VCE PHYSICS UNIT 3 NOTES: 2009
Motion in one and two dimensions
Copyright: AIP (Vic Branch) Education Committee Motion in one and two Dimensions
• apply Newton’s laws of motion to situations involving two or more coplanar forces acting along a straight line and in two dimensions;
Definitions of important concepts
Displacement  Change in position, that is, where an object is in relation to some reference point. It is measured in metres (m), and its symbol is usually , x, or s or sometimes d.
Velocity – Rate of Change in Displacement or Change in Displacement / Time taken. It is measured in metres per second (m/s or ms^{1}), and its symbol is v. Symbolically, the definition is v = s / t.
Acceleration – Rate of change of Velocity or Change in Velocity / Time taken. It is measured in metres per second squared (m/s^{2 } or ms^{2}), and its symbol is a. Symbolically the definition is a = v / t.
These three quantities are vectors, that is, they need both magnitude (or size) and direction to fully specify them. For example, in a country trip, Echuca’s displacement from Melbourne is 400 km North, the average velocity from the trip from Melbourne to Echuca was 85 km per hr or 24 m/s North, the acceleration away from the traffic lights was 5.5 m/s^{2} North.
The relationships between the three quantities can be described graphically when graphed against time.
Disp Vel Accel’n
Time Time Time
To go from left to right: Use the gradient. The gradient of disp vs time graph gives the vel vs time graph, and the gradient of the vel vs time graph gives the accel’n vs time graph.
To go from right to left: The “area under the graph” The area under the accel’n vs time graph gives vel vs time graph, and the area under the vel vs time graph give the disp vs time graph.
A velocity vs time graph can be used to show the difference between the instantaneous velocity and average velocity. The instantaneous velocity is the velocity at an instant in time and is obtained from the gradient from the tangent at that instant, t_{3}, while the average velocity is the area under the graph from t_{1} to t_{2} divided by the time difference, t = t_{2} – t_{1}.
Vel
t_{1} t_{2} t_{3} Time
For a constant acceleration, a set of formulas can be used to solve problems. These use the following quantities and their symbols.
Vel v = u + at
v s = (u + v)t/2
s = ut + ½ at^{2}
u v^{2} = u^{2} + 2as
s = vt – ½ at^{2}
t Time
Initial vel (u), final velocity (v), acceleration (a), time taken (t) and displacement (s).
Energy in its different forms is a useful means of analysing Mechanics problems. The forms of energy include:
Kinetic Energy is the energy an object has due its motion. It is calculated from the definition:
KE = ½ mv^{2}.
The mass must be in kilograms, and the velocity in metres per second, and KE is measured in joules.
Gravitational Potential Energy is the energy an object has due to its position, as with all forms of potential energy. In this case, the position is the position in a gravitational field, which can be measured from any reference point, but usually the surface of the earth. It is calculated from the definition:
Grav Pot’l E = mgh
The mass must be in kilograms, the value of ‘g’ is the acceleration due to gravity, either 10 or 9.8 m/s^{2}, and ‘h’ is the height above the reference point. Again the unit of energy is joules.
Elastic Potential Energy is the energy stored in a compressed or stretched material.
Energy cannot be created or destroyed, it is just converted from one form to another. Other forms of energy include Light, Sound and Thermal energy. This means that the total amount of energy is constant or that Energy is conserved.
In the case of an object moving in a gravitational field, such as a roller coaster, the total of Gravitational Potential Energy and Kinetic Energy is constant, if frictional losses to thermal energy are ignored. This means that as the roller coaster comes down and loses Grav PE, it gains speed with the increase in KE.
Isaac Newton proposed three laws to explain the how and why of motion.
First Law: An object will remain at rest or at a constant speed in the same direction until acted upon a net force.
Second Law A net force on an object produces an acceleration equal to the force / mass.
