# Universal Gravitation

## Gravitational Field Strength

Gravitational Field Strength

### All objects with mass have an associated Gravitational field. This Gravitational field will interact with any other masses within the field to apply a Force of attraction.

All objects with mass have an associated Gravitational field. This Gravitational field will interact with any other masses within the field to apply a Force of attraction.

### The larger the mass, the larger the Force the Gravitational field can apply. The measure of the size of this attractive Force is known as the Gravitational field strength (g).

The larger the mass, the larger the Force the Gravitational field can apply. The measure of the size of this attractive Force is known as the Gravitational field strength (g).

### The Gravitational field strength has units of Nkg^{-1} and is defined as :-

The Gravitational field strength has units of Nkg

^{-1}and is defined as :-

### 'The Force exerted upon a mass of 1 kg at a certain point within a Gravitational field.'

'The Force exerted upon a mass of 1 kg at a certain point within a Gravitational field.'

### The Gravitational field strength depends on two variables:-

The Gravitational field strength depends on two variables:-

### 1. The Mass of the object.

1. The Mass of the object.

### 2. The Distance from the object.

2. The Distance from the object.

### The table below shows the Gravitational field strength at the surface of several celestial bodies:-

The table below shows the Gravitational field strength at the surface of several celestial bodies:-

### The above table shows clearly that mass and distance have an effect of the Gravitational field strength. This can be seen most clearly with the gas giant Saturn.

The above table shows clearly that mass and distance have an effect of the Gravitational field strength. This can be seen most clearly with the gas giant Saturn.

### Saturn has a mass 95 times larger than the Earth, but as it has a volume 100 times larger the Gravitational field strength at the surface of Saturn is actually lower than on the surface of the Earth.

Saturn has a mass 95 times larger than the Earth, but as it has a volume 100 times larger the Gravitational field strength at the surface of Saturn is actually lower than on the surface of the Earth.

## Gravitational Field Strength and Altitude

Gravitational Field Strength and Altitude

### As the distance from an object increases, the Gravitational field strength gets smaller. Due to this, the value of 'g' is reduced at higher altitude. The graph below shows how the value of 'g' varies in relation to Height:-

As the distance from an object increases, the Gravitational field strength gets smaller. Due to this, the value of 'g' is reduced at higher altitude. The graph below shows how the value of 'g' varies in relation to Height:-

### The above relationship shown is an inverse square law relationship. As the distance from the mass is doubled, the value of 'g' is reduced to a quarter.

The above relationship shown is an inverse square law relationship. As the distance from the mass is doubled, the value of 'g' is reduced to a quarter.

### Note - The value of 'g' below the surface reduces due to the fact that the deeper into a planet the object is, the less mass below to contribute to 'g'. This gives a value of 'g' of zero at the planet's core.

Note - The value of 'g' below the surface reduces due to the fact that the deeper into a planet the object is, the less mass below to contribute to 'g'. This gives a value of 'g' of zero at the planet's core.

## Newton's Law of Universal Gravitation

Newton's Law of Universal Gravitation

### As has been shown previously, the Gravitational field strength varies greatly between locations. In 1687, Isaac Newton published 'The Principia', containing his work on Gravity. Within this publication, Newton had devised a general rule which would allow the calculation of the Force of attraction due to Gravity between any two objects at any separation. This formula is called Newton's Law of Universal Gravitation :-

As has been shown previously, the Gravitational field strength varies greatly between locations. In 1687, Isaac Newton published 'The Principia', containing his work on Gravity. Within this publication, Newton had devised a general rule which would allow the calculation of the Force of attraction due to Gravity between any two objects at any separation. This formula is called Newton's Law of Universal Gravitation :-

### Where :-

Where :-

### F = Force between the two masses (N)

F = Force between the two masses (N)

### G = Gravitational Constant = 6.67x10^{-11} m^{3}kg^{-1}s^{-2 }

G = Gravitational Constant = 6.67x10

^{-11}m

^{3}kg

^{-1}s

^{-2 }

### m_{1} = Mass 1 (kg)

m

_{1}= Mass 1 (kg)

### m_{2} = Mass 2 (kg)

m

_{2}= Mass 2 (kg)

### r = The separation between Masses (center to center)

r = The separation between Masses (center to center)

### Note - The above formula works well for all classical mechanics problems, everything that Newton could measure in the 17th century. It is now known, however, that the above formula does not work when dealing with very strong Gravitational fields. In these situations, Einstein's theory of General Relativity must be used.

Note - The above formula works well for all classical mechanics problems, everything that Newton could measure in the 17th century. It is now known, however, that the above formula does not work when dealing with very strong Gravitational fields. In these situations, Einstein's theory of General Relativity must be used.

