

### Gravity

The acceleration due to gravity on the Moon is about 1/6 that of on the Earth:

g_{E}= GM_{E}/R_{E}^{2}

g_{M}= GM_{M}/R_{M}^{2}

g_{E}/g_{M}= (GM_{E}/R_{E}^{2})/(GM_{M}/R_{M}^{2})

g_{E}/g_{M}= (M_{E}/M_{M})(R_{E}^{2}/R_{M}^{2})

g= (81.3)(13.4) = 6_{E}/g_{M}

or

g=_{M}g/6_{E}

## Summary

The acceleration due to gravity, **g**, is considered a constant and comes from the *Universal Gravitation Equation*, calculated at the Earth's surface. By substituting in values for the mass and radius of the Earth, you can find the value of **g**.

Think clearly and logically

## Resources and references

### Websites

**Acceleration Due to Gravity** - TutorVista.com

**The Value of g** - Physics Classroom

**Acceleration Due to Gravity** - Haverford College

**The Acceleration of Gravity** - Physics Classroom

### Books

**Top-rated books on Simple Gravity Science**

**Top-rated books on Advanced Gravity Physics**

## Questions and comments

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## Gravity on Earth versus on Moon