Third Law If object A exerts a force on an object B, then object B exerts an equal and opposite force on object A. F_{by A on B} =  F_{by B on A}. Note: These two forces act on different objects.
Weight is the gravitational force that the Earth exerts on all masses. Close to the Earth, the size of the force on an object can be calculated by multiplying its mass by (the acceleration due to gravity), that is :
W = mg, where g =9.8 Newton/kg or m/s^{2}. It also acceptable to approximate this to 10 N/kg.
This force acts vertically downwards and can be considered to act through the centre of the object, that is, the geometric centre of mass.
Friction is a force that acts on an object when it moves or attempts to move across the surface of another object. The force acts in the opposite direction to the way the object is moving or attempting to move. While the object is not moving the friction force balances the pulling force up to a limiting value, whereupon the object moves and the friction holds a steady value.
The cause of Friction is the roughness of the surfaces in contact. At the microscopic level, the surfaces are not flat and “hills” of one surface fit in to the “valleys’ of the other surface and resist motion. It is only when the force is large enough that the “hills” of one surface are pushed forward across the tops of the “hills” of the other surface.
Friction is usually considered as a force that slows objects down, like drag or air resistance. Yet in situations, such as walking, friction is necessary for movement to occur.
Normal Reaction is the force a surface exerts on an object when that object is being pulled into it, typically by the Earth, that is, the object’s weight. This Force, the Reaction Force, acts at right angles to the surface; that is why it is called the “Normal” Reaction.
The cause of Normal Reaction is that the Weight force pulls the box into the surface. This squashes the atomic layers in the top of the surface. These atomic layers push back providing an upward force. It is an identical situation to sitting on a soft couch, your weight pulls you down, until the cushion is compressed to such an extent that the size of the upwards force balances your weight force.
Normal Reaction: Force by Floor on Box
Weight: Force by Earth on Box
Note: These two forces are not an example of Newton’s Third Law: They act on the same object, and they are not always equal and opposite (see below).
Combining the forces
Consider the forces on the moving car below. The car is rear wheel drive with the engine making the rear wheels turn. The front wheels are turned by the road.
Reaction Reaction: Force by Road on Car
Friction: Force by Road on Car
Friction: Force by Road on Car
Weight: Force by Earth on Car
Rear wheel
The rear wheels are turned by the engine.
So the tyre pushes back on the road.
By Newton’s 3^{rd} law, the road pushes forward on the tyre,
And the car moves forward.
Force by Tyre on Road Force by Road on Tyre
(from the engine) (The Driving Force on the Car)
On the Front wheel the friction of the road makes the tyre rotate, and reduces the effect of the driving force. Even though the horizontal forces on the car are both friction, they don’t cancel. The forward friction force on the back wheels is far greater than that on the front wheels, and the car goes forward.
Connected Bodies
These laws of Motion can be applied to a set of problems called “Connected Bodies”, that is, problems where one object is pulling or pushing another. Typical examples are a car pulling a caravan, a mass hanging over an edge pulls another along the bench. An example of pushing is in a railway shunting yard, where a train engine pushes carriages along the track.
Questions that can be asked are: What forces acting on each member of the system? What is the overall acceleration of the combined system?
Example: Trailer Car
Tension
Force by Force by
Car on Trailer Trailer on Car
Friction Driving Force
Force by road on tyre (Force by road on tyre)
Weight and Reaction forces have been omitted from the diagram. Air resistance on the car and trailer and road friction on the tyres of the car have been ignored.
The strategy for solving such questions is:

Label all forces acting

The whole system and each mass of the system accelerates at the same rate

Apply Newton’s 2^{nd} Law: Net Force = Mass x Accel’n applies to each mass in the system and to the whole system as well.

Forces between masses in the system are examples of Newton’s 3^{rd} Law.
The car is slowed down by the trailer because the trailer is pulling backwards on it. The trailer is pulled forward by the car. The Tension forces in the tow bar are an example of Newton’s 3^{rd} Law.
The mass of the car is 1500 kg and the trailer is 500kg. The magnitude of the Driving Force is 2000 N, while the Road Friction on the trailer is 800 N.

Find the acceleration of the car and trailer.

Find the force the car exerts on the trailer

Find the force the trailer exerts on the car.
CarTrailer system: Total Mass = 1500 + 500 = 2000 kg. Net Force = 2000 – 800 = 1200 N.
Accel’n = Net Force/ Mass = 1200 / 2000 = 0.60 m/s^{2}.
Trailer system: Mass = 500 kg. Net Force = T – 800. Accel’n = 0.6 m/s^{2}
Net Force = Mass x Accel’n. T – 800 = 500 x 0.6. T = 800 +300 = 1100 N
Car system to find force of Trailer on Car apply Newton’s 3^{rd} law = 1100 N, or
the long way: Mass = 1500 kg, Net force = 2000 – T, accel’n = 0.6 m/s^{2}
Net Force = mass x Accel’n . 2000 – T = 1500 x 0.6, T = 2000 – 900 = 1100 N
Forces in two Dimensions
Newton’s 2^{nd} Law of motion applies to the Net Force on an object. When there are two forces acting on an object at an angle to each other the object nevertheless moves in one direction. What is this direction? and how does it relate to the two forces?
In such a case, to find the Net Force the vector nature of force comes into play. If a force of 40 N in a Northerly direction is combined with a force of 30 N in an easterly direction, then the net force is given by the following calculation:
30 N
40 N
50 N at an angle N 37^{0} E
The object will move in a direction N 37^{0} E.
If a box is being pulled along the ground by a rope at an angel to the horizontal, then the pulling force will have both a vertical and horizontal component.
P
N
Vert: P sin + N = mg
Horiz: Pcos– Fr = Net Force = ma
Friction
mg
• analyse uniform circular motion of an object in a horizontal plane (F_{net} = mv^{2}/R) such as a vehicle moving around a circular road, a vehicle moving around a banked track; and object on the end of a string;
Many objects move in circles at a constant speed, planets, moons, fun rides, cornering cars. The following analysis can apply to all these situations.
V_{1} V_{1}
Velocity V V
Vectors: V_{2} V_{2}
Always
At a tangent The change of velocity is
To the circle directed inwards towards
the centre of the circle.
The speed is constant around the circle, and the acceleration is also constant in magnitude and always directed towards the centre.
This means that in every instance of circular motion, the net force is also directed towards the centre.
In analysing circular motion, the first task is to describe the kinematics of the motion, that is, the relationships between displacement, speed, acceleration and time.
The displacement measures of circular motion are the Radius, r, and the Circumference. The time measures are the Period, T, the time for one revolution, and the Frequency, f, the number of revolutions per second, with the unit of Hertz.
An expression for the speed can now be obtained.
Speed = Distance / Time = Circumference / Period. Speed, v = 2r/T = = 2fr.
To determine the acceleration, it can be seen that the velocity vector sweeps out a full circle in the time of one period, so by analogy with the definition of speed in a circle, the acceleration can be given by:
acceleration, a = 2v/T, additional expressions for the acceleration can be obtained by substituting from the expressions for the speed, v.
a = 2v/T = v^{2}/r = 4^{2}r/T^{2} = =4^{2}f^{2}r
This acceleration is also called the centripetal acceleration, because the accelerating is “centreseeking”.
The expressions for the acceleration can now be incorporated in Newton’s 2^{nd} Law.
Net Force, F = ma = mv^{2}/r = m.4^{2}r/T^{2} = =m.4^{2}f^{2}r
Typical Problems
Car turning left, viewed from back Car on a banked road
N N cos = mg
a N N Friction = net force, ma a N sin = mv^{2}/r
= mv^{2}/r tan = v^{2}/(gr)
Friction
Mg
mg
For the object in the rear window
Tension, T
a T sin = Net Force = ma = mv^{2}/r
T cos = mg
Weight, mg

