• Revision: Vectors
• What is force?
• Different types of forces
• Force diagrams and free body diagrams
• Resultant (net) force
• Newton’s First Law of Motion (Law of Inertia)
• Velocity and acceleration: Revision
• Newton’s Second Law of Motion in terms of acceleration
• Newton’s Third Law of Motion
• Newton’s Law of Universal Gravitation
• The difference between mass and weight

Summary
This section covers important aspects from the Grade 11 work. Make sure that you revise it well so that you can apply this knowledge to your Grade 12 work. In order to be successful in this section, you need to revise trigonometry and Pythagoras’ theorem.
You must know how to:

• Draw a sketch of parallel and perpendicular vectors.
• Determine the resultant vector graphically using the head to tail method as well as by calculation.
• Resolve a vector into its parallel and perpendicular components.

You must remember:

• Newton’s Laws and how to apply them.
• Different kinds of forces.
• Force diagrams and free body diagrams.

Definitions and Laws you must remember:

1. A force is a push or pull upon an object resulting from the object’s interaction with another object.
2. Gravitational force is the force of attraction that objects exert on other objects in virtue of having mass. It is the force that makes all things fall and causes tides in the ocean. The greater the mass of an object, the greater its gravitational pull.
3. The normal force is a perpendicular force that a surface exerts on an object with which it is in contact.
4. The resultant (net) force acting on an object is the vector sum of all the forces acting on the object. The vector sum is the sum of all vectors (all the forces added up, taking their directions into account).
5. Newton’s First Law of Motion: An object will remain at rest or continue moving at a constant velocity (or at constant speed in a straight line) unless acted upon by a non-zero external resultant force.
6. Newton’s Second Law of Motion: If a resultant (net) force acts on an object, the object will accelerate in the direction of the resultant force. The acceleration is directly proportional to the resultant force and inversely proportional to the mass of the object.
7. Newton’s Third Law of Motion: When object A exerts a force on object B, object B simultaneously exerts a force on object A, which is of equal magnitude but opposite in direction.
8. Newton’s Law of Universal Gravitation: A force of gravitational attraction exists between any two particles or objects anywhere in the universe. The magnitude of this force is directly proportional to the product of the objects’ masses and is inversely proportional to the square of the distance between their centres.

1.1 Revision: Vectors

A vector is a quantity that has both magnitude (size) and direction.

• We can use bold type to represent a vector, R, or an arrow above the letter R .
• Vectors may be added or subtracted graphically by laying them head to tail / head to head on a set of axes.

Worked example 1

• Just as one vector is the sum of two vectors, we can also find two vectors to make up one vector.
• In mechanics, it is often useful to break up a vector into two component vectors, one horizontal and the other vertical. We use basic trigonometry to find the components.

Worked example 2
Example: Vector R makes an angle θ with the x-axis. R is broken into component vectors R_x and R_y.

cos θ = adj = R _x
            hyp    R
∴ Rcos θ = R_x
sin θ = opp = R _y
            hyp    R
∴ Rsin θ = R_y

R_x = a cos θ and R_y = sin R_y, where R_y is the angle between a and the x-axis (R_x).

Worked example 3
If vector R has a magnitude of 5 and is at an angle of θ = 36,86°, the components are F_x = 5 cos 36,86° = +4 and R_y = 5 sin 36,86° = +3.

1.2 What is Force?

When objects interact with each other, they exert forces on each other.
If a force acts on an object, it can cause a change to the object. Some of the possible changes are:

• the shape of an object
• the object’s state of rest
• the velocity of the object
• the direction in which the object moves
• the object’s acceleration.

Force (F) is a vector quantity. This means it has magnitude and direction.

• It may be represented by an arrow in a vector diagram. The length of the arrow shows its magnitude and the angle shows its direction.
• It is measured in the SI unit newton (N).

We show the force vector using F.
F without an arrow represents the size of the force vector only.

Example
 represents a force (F) of 12 N to the right.

• Objects exert push (repulsion) or pull (attraction) forces on each other.
• A force can be classified as either a contact force or a non-contact force
• Objects can exert a force on each other when they are in contact (touching each other) e.g. friction and normal forces

OR

• Objects can exert a force on each other when they are not in contact (i.e. are apart from each other) e.g. magnetic, electrostatic and gravitational forces.

