Reference:Nelson Physics VCE Units 3 & 4Chapter 15Page 395

G.1 NEWTON'S LAW OF UNIVERSAL GRAVITATION

Using his own work and that of Johannes Kepler, Sir Isaac Newton proposed that there was a force of attraction between all objects, simply because they had mass.Further more he proposed that the force of attraction was directly proportional to the mass of each object and inversely proportional to the square of the distance between them.

Newton called this the force of universal gravitation, and is given by:

whereFis the magnitude of the gravitational force.

m₁, m₂are the masses involved.

ris the distance between the centres of the masses.

Gis a constant.

The constant involved is the same for all masses.

It is called the Universal Gravitational Constant.

It has a value of 6.67 x 10^-11 N m² Kg^-2

Example.

Consider two people standing 1 m apart.One has a mass of 65 Kg and the other 85 Kg.Calculate the force of attraction between them.

G = 6.67 x 10^-11 N m² Kg^-2r = 1 m

m₁ = 65 Kgm₂ = 85 kg

F =6.67 x 10^-11 x 65 x 85

1²

=3.69 x 10^-7 N(hardly a very significant force)

G.2 GRAVITATIONAL FIELD

Gravitational field is the gravitational equivalent of a magnetic field.Thus any region where there is a gravitational effect there is a gravitational field.

Gravitational field has a direction, it is in the same direction as the gravitational force.On Earth it is towards the centre of the Earth.

The strength of the gravitational field is defined as the gravitational force per unit mass:

i.e.

Consider two masses M and m.Let us calculate the gravitational field experienced by m.

The force on mass m is:

The gravitational field strength is:

\

Example.

What is the gravitational field strength 100 km above the surface of the moon.

Mass of Moon is 7.34 x 10²² Kg, Radius of Moon is 1.74 x 10⁶ m.

G = 6.67 x 10^-11

M = 7.34 x 10²²

r = (1.74 x 10⁶ + 100000) = 1.84 x 10⁶)

g = 6.67 x 10^-11 x 7.34 x 10²²

(1.84 x 10⁶)²

=1.45 N Kg^-1

Problem Set #1:TextPage 424Questions 1 – 15

G.2.1 Planetary Motion

Kepler’s Laws for planetary motion:

(a planet has average radial distance r from the sun and period T)

1.Planets move in elliptical orbits (a circle is a special ellipse)

2.The area swept out by the radial vector in a given time is constant for a given planet.

3.For each planet, orbiting the same central body, the ratio = constant.

G.3 WEIGHTLESSNESS

The dictionary definition of weightless is to have no weight.Since weight is the force due to gravity, the object must be in a region where there is no gravitational field.Is there such a place?

as r ®¥ ,g ® 0

So it would appear that it is not possible to be truly weightless.

When we refer to weightlessness we are referring to a sensation.When you are in a lift and it starts to go down you feel as though you have become lighter, you felt less force through your feet.If the lift were to fall with an acceleration of g, you would feel no force through your feet.The is the sensation known as weightlessness.

An astronaut feels this sensation when in a space ship, since both he/she and the space ship are falling at the same rate.

·Satellites are in free fall around the body they orbit.

·Astronauts inside orbiting satellites are also in free fall.

·The sensation of weight depends on the size of the normal reaction force. 

·In free fall the normal reaction force is zero.

·Astronauts have the sensation of weightlessness .

G.4 SATELLITES

A satellite is defined as a smaller object orbiting around a larger one.Satellites can be natural (such as the moon) or artificial (such as communications satellites).In all cases the satellite is held in orbit only by gravitational forces.

G.5 CIRCULAR MOTION AND GRAVITY

Let us consider a satellite orbiting the Earth.The path of the orbit will be almost circular.

Thusa=v²applies

r

as doesF=mv²

r

In this case the force is the force given by Newton's universal law of gravitation.(more later)

G.5.1 The Period of Circular Motion

Our satellite travels at uniform velocity (v) in a circle of radius (r), the distance travelled in one complete orbit is given by:

v=d

t

ort=d(eqn1)

v

the distance is given by:

d=2 p r(eqn 2)

putting the two equations together we get:

However sincea = v²by rearranging and substituting we get:

r

and

G..5.2 The Speed of a Satellite

For a satellite to orbit the Earth it must have some minimum velocity, otherwise it will fall down to Earth.

A too slow, C too fast, B just right.

Looking at the force on the satellite.

The centripetal force=The force of gravity

\

whereM is the mass of the central body.

r is the radius of the orbit.

G.5.2 An Interesting Result

and

and

Combining we get

Kepler's third law

Problem Set #2:TextPage 425Questions 16 – 39

G.6 GRAVITATIONAL POTENTIAL ENERGY

Previously we have looked at small changes in the height of an object above the ground.In this case g is effectively constant and we use,

U_g = m gDh

However as we saw in section G.2, Newton's Law of Universal Gravitation tells us that g is not constant but varies inversely with the distance of separation squared.The force distance graph for a small mass (m) as it moves away from a large mass (M) is shown below:

As we know work done = area under the graph.In this case the work done = change in potential energy.

Which simplifies to

When we are talking about gravitational potential energy, zero potential energy is defined as occurring when the separation is infinite.So at some point, a distance r from the large mass (M) the potential energy is found by:

What about the minus sign?It represents the fact that work must be done to move an object away from the Earth's surface (increase in U_g).If on the other hand two objects are moving towards each other then the gravitational potential energy will decrease, being transformed into some other kind.Probably kinetic.

Problem Set #3:TextPage 430Questions 40 – 43