Apply Newton’s second law of circular motion in a vertical plane, consider forces at the highest and lowest position only
In the case of circular motion in a vertical plane, particularly when restricted to the highest and lowest positions, the forces contributing to the Net Force on an object are its weight and the reaction force from the surface it is in contact with. The weight force is always downwards, but the reaction force could be either up or down.
Two common scenarios are: i) a car going over a crest or though a dip, and ii) a roller coaster ride.
i) A car going over a crest A car going through a dip
N N a
mg – N = ma = mv^{2}/r N – mg = mv^{2}/r
mg mg > N mg mg < N
a when N = 0, v^{2} = gr.
The reaction force, N, will change as the car’s speed changes. In the diagram on the left, if the car travels at the right speed, then N can equal zero. This means that the ground will exert no force on the car, the car will be, in effect, “feel weightless”. In the diagram on the right, the reaction force provides the driving force towards the centre of the circle. N > mg, so the driver feels heavier. In both these cases, the weight force is pulling the car down into the road, so the reaction force must be either up, or zero for the right speed, but never downwards.
ii) Roller coaster at bottom of loop Roller coaster at top of loop, track prevents trolley falling.
slow fast
N a N
A a
Mg N
N – mg = ma = mv^{2}/r mg mg
mg < N mg – N = ma = mv^{2}/r mg + N = mv^{2}/r
Same as car going through dip. Apparent weightlessness is possible, Restraining rail keeps
When N=0, mg = mv^{2}/r, faster trolley on rails by
v^{2} = gr providing a downwards
Same as car going over crest reaction force
A falling roller coaster from an energy perspective
A roller coaster on a circular track does not travel at a constant speed, it is accelerated by the force of gravity.
As the roller coaster falls under gravity, it loses
v Grav Pot’l Energy, and gains Kinetic Energy.
2r If the coaster was travelling at speed to be temporarily feeling
weightless at the top, then by the bottom, the speed would be
v much greater, and a large reaction force, making the
passengers feel 6g’s.

investigate and analyse the motion of projectiles near the Earth’s surface including a qualitative description of the effects of air resistance;
To describe projectile motion, if air resistance is initially ignored, two principles apply:

Horizontally, the horizontal component of the velocity is constant, this means the relationship to use is Distance = Horiz Speed x Time

Vertically, the acceleration is a constant at the value of the acceleration due to gravity, g. The relationships to use are the equations of uniform accelerated motion, such as v = u + at, etc.
The combination of these two constraints produces a parabolic path.
If an object is projected at an angle, , with speed, V, then the horizontal component of = V cos ,
The initial vertical component = V sin .
Questions that can be asked include:

Time of Flight (equals twice the time to max height, vert vel = zero)

The Range (equals the time of flight x horizontal speed)

Maximum Height (equals Displacement when vert vel = zero)
Projectile Motion With Air Resistance: Air Resistance is a force that acts in the opposite direction to the velocity and increases with the magnitude of the velocity. It has a downward component adding to gravity as the object goes up, but opposes gravity as the object returns to ground.
This means that:

The maximum height and time to maximum height are decreased,

The time to return to ground longer than the time to go up,

The range is reduced.

