Isaac Newton and his Universal Law of Gravity

1.9 Isaac Newton and his Universal Law of Gravity

Isaac Newton was born in the manor house of Woolsthorpe, near Grantham, in 1642, the same year Galileo died. His father, also called Isaac Newton, had died before his birth and his mother, Hannah, married the minister of a nearby church when Isaac was 2 years old. Isaac was left in the care of his grandmother and effec- tively treated as an orphan. He attended the Free Grammar School in Grantham but showed little promise in academic work and his school reports described him as ‘idle’ and ‘inattentive’. His mother later took Isaac away from school to manage her property and land, but he soon showed that he had little talent and no interest in managing an estate.

An uncle persuaded his mother that Isaac should prepare for entering university and so, in 1660, he was allowed to return to the Free Grammar School in Grantham to complete his school education. He lodged with the school’s headmaster who gave Isaac private tuition and he was able to enter Trinity College, Cambridge in 1661 as a somewhat more mature student than most of his contemporaries. He received his bachelor’s degree in April 1665 but then had to return home when the University was closed as a result of the Plague. It was there, in a period of 2 years and whilst he was still under 25, that his genius became apparent.

26 Introduction to Astronomy and Cosmology

There is a story (which is probably apocryphal) that Newton was sitting under the apple tree in the garden of Woolsthorpe Manor. He might well have been able to see the fi rst or last quarter Moon in the sky. It is said that an apple dropped on his head (or thudded to the ground beside him) and this made him wonder why the Moon did not fall towards the Earth as well.

Newton’s moment of genius was to realise that the Moon was falling towards the Earth! He was aware of Galileo’s work relating to the trajectories of projectiles and, in his great work Principia published in 1686, he considered what would happen if one fi red a cannon ball horizontally from the top of a high mountain where air resistance could be ignored. The cannon ball would follow a parabolic path to the ground. As the cannon ball was fi red with greater and greater velocity it would land further and further away from the mountain. As the landing point becomes further away the curvature of the Earth must be considered. In a more popular work published in the 1680s called A Treatise of the System of the World,

he included Figure 1.17. The mountain is impossibly high in order for it to reach above the Earth’s atmosphere. However, this is a thought experiment, not a real one. One can see from this that, if the velocity is gradually increased there would come a point when the cannon ball would never land – and would be in an orbit around the Earth.

Astronomy, an Observational Science

Let us fi rst treat this experiment quantitatively and use modern day values and units to extend Newton’s arguments to encompass the Moon. You can calculate that the surface of the Earth falls below a fl at horizontal line by approximately 5 m over a distance of 8 km. If one drops a mass from rest at the surface of the Earth,

it will drop a height of ½gt 2 in a time t, where g is the acceleration due to gravity at the Earth’s surface. The value of g is 9.8 m s ⫺2 so the fall would be 4.9 m. So, if we fi red the cannon ball with a speed of about 8000 m s ⫺1 , its fall after 8 km would

match the falling away of the Earth’s surface and the cannon ball would remain in orbit.

Newton applied the same logic to the motion of the Moon, realising that, if the gravitational attraction between the Earth and Moon caused it to fall by just the right amount, it too would remain in orbit around the Earth. He knew enough about the Moon to be able to calculate the value of the acceleration of gravity at the distance of the Moon. For this, he needed to know the radius of the Moon’s orbit (assumed to be circular) about the centre of the Earth, and also the period of its orbit around the Earth. Newton used a value of the radius of the Moon’s orbit of

384 000 km and a period of 27.32 days or 2.36 ⫻ 10 6 s. Referring to Figure 1.18 (which is not to scale) one can easily calculate the distance, L (the length AB), and direction (tangential to the radius vector) that the Moon would travel in 1 s if sud- denly there were no gravitational attraction between the Earth and the Moon.

28 Introduction to Astronomy and Cosmology

Using the small angle approximation, where sin θ ⫽ θ (in radians), then L ⫽ R ⫻ θ, so

θ ⫽ (1/2.36 ⫻ 10 6 ) ⫻ 2 ⫻ π ⫽ 2.66 ⫻ 10 ⫺6 rad. Giving

L ⫽ 1.022 km.