## Example 1 -

Example 1 -

### Gravitational Attraction to the Earth

Gravitational Attraction to the Earth

### A pupil of mass 50 kg is sitting on the Earth surface and is attracted to the Earth which has a Mass of 5.97x10^{24} kg. What is the Gravitational Force that the pupil experiences if the radius of the Earth is 6.38x10^{6} m?

A pupil of mass 50 kg is sitting on the Earth surface and is attracted to the Earth which has a Mass of 5.97x10

^{24}kg. What is the Gravitational Force that the pupil experiences if the radius of the Earth is 6.38x10

^{6}m?

### G = 6.67x10^{-11} m^{3}kg^{-1}s^{-2}

G = 6.67x10

^{-11}m

^{3}kg

^{-1}s

^{-2}

### m_{1} = 5.97x10^{24} kg

m

_{1}= 5.97x10

^{24}kg

### m_{2} = 50 kg

m

_{2}= 50 kg

### r = 6.38x10^{6} m

r = 6.38x10

^{6}m

### F = (6.67x10^{-11} x 5.97x10^{24} x 50) / (6.38x10^{6} x 6.38x10^{6})

F = (6.67x10

^{-11}x 5.97x10

^{24}x 50) / (6.38x10

^{6}x 6.38x10

^{6})

### F = 490 N

F = 490 N

### or

or

### W = mg

W = mg

### W = 50 x 9.8

W = 50 x 9.8

### W = 490 N

W = 490 N

### Note - If the Weight of the pupil is calculated, the same value is found for the Force. This allows the calculation of 'g' at any given point as a rearrangement of w = mg.

Note - If the Weight of the pupil is calculated, the same value is found for the Force. This allows the calculation of 'g' at any given point as a rearrangement of w = mg.

## Example 2 -

Example 2 -

### Gravitational Attraction between pupils

Gravitational Attraction between pupils

### What is the Gravitational Force between two pupils sitting 2 m apart if both pupils have a mass of 50 kg ?

What is the Gravitational Force between two pupils sitting 2 m apart if both pupils have a mass of 50 kg ?

### G = 6.67x10^{-11} m^{3}kg^{-1}s^{-2}

G = 6.67x10

^{-11}m

^{3}kg

^{-1}s

^{-2}

### m_{1} = 50 kg

m

_{1}= 50 kg

### m_{2} = 50 kg

m

_{2}= 50 kg

### r = 2 m

r = 2 m

### F = (6.67x01^{-11} x 50 x 50) / (2 x 2)

F = (6.67x01

^{-11}x 50 x 50) / (2 x 2)

### F = 4.17x10^{-8} N

F = 4.17x10

^{-8}N

### F = 41.7 nN

F = 41.7 nN

### Note - as can be seen in the above example, the Gravitational attraction between small objects is negligible but non-zero.

Note - as can be seen in the above example, the Gravitational attraction between small objects is negligible but non-zero.

## The Schiehallion Experiment

The Schiehallion Experiment

### Newton throughout his work considered ways to use his Law of Universal Gravitation in order to measure the Mass of the Earth, but could not devise a suitable method. In 1774 Nevil Maskelyne, the Astronomer Royal devised an experiment to use the Gravitational attraction of a mountain to find an approximate value for the mass of the Earth.

Newton throughout his work considered ways to use his Law of Universal Gravitation in order to measure the Mass of the Earth, but could not devise a suitable method. In 1774 Nevil Maskelyne, the Astronomer Royal devised an experiment to use the Gravitational attraction of a mountain to find an approximate value for the mass of the Earth.

### This experiment was performed using the Munro Schiehallion in Perthshire, Scotland.

This experiment was performed using the Munro Schiehallion in Perthshire, Scotland.

### By measuring the deflection of a plumb-line by the mountain and by then accurately measuring the volume of the mountain to find its total Mass ( using the average density of its rock ), the Gravitational attraction of the mountain could be calculated. This could then be scaled-up to estimate the average density and therefore the Mass of the Earth.

By measuring the deflection of a plumb-line by the mountain and by then accurately measuring the volume of the mountain to find its total Mass ( using the average density of its rock ), the Gravitational attraction of the mountain could be calculated. This could then be scaled-up to estimate the average density and therefore the Mass of the Earth.

### The results from the experiment gave a value of 4500 kgm^{-3}. Compared to the modern value of 5515 kgm^{-3} this experiment gave an result with an error of less than 20 %, an exceptionally close value given the equipment being used.

The results from the experiment gave a value of 4500 kgm

^{-3}. Compared to the modern value of 5515 kgm

^{-3}this experiment gave an result with an error of less than 20 %, an exceptionally close value given the equipment being used.

### The video below shows a summary of the Schiehallion Experiment.

The video below shows a summary of the Schiehallion Experiment.