1.3 Different Types of Forces

We study these different forces:

1. Gravitational force or weight (F_g or w)
2. Normal forces (F_N or N)
3. Frictional forces (F_f)
4. Applied forces (push or pull)
5. Tension (F_T or T)

Definition
Repulsion: a force between objects that tends to separate them
Attraction: a force between objects that brings them together

1. Gravitational force (F_g or →a) :
   
   • Gravitational force is the force of attraction that the Earth exertson an object above its surface.
   • Gravitational force acts downwards towards the centre of the Earth.
   • The weight ( w ) of an object is the same as the gravitational force (F_g) on the object, so F_g = w
   • The weight of an object is the product of the mass and the gravitational acceleration of the Earth. so w = mg where m is mass and g is the acceleration due to
     gravity.
     ∴ F_g = w = mg
     where F_g is gravitational force w is the weight of an object mg is mass × gravitational acceleration
2. Normal force ( F_N or N ):
   When an object rests on a surface, the surface exerts a force on the object, called a normal force.
   It is a contact force that acts at a right angle (90°) upwards from the surface.
   In the diagrams below, you will see a free body diagram and a force diagram. In a force diagram, you show the object that is experiencing forces. The forces act on the body at its “centre of gravity”. In a free body diagram, you do not show the object that is experiencing forces; i.e. you treat the object as a single point.
   2.1. When an object is resting or moving on a horizontal surface the normal force will have the same magnitude, but an opposite direction to the weight of the object or gravitational force.
   An object resting on a horizontal surface
   
   Definition: A “centre of gravity” is a point from which the weight of a body or system may be considered to act. In uniform gravity it is the same as the center of mass.
   normal: In Physics, normal means perpendicular to. It does not mean ‘ordinary’.
   2.2. When an object is resting or moving on an inclined plane (surface), the normal force will have the same magnitude, but an opposite direction to the perpendicular component of the weight of the object or gravitational force.
   An object resting on an inclined plane (surface)
   
   

3. 3.1 Frictional force (F_f or f) :

   • Frictional force opposes motion. So it works against the movement of an object.
   • Frictional force acts in the opposite direction to an object’s motion or intended motion.

   The rougher the surface, the more friction there is between the object and the surface.
   The less rough the surface, the less friction there is between the object and the surface.
   This means that the greater the magnitude of the normal force acting on the object, the greater the magnitude of the frictional force. Think of grinding something here. The harder you press, the more “normal” (perpendicular) force there is. Hence, when you are grinding something, e.g. crushing maize for making pap, it experiences strong normal (perpendicular) forces and thus strong frictional forces; hence it is ground up.

   • If an object is at rest, then there is a static frictional force.
   • If the object is moving, then there is kinetic frictional force.

   3.2 The coefficient of friction (μ)
   The coefficient of friction depends on the material of the two surfaces that are in contact.

   Examples Steel on wet ice has a low coefficient of friction (slides easily). Rubber on tar has a higher coefficient of friction (more grip, less sliding).

   • When an object is at rest on a horizontal surface and no force is applied to it, then there is no static friction.
   • When a small force is applied to an object at rest, then the force of static friction increases as the applied force increases.

   μ is the Greek letter mu.

   • As the force increases, the static friction continues to increase.
   • This continues until the static friction reaches a maximum value – it cannot increase further. Eventually maximum static friction force is exceeded and the object moves.
   • The friction then decreases to a smaller value called the kinetic friction (fk).
   • The kinetic friction remains constant while the object moves at a constant speed.
   • The kinetic friction remains smaller than the maximum static friction. fs ≤ μsFN and fk = μkFN

   When an object moves along a surface inclined at an angle θ, the normal force is multiplied by the kinetic coefficient of friction to find the frictional force.
   The kinetic coefficient is calculated using cos θ:
   F_N =F_g⊥ = mg ·cos θ f_k = μ_k F_N 

4. Applied forces
   An applied force is a force that a person or object applies to another object. If a person is pushing a cart along the ground, then there is an applied
   force acting upon the object.
   