As a result of the mutual attraction of the Earth and the Moon, the Moon will actually follow the curved path AC which can be thought of as being made up of the straight line motion AB and a fall BC, the distance fallen by the Moon in 1 s.

Let the distance from the centre of the Earth, point E, to A be R and from E to B

be D. Finally let the distance from B to C be d. As the orbit is circular, the distance from E to C is also R.

With this notation, d ⫽ D ⫺ R In the right-angled triangle ABE, R/D ⫽ cos θ, or

D ⫽ R/cos θ.

Hence, d ⫽ D ⫺ R ⫽ R/cos θ ⫺ R ⫽ R [(1/cos θ) ⫺ 1].

Now θ is a very small angle, so we may write:

cos θ ⫽ 1 ⫺ (θ 2 /2)

where θ is in radians. Using the binomial theorem to fi nd 1/cosθ, and ignoring all but the fi rst two terms (θ is very small) we get:

1/cos θ ⫽ 1 ⫹ (θ 2 /2).

Substituting in the expression for BC, we obtain:

d ⫽R⫻θ 2 /2

⫽ [3.84 ⫻ 10 8 m ⫻ (2.66 ⫻ 10 ⫺6 ) 2 ] /2

⫽ 1.36 ⫻ 10 ⫺3 m. So, in 1 s, the Moon falls vertically towards the Earth by just over 1 mm.

Astronomy, an Observational Science

Let us assume that the acceleration due to gravity at the distance of the Moon is g m , then this fall would be equal to ½g m t 2 so that g m is 0.00272 m s ⫺2 . This is considerably less than the value of 9.81 at the Earth’s surface, so the force of gravity must fall off as the distance between the two objects increases. The value of g at the distance of the Moon compared with that at the Earth’s surface was 0.00272/9.81 ⫽ 2.77 ⫻ 10 ⫺4 . This is a ratio of 1/3606. Newton

knew that the radius of the Earth was 6400 km, so that the Moon, at a dis- tance of 384 000 km, was precisely 60 times further away from the centre of the Earth than the Earth’s surface. Hence, the value of g at the distance of the Moon had fallen almost precisely by the ratio of the distances from the centre of the Earth squared!

This led Newton to his famous inverse square law: the force of gravitational attraction between two bodies decreases with increasing distance between them as the inverse of the square of that distance.

However, Newton had a problem: he felt that he could not publish his law until

he could prove that the gravitational pull exerted by a spherical body was pre- cisely the same as if all the mass were concentrated at its centre. This can only be proved by calculus and it took Newton a while to develop the ideas of calculus, which he called ‘ fl uxions’. It was only then that he felt confi dent enough to pres- ent his theory to the world.

The proof is not diffi cult, but rather long: as a sphere can be regarded as set of thin nested shells, the required proof is to show that the gravitational attrac- tion of a thin shell is as if all its mass were concentrated at its geometrical centre. You will thus see that there is no requirement that the body is uniformly dense (the Earth certainly is not) but it does require that the density at a given distance from the centre is constant – the body must have a spherically symmetric mass distribution.

Newton realized that the force of gravity must also be directly proportional to the object’s mass. Also, based on his third law of motion, he knew that when the Earth exerts its gravitational force on an object, such as the Moon, that object must exert an equal and opposite force on the Earth. He thus reasoned that, due to this symmetry, the magnitude of the force of gravity must be proportional to both the masses.

His law thus stated that the force, F, between two bodies is directly proportional to the product of their masses and inversely proportional to the distance between their centres. This can be written as:

F αM 1 M 2 /d 2

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where M 1 and M 2 are the masses of the two bodies and d is their separation. Thus one can write:

F ⫽G⫻M M /d 2 1 2

which is Newton’s Universal Law of Gravitation , where G is the constant of proportionality called the universal constant of gravitation. Why Universal? Using his second law (force ⫽ mass ⫻ acceleration) and his Law of Gravity, Newton was able to deduce Kepler’s third law of planetary motion. This deduction showed him that his law was valid throughout the whole of the then known Solar System. To him that was Universal!