Apply the concept of work done by a constant force

Work done = constant force x distance moved in direction of net force

work done = area under force  distance graph;;
Work Done by a constant force is defined as equal to the size of the force x the displacement, (W= F x d). It has the units of Newton.metres or joules.
What is the effect of the Work Done by this force?
As the units for Work Done would indicate the effect is energy related. In fact, the Work Done equals a change in the Energy of the mass.
When considering the Net Force, and only in that case, the Work Done equal the change in Kinetic Energy, as can be shown using the equation ‘v^{2} = u^{2} + 2as”.
If both sides are multiplied by the mass, m, and then divided by 2, you get:
½ mv^{2 } = ½ mu^{2 } + mas
or rearranging: mas = ½ mv^{2 }  ½ mu^{2 } , which is Force x Disp = Change in Kinetic Energy.
Note: This only applies for the Net Force. In the case below:
P (20 N)
M = 2.0 kg
Friction (5.0 N)
4.0 m
In this case: The gain in KE is (20 – 5.0) x 4.0 = 60 Joules. While the energy supplied by the person pulling the mass supplies 20 x 4.0 = 80 joules. The Friction force explains the lost energy 5.0 x 4.0 = 20 joules which heats up the mass and the surface on which it moves.
This situation can be represented on a graph, a Force vs distance graph.
If the force varies over the distance, then Work done = F x d does not apply, the the relationshop:
Work done = Area under a Force vs distance graph still does.
To determine the Work done, you can either count squares using the grid and scales on the axes, or divide the shape up into triangles and rectangles and calculate the area of each. In either case an approximate answer is acceptable.
If a mass was being raised vertically at a constant speed, the size of the listing force would equal the weight force, and the work done by the lifting force would equila the gain in gravitational potential energy.
• apply laws of energy and momentum conservation in isolated systems
• analyse impulse (momentum transfer) in an isolated system, for collisions between objects moving along a straight line (Ft = mv);
Impulse And Momentum
Newton’s Laws of Motion enable the development of new concepts, such as impulse and momentum. Impulse is defined as Force x Time or the Area under a Force vs Time graph. Therefore Impulse has units of N.sec, and symbol I. Momentum is defined as mass x velocity, it has units of kg.m.sec^{1}, and has symbol, p. Both quantities are vectors.
Using Newton’s 2nd Law, F = ma, the expression can be rearranged to produce:
Impulse = Force x Time = mass x (v/t) x t = mass x v = (mv) = p = Change of Momentum
In a collision in an isolated system*, which means that there is no other force acting on the objects colliding, two objects interact and exchange momentum and energy. Newton’s Third Law can be applied to collisions. In a collison each object exerts a force on the each other. These two forces are equal and opposite.
F_{by A on B} =  F_{by B on A}
If the collision lasts a time, t, then F_{by A on B} . t, =  F_{B on A} . t,
but this equals the impulse, so Impulse _{by A on B} =  Impulse _{by B on A},
and as above, Impulse = Change of Momentum, so p_{B} =  p_{A}
but p = Final momentum – Initial momentum,
so,
Final Mom of B – Initial Mom of B =  (Final Mom of A – Initial Mom of A)
Rearranging gives:
Initial Mom of A + Initial Mom of B = Final Mom of A + Final Mom of B
That is, Momentum is conserved and is conserved in every single collision.
When a car crashes into a tree embedded in the earth, the impulse the tree exerts on the car is equal and opposite to the impulse the car exerts on the tree and the earth it is attached to. In this case the momentum is not destroyed, it cannot be destroyed, it is always conserved. In this case the car’s momentum has been transferred to the earth.
It is the force during the impact that does damage to the car and its occupants. The impulse on the car and its occupants and their change in momentum is determined by how fast they were initially travelling, but design changes to the car can reduce the force of impact.
The change in momentum is unchanged, the impulse is unchanged, but, by the definition of impulse, if the duration of impact can be increased, the size of the force can be reduced. This is the purpose of bumper bars, crumple zones, air bags and crash helmets. Seat belts are designed to keep you in the car and let the crumple zone do its work.
* An example of a collision in a system that is not isolated is a collision between a golf ball and a bird in flight because during the collision weight forces are acting and so momentum would not be conserved.
• apply laws of energy and momentum conservation in isolated systems
• analyse transformations of energy between: kinetic energy, strain potential energy, gravitational potential energy and energy dissipated to the environment considered as a combination of heat, sound and deformation of material
 kinetic energy, i.e. ½ mv^{2}, elastic and inelastic collisions in terms of conservation of kinetic energy
Collisions can be analysed from the point of view of Kinetic Energy, as well as Momentum. The difference is though, that the Conservation of Momentum applies to every single collision, but the Conservation of Kinetic Energy only applies to small number of collisions.