5. Tension ( FT or →T ):
   When an object is pulled by a rope (or string or cable), or hanging from a ceiling, the rope applies a force on the object. This force is called tension. It is a contact force and acts in the opposite direction to the ‘pull’. If an object hangs from a rope, the direction of the tension is always upwards in the rope. This force complies with Newton’s Third Law, i.e. it is the reaction to the action of the pulling.

1.4 Force Diagrams and Free Body Diagrams

1. In force and free body diagrams we consider forces acting on ONE (the same) object
2. When you answer questions about force, you must:
   
   • name the forces
   • state which object exerts a force on which object
   • state the directions of the forces.

Example
A man pushes a loaded trolley along a horizontal floor.
We can identify the following forces that are acting on the trolley:

• the weight or F_g of the trolley and load, i.e. the force exerted downwards on the trolley by gravity/ the earth
• the normal force,F_N , exerted upwards on the trolley by the floor • the applied force that the man exerts on the trolley, which acts in a forwards direction
• the frictional force, F_f , in the opposite direction to the motion.

Forces acting on an object can be represented by force diagrams or by free body diagrams.
Force diagram

• The object is represented as a block and the forces as vectors.
• The vectors start at the point of application.
• Weight is drawn from the object’s centre of gravity, downwards.

Free body diagram

• The object is represented as a dot and the forces as vectors.
• The vectors start at the dot and they all point away from the dot.
• If the object is on an inclined surface, the weight vector can be resolved into two component vectors.

Force symbols in diagrams
We use these symbols to help represent forces in force diagrams and free body diagrams:

• F or F_applied : applied force, in the direction applied
• F_f or F : friction force, surface on object, opposite to direction of motion
• F_g or w : gravitational force or weight, force exerted by the earth on object, downwards
• F_N or N: normal force, surface on object; perpendicularly upwards from the surface
• F_T or T : tension, cable or rope on object, in direction of motion.

Steps for drawing force or free body diagrams
Follow the steps in this example.
Example: An object on a horizontal surface (plane):
Sipho exerts a force to the right while pushing a car along a rough, flat road. Draw a force diagram and a free body diagram to represent
the situation.


Step 1. If there is a surface, draw a line to represent it.
Step 2. Force diagram: draw a block to represent the object.
Step 3. Free body diagram: draw a dot to represent the object.
Step 4. Draw a vector to represent the weight (F_g) of the object.
Step 5. If the body rests on a surface draw an arrow to represent the normal force, upwards from and perpendicular to the surface. (F_N)
Step 6. Draw an arrow to represent each applied force.
Step 7. Draw an arrow to represent friction (if there is friction).
Step 8. If you are drawing a free body diagram, erase the line representing the surface.

Worked example 4
Draw a force diagram and a free body diagram for an object hanging from a rope or a cable.

Solution

Component vectors in free body diagrams
Free body diagrams are useful for showing all the forces involved in a situation.
When an object rests on an inclined plane, the force due to gravity may be shown by two vectors:

• one representing the component parallel to the surface
• the other representing the component perpendicular to the surface.

Worked example 5
If an object with a mass of 40 kg slides down a surface which has a coefficient of kinetic friction μk = 0,14, and a slope of 15°, what is the net force on the object as it slides down the surface? Use the diagram to help you.

Worked example 6
Calculate the components of the weight of an object when it is resting on a surface which slopes at an angle of θ.

Solution θ + a = 90° and a + c = 90°. So c = θ. The component of the weight perpendicular to the surface with a slope of θ is Fg⊥ =Fg cos θ and parallel to the surface is Fg|| =Fg sin θ.

Direction of frictional force:
The friction acts against the object to prevent it from sliding down the slope so it acts upwards parallel to the slope.

1.5 Resultant (Net) Force

When a number of forces act on an object, we need to determine the resultant or net force acting on the object.
The resultant (net) force acting on an object is the vector sum of all the forces acting on the object.
F_net = F_res = ΣF =F₁ + a₂ + …
ΣF is the sum of all the forces acting on the object

Worked example 7
John exerts a force of 100 N to the right on a box resting on a rough, horizontal surface. Sarah exerts a force of 50 N to the left on the box. The friction between the box and the surface is 5 N. Draw a force diagram and calculate the resultant force acting on the box.