1.9.1 Derivation of Kepler’s third law

To simplify the derivation, it is assumed that the orbit of the planet is circular. The acceleration that must act on the object to keep it in a circular orbit around

a body is given by:

a ⫽v 2 /r

where v is the object’s speed in its orbit and r is the radius of the circular orbit. Also, stating Newton’s Second Law:

F ⫽ma

where F is force, m is mass and a is acceleration. So the inward force to act on the planet to overcome its inertia and keep it in a circular orbit is:

m v 2 p /r.

This force is provided by the gravitational force between the Sun and the planet so:

m p v 2 /r ⫽ G m s m p /r 2 .

Astronomy, an Observational Science

Cancelling m p and r, we get:

v 2 ⫽Gm s /r

The period P of the orbit is simply 2πr/v, so v ⫽ 2πr/P.

Thus, 4π 2 r 2 /P 2 ⫽Gm s /r Giving: 4π 2 r 3 ⫽Gm s P 2

Dividing both sides by Gm s and swapping sides gives:

P 2 ⫽ (4π 2 /Gm )r 3 s .

As the part in brackets is a constant, P 2 is proportional to r 3 – Kepler’s Third Law! Over 300 years after the publication of his great theory, it is diffi cult to be pre- cise as to how he developed the theory in his mind. The outline above is, I am sure, how Newton would like us to believe it happened – observations leading to an inverse square law that was proven by showing that Kepler’s Third Law could be derived from it. However, it could have happened in an inverse man- ner. From Newton’s Second Law and Kepler’s Third Law one can show that the forces between the Sun and planets must follow an inverse square law. From this, Newton could calculate the value of g at the distance of the Moon, and hence the period of its orbit by working backwards through the calculation above. This observation would then show that a prediction of the law was true. It does not really matter which was the case, Newton’s Universal Law of Gravitation was an outstanding achievement.

Newton derived a value of G by estimating the mass of the Earth assuming it had an average density of 5400 kg m ⫺3 . He suspected that the Earth increases in density with increasing depth and had simply doubled the value of 2700 kg m ⫺3 that is measured at the surface of the Earth. (This was a pretty good – and very lucky – estimate, as it is actually 5520 kg m ⫺3 !)

Again, using modern values and units, let us carry out this calculation. We derived a value for the acceleration due to gravity at the distance of the

Moon, g m , above:

g m ⫽ 0.00272 m s ⫺2 .

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From Newton’s Second Law (force ⫽ mass ⫻ acceleration), the force acting on the Moon must have been:

M m ⫻ 0.00272 kg m s ⫺2 ⫽M m ⫻ 0.00272 N Equating this with the force between them, as calculated from his law of gravity:

G ⫻M e M m /d 2 ⫽M m ⫻ 0.00272,

Giving

G ⫽ 0.00272 ⫻ d 2 /M e

(Notice that the Moon’s mass cancels out.) With

d ⫽ 384 000 km ⫽ 3.84 ⫻ 10 8 m

and

M e ⫽ 5400 ⫻ 4/3 ⫻ π ⫻ (6.4 ⫻ 10 6 ) 3 kg

⫽ 5.93 ⫻ 10 24 kg

G ⫽ 0.00272 ⫻ (3.84 ⫻ 10 8 ) 2 /5.93 ⫻ 10 24 Nm 2 kg ⫺2

⫽ 4.0 ⫻ 10 14 /5.93 ⫻ 10 24 Nm 2 kg ⫺2 ⫽ 6.76 ⫻ 10 ⫺11 Nm 2 kg ⫺2

Due to his lucky estimate of the mean density of the Earth, this was a very good result – the now accepted value of G being 6.67 ⫻ 10 2 ⫺11 Nm kg ⫺2 .