Also during a collision the total momentum is still conserved, whereas Kinetic Energy goes into storage as Potential Energy, and none, some or all of that energy may be returned at Kinetic Energy.
The collisions in which Kinetic Energy is conserved are called Elastic Collisions because the force of interaction between the colliding masses only depends on the separation of the masses, not on whether the masses are approaching each other or moving apart.
Consider two gliders, one heavy , one light , with repelling magnets approaching each other on an air track. This is very close to an Elastic Collision. The total Momentum and Total Energy are represented by .
The heavy glider pushes the light glider back in the opposite direction.
Momentum Kinetic Energy
Time Time
Force
The graphs as the masses approach and recede
are identical.
Separation
Consider a truck and a car approach each other, collide and move off together.
Momentum Kinetic Energy
E
Time Time
Force
The area in between the two graphs equals the
amount of energy lost, E.
Separation
• analyse transformations of energy between: kinetic energy, strain potential energy, gravitational potential energy and energy dissipated to the environment considered as a combination of heat, sound and deformation of material
 strain potential energy, i.e. area under force distance graph including ideal springs obeying Hooke’s law, ½ kx^{2},
As springs are stretched, they become harder to stretch. If masses are successively added to a spring, a Force vs Extension graph can be drawn, where the Force is the weight of the attached masses, and the Extension is the increase over the natural, unstretched length. The graph is usually either of the two graphs below.
Force Force
Extension Extension
The Extension is proportional to the Applied Force. In some cases, the graph goes through the origin, in other cases, it seems that a small force is required before the spring begins to stretch.
In the case on the left, relationship can be expressed as a formula:
Force = Spring Constant x Extension or F =kx, this is known as Hooke’s Law
The Spring Constant has symbol, k, and the units are Newtons/metre.
In the case of the graph on the right, the Spring Constant would still be the gradient of the graph, rather than the value of the force at a point divided by the value of the extension.
Because the graph is a Force – Displacement graph, the area under the graph represents Work Done. As the spring is stretched energy is stored in the spring. If a stretched spring was pulled down and released, the spring would contract suddenly, giving kinetic energy to the mass attached to the end. This means there is energy stored in the spring, elastic potential energy.
The amount of the elastic potential energy stored in the stretched spring is therefore given by the area under the graph and a formula can be generated for the graph on the left above.
Elastic Potential Energy = ½ F . x = ½ kx^{2}
The unit of Elastic Potential Energy is “joules”, and the quantity usually has the symbol, E_{p}
For graphs such as the one on the right above, the elastic potential energy is easiest calculated from the “Area under the Graph”.
• apply gravitational filed and gravitational force concepts g = GM/r^{2} and F = GM_{1}M_{2}/r^{2};
• model satellite motion (artificial, moon, planet as uniform circular orbital motion, ( a = v^{2}/r = 4^{2}r/T^{2} );
Most satellites of the Earth, including the Moon travel in circular orbits, so the relationships and ideas of circular motion described above apply to this topic. Of importance is the realisation that while the velocity is along the tangent, the acceleration is inwards towards the centre, and so must the net force.
Newton’s Law of Universal Gravitation was developed, not so much from seeing an apple fall from a tree, but the speculation that the force from the earth that acted over a few metres to pull the apple off the tree, may also be the force to pull the moon inwards as it moved in its circular orbit.
Earth
a = 9.8 m/s^{2} a = ?
R_{E} 60 x R_{E}
Moon
R_{E} = 6.38 x 10^{6 }m T = 28 days
R of orbit = ??
Newton calculated the acceleration of the moon using his relationship, a = 4^{2}r/T^{2}, and compared this with the acceleration due to gravity at the surface of the earth. Newton found that while the moon is about 60 earth radii away, the acceleration due to gravity at the moon’s distance is 1/3600 of the acceleration at the earth’s surface.
This result is evidence that:
the acceleration due to gravity is inversely proportional to the square of the distance.
acceleration 1/r^{2}
Using his 2^{nd} Law of motion, Force _{Earth on Moon} Mass _{Moon}/r^{2}
Using his 3^{rd} Law of motion, Force _{Earth on Moon} =  Force_{ Moon on Earth}, which is Mass _{Earth }/r^{2}
This means that the gravitational attraction between the Earth and the Moon is given by:
Force Mass_{ Earth} Mass _{Moon}/r^{2}
Newton was able to write this as an equation:
Force = G Mass_{ Earth} Mass _{Moon}/r^{2}
Where G is the Universal Gravitational Constant.
But Newton had no way of determining the value of G, as the Masses of the Earth and Moon could not be determined.
It took another 100 years until Cavendish measured in the laboratory the force between two known masses, that the value of G could be determined. The accepted value today is 6.67 x 10^{11} Nm^{2} kg^{2} .
The Law of Gravitation can now be combined with the expressions for circular motion to solve satellite problems. For a satellite in orbit, the gravitational force provides the centripetal acceleration.