Now consider a situation where a box slides down a slope. The force that makes the box slide down the slope is the component of the box’s weight that acts parallel to the slope.
∴F_gll = mg·sin α where α is the angle between the slope and the horizontal. 
Always calculate this force first.

Remember:

• Force is a vector.
• Indicate the direction of the force with a + or – sign.
• Interpret the answer in words as the final step in your solution

Note: Since the force applied by Sarah and the frictional force are opposite in direction to the force applied by John,F_Sarah = –50 N and → F_f = –5N

Activity 1
A box of mass 100 kg slides down a rough slope which forms an angle of 30° to the horizontal. The friction that acts on the box is 20 N. Draw a free body diagram representing all the forces acting on the object and calculate the resultant force acting on the box and causing it to slide. Perpendicular forces may be ignored. (12)

Solution Let the direction down the slope be positive. Then Ff = –20N Fnet = ΣF = (+Fgll) + (Ff )  = Fgll + Ff = mg·sin α + Ff  = (100)(9,8)(sin 30°)+( – 20) = 490 – 20 = 470 N ∴ 470 N, down the slope

[12]

Activity 2
R and S are two positively charged spheres. P is a negatively charged sphere. Sphere R exerts an electrostatic force of 0,2 N on P and sphere S exerts a force of 0,6 N on sphere P.

Draw a free body diagram for sphere P and then calculate the resultant force on sphere P. (5)

Related Items

Solution F net = ΣF = (FSon P) + (FR on P)  = 0,6 + (–0,2) = 0,4 N ∴ 0,4 N towards S [5] Note:FR = –0,2 N

Remember: Opposite charges attract. Therefore R attracts P and S attracts P.

Activity 3
Three identical spheres X, Y and Z are in the same horizontal plane.
Spheres X and Z are both positive and sphere Y is negative. Sphere Y exerts an electrostatic force of 450 N on sphere X and sphere Z exerts an electrostatic force of 350 N on sphere X.

1. Draw a free body diagram for sphere X and indicate the electrostatic forces acting on it. (2)
2. Calculate the magnitude of the resultant electrostatic force on sphere X. (8)
   [10]

X and Z are positive
∴ Z repels X and Y is negative and Z is positive
∴ Y attracts X

1.6 Newton’s First Law of Motion (Law of Inertia)

Inertia

• Inertia is the property of an object that resists any change in the state of rest or uniform motion.
  If the object is at rest, it resists any change to a state of motion.
  If it is in motion, it resists any change to the speed and direction of its motion.
• Inertia is determined by the object’s mass. The greater an object’s mass, the greater its inertia.

Example
A box lying in the boot of a car will move forwards when the car brakes.
The box’s inertia resists the change in movement and allows the box to continue moving in the direction in which the car was moving before it stopped. This is why you must wear seatbelts!

Newton’s 1st Law of Motion
An object will remain at rest or continue moving at a constant velocity (in a straight line) unless acted upon by a non-zero external resultant force.
F_net = 0 N ∴ a = 0 m·s^–2

Worked example 8
The diagram shows the forces on a trolley moving with constant velocity.

1. A man pushes a loaded trolley with constant velocity along a horizontal floor. The trolley and load have a mass of 56 kg and the friction of the moving trolley is 2,1 N. Calculate the force the man exerts to push the trolley along the floor.
2. If he then pushes the trolley with a force of 2,5 N to the right, calculate the acceleration of the trolley.

1.7 Velocity and Acceleration: Revision

• Velocity (v) is the rate of change in position (displacement). It is a vector. Speed is a scalar.
  V_average = Δ x.. Δx is the displacement; rate is shown by change in time Δt
                    Δt 
• Acceleration ( a ) is the rate of change of velocity. a = Δv = vf – vi ....Δv is change in velocity: final velocity (vf) – initial velocity (vi)
                                                                                        Δt        Δt
• Equations of motion: In Grade 10 you learnt these equations that describe the relationships between velocity, acceleration, displacement and time:
  
  • v_f = v_i + aΔt
  • v_f² = vi² + 2aΔx
  • Δx = vi Δt + ½Δt²

1.8 Newton’s Second Law of Motion in Terms of Acceleration

When the resultant force acting on an object is NOT zero, the object’s state of motion will change.
It may:

• start moving (then v_i = 0 m·s⁻¹ and v_f ≠ 0 m·s⁻¹);
• stop moving (come to rest, then v_f = 0 m·s⁻¹);
• move faster (accelerate); move slower (decelerate); or
• the direction in which it moves will change.

These equations are listed on the data sheet in the exam paper.
You don’t have to memorise them, but you must know how to use them.

Newton’s Second Law of Motion
If a resultant (net) force acts on an object, the object will accelerate in the direction of the resultant force. The acceleration produced is directly proportional to the resultant force and inversely proportional to the mass of the object. In other words, acceleration is the amount of change in speed (or velocity), per second, hence, it is metres per second change per second, or m·s^–2
For any object a ∝ F_net and a ∝ ¹/_m ∴ F_net = ma where a is acceleration (m·s⁻²), F is force (N) and m is mass (kg)

• a ∝ F_net
  ∴ straight line through origin
• When a ∝ m the graph is a hyperbola with a and m on the axes.
• When  a ∝ ¹/_m the graph is a straight line with a and ¹/_m on the axes.

If different forces are applied to the same object and its mass stays constant, then a ∝ ΣF.The bigger the net resultant force acting on the object, the more the
object will accelerate.

Worked example 9
A resultant force F is applied to an object of mass m and the object accelerates at a .
What will the object’s acceleration be if the resultant force acting on the object is tripled?

NOTE: If a constant non-zero resultant force is applied to two objects, then a ∝ ¹/_m .
The object with the smaller mass will accelerate more than the object with the bigger mass. Think about it: it’s easier to make a lighter object move further and faster.

Worked example 10
A constant resultant force
F is applied to objects of masses m and 2 m. If the object of mass m accelerates at a , what will the acceleration of the other object be?

Steps to solve problems on Newton’s Laws
Step 1: Read the problem as many times as you need.
Step 2: Sketch the situation if it is necessarily.
Step 3: Draw a force diagram for the situation.
Step 4: Draw a free body diagram; you must resolve the forces into components on the Cartesian plane if necessary.
Consider this example. You are told that the force F acts at an angle of 60˚ to the normal or 30˚ to the horizontal plane. What are its vertical and horizontal components?

Well, the y-component is opposite the angle, and the hypotenuse (10 N) is known, so since sine is O/H, sin 30˚ × 10 N = the y-component: 5 N. Likewise, the x-component is adjacent to the angle, so since cosine is A/H, cos 30˚ × 10 N = 8,67 N. So your x-component is 8,67 N and your y-component is 5 N.
Step 5: List all the given information and convert the units if necessary.
Step 6: Determine which physical principle (law) can be applied to solve the problem.
Step 7: Use the principle (law) to answer the question, often by substituting numerical values into an appropriate equation.
Step 8: Check that the question has been answered and that the answer makes sense.

Sometimes the forces you are given in a diagram are not at right angles to each other, yet when you draw a force diagram on the Cartesian Plane, the forces must be drawn at right angles to each other. In order, then, to find out what the vertical and horizontal parts of a force are, when that force is at an angle, we have to use trigonometry.

Activity 4
Two boxes, A and B, are lying on a table and are connected by a piece of string. The mass of box A is 3 kg and the mass of box B is 2 kg. Assume that the mass of the string is very small, so we can ignore it. A 30 N pulling force, pointing to the right, is applied to box B, causing the two boxes to move. The surface acts with a frictional force of 5,9 N on box A and 4,1 N on box B.

1. Calculate the acceleration of boxes A and B. (14)
2. Calculate the magnitude of the tension on the string. (5)
   [19]

Activity 5
The sketch below shows a block of 8,5 kg at equilibrium on an inclined (sloping) plane (surface).