G Mass_{ Earth} Mass _{Satellite }/R^{2} = Mass _{Satellite} x (4^{2}R / T^{2})
G Mass_{ Earth} /R^{2} = 4^{2}R / T^{2}
G Mass_{ Earth} /4^{2 } = R^{3} / T^{2}
A property of the Central Body = A property of the Satellite
This means that R^{3} / T^{2} has the same value for all satellites going around the Earth, including the Moon. This proves Kepler’s 3^{rd} Law who found that R^{3} / T^{2} was the same for all planets going around the Sun.
For a geostationary satellite, the period of the satellite is the same as the period of rotation of the Earth, that is, 24 hours. This way the satellite will permanently stay in the same location in the sky. The radius of the orbit can be determined from the above equation.
Alternatively, the radius could be obtained by comparison with the Moon.
(R^{3} / T^{2}) _{Satellite} = (R^{3} / T^{2})_{ Moon}
A useful concept to explain action at a distance is the concept of a gravitational field, similar to a magnetic or an electric field. That is, there is something about the space around a large mass such that if an object is placed at a point in that space it will experience a force. The size of this force equals the size of the gravitational field at that point times the mass of the object.
The Gravitational Field is a property of the space around a mass regardless of whether anything is there.
Newton’s Law of Universal Gravitation can be rearranged to give an expression for the Gravitational Field.
Force = G Mass_{ Earth} Mass _{Object }/r^{2} = (G Mass_{ Earth} /r^{2} ) x Mass _{Object} = (Grav Field at r) x Mass _{Object}
Gravitational Field at r = G Mass_{ Earth} /r^{2}
g = GM/r^{2}
Force = Gravitational Field of Earth x Mass of object
or in general, F =mg,
where ‘g’ equals the gravitational field strength (Newtons/ kg) or the alternative expression, the acceleration due to gravity (m/sec^{2}).
Substituting in Mass of Earth for “M”, Radius of Earth for “r”, should give a value of 9.8 N/kg for “g”. Try this. Similarly, at an altitude of one Earth radius above the surface of the Earth, the value of “g” should be: g = 9.8 / (1 + 1)^{2} = 9.8 / 2^{2} =9.8 / 4 = 2.45 m/s^{2}.
• analyse transformations of energy between kinetic energy ,…, gravitational potential energy, …
 gravitational potential energy, … from area under forcedistance and area under fielddistance graphs multiplied by mass;
A graph of gravitational force – distance would look like the graph below. However because the force must between two specific masses, each pair of masses will have a different graph. Even if one of the masses is the Earth, each satellite would need a different graph.
Because the graph is a Force – Distance graph the area under the graph represents Work Done by or against the Gravitational Force. Work Done results in a Change in Kinetic Energy.
In the above case, the shaded area represents the amount of kinetic energy a 1.0 kg mass would need to leave the surface of the earth and rise to an altitude of one Earth radius. Alternatively, it also represents the amount of kinetic energy gained if the 1 kg mass was dropped to the earth from an altitude of one Earth radius. Such an area can be calculated either by breaking the shape into simpler shapes such as triangles and rectangles. Alternatively, with a grid, small squares could be counted.
A different graph will be needed for each mass. One way to overcome this problem is to use a graph of the Gravitational Field against Distance. Such a graph would be the same for all masses, but the area would have units of Joules / kg. To obtain the actual Kinetic Energy the area is multiplied by the mass of the object.
• apply the concepts of weight (W = mg), apparent weight (reaction force, N), weightlessness (W = 0) and apparent weightlessness (N = 0);
Weight is the Gravitational Force exerted by the Earth on a mass.
The Apparent Weight is the force an object experiences from the surface it is in contact with. When you sit on a chair, the earth pulls you down into the chair, while the squashed chair pushes up on you.
Force by chair on You (Reaction Force, N)
Force by Chair on You
Force by Earth on You (W = mg)
Force by You on Chair
Force by You on Earth (W = Mg’)
By Newton’s 3^{rd} Law, when the chair exerts an upward force on you, your body exerts a downward force on the chair. This force comes from the compression of the spine as the earth pulls you into the chair. This is the force that tells you that you have weight.
If you were sitting on the chair in a lift that is accelerating upwards, you would feel heavier because the Force by the Chair on You is greater than your Weight force to provide a net force upwards.
In the case of an astronaut in the space shuttle, both the astronaut and the shuttle are accelerating towards to earth with the value of the acceleration due to gravity at that distance. Therefore the shuttle exerts a zero reaction force on the astronaut (N = 0). With no contact force on the body, the body feels no compression, and so the astronaut feels weightless.
Real weightlessness occurs when the strength of the gravitational field is zero, i.e. g = zero, so W = 0. For the gravitational field to be zero, you need to be an infinite distance from any matter or galaxy, so there is no point within our galaxy, the Milky Way, where you would be really weightless.