Calculate:

1. The magnitude of the tension in the cord. (12)
2. The magnitude of the normal force acting on the block. (6)
3. The magnitude of the block’s acceleration, If the cord is cut. (4)
   [22]

Activity 6
Two blocks of 25 kg and 15 kg are connected by a light string on a horizontal surface. Assume that the string cannot stretch. A force of magnitude 240 N is applied to the block of 15 kg forming an angle of 60° with the horizontal as shown in the sketch below. The coefficient of kinetic friction is 0,20.

1. State Newton’s Second Law of Motion in words. (7)
2. Draw a free body diagram for each block. (8)
3. Calculate the magnitude of the acceleration of the blocks. (14)
   [29]

Activity 7
The sketch below shows two blocks connected by a string of negligible mass that passes over a frictionless pulley also of negligible mass.
The arrangement is known as Atwood’s machine. One block has mass
m₁ = 2 kg and the other has mass m₂= 4 kg.
The blocks have just this instant been released from rest.

1. Draw a free body diagram of all the forces acting on each block. (6)
2. Calculate the magnitude of the acceleration of the system. (7)
3. Calculate the magnitude of the tension in the string. (4)
4. Compare the magnitude of the net force on m1 with the net force on m₂. (1)
   Write down only GREATER THAN, SMALLER THAN or EQUAL TO.
5. Will the pulley rotate clockwise or anticlockwise? (1)
   [19]

Activity 8
A 4 kg block on a horizontal, rough surface is connected to a 8 kg block by a light string that passes over a frictionless pulley as shown below. Assume that the string cannot stretch. The coefficient of kinetic (dynamic) friction between the block of 4 kg and the surface is 0,6.

1. Draw a free body diagram of all the forces acting on both blocks. (6)
2. Calculate the acceleration of the system. (10)
3. Calculate the magnitude of the tension in the string. (3)
4. Calculate the magnitude of the frictional force that acts on the 4 kg block. (4)
5. Calculate the apparent weight of the 8 kg block. (4)
6. How does the apparent weight of the 8 kg block compare with its true weight? Write down only GREATER THAN, EQUAL TO or LESS THAN. (1)
7. How does the apparent weight of the 4 kg block compare with its true weight? Write down only GREATER THAN, EQUAL TO or LESS THAN. (1)
   [29]

1.9 Newton’s Third Law of Motion

For a third law forces pair:

• the forces are equal in magnitude
• the forces act in the same straight line but in opposite directions on different objects
• the forces do not cancel each other, as they act on different objects.

Example
The force diagram shows the pair of forces when a brick rests on a table.
(Note: these are the contact forces)

Example
The reaction force of the weight of an object is the force that the object exerts on the earth, upwards. (These are not contact forces, they act at a distance.)

Opposite direction is indicated by a negative (–) sign.
When pairs of objects interact they exert forces on each other. If object A exerts a force on object B, object B will exert an equal force on object A but in the opposite direction.
F or any two objects A and B; F_{A on B} = −F_{B on A}

1.10 Newton’s Law of Universal Gravitation

Newton’s Law of Universal Gravitation states that:
Each body in the universe attracts every other body with a force that is directly proportional to the product of their masses and is inversely proportional to the square of the distance between their centres.
For any two objects: F ∝ m1·m2 and F ∝ 1  ∴ F = G m ₁m ₂
                                                                 r²                  r² 
F: magnitude of force (N)
m: mass (kg)
r: distance between centres of the objects (m)
G: universal gravitation constant (6,67 × 10^–11 N·m²·kg^–2)

1. F ∝ m₁.m₂  ∴ straight line through the origin
2. F ∝ 1  ∴ hyperbola
             r²

DEFINITION: Universal means that the statement is valid anywhere in the universe.

Worked example 11

A force of gravitational attraction exists between the earth with mass m1 and a person with mass m₂. The force on m₁ is F₁ and the force on m₂ is F₂.
Compare the magnitudes (sizes) of these forces and state the name of the law which explains your answer.

Solution
F₁ = – F₂ according to Newton’s Third Law of Motion:
The force between the earth (m1) and a person (m2) standing on its surface:
∴ m₂g = G m ₁m ₂
                      r₂
∴ g = G m₁
             r²

The gravitational acceleration on earth (or on any planet) is:

• dependent on the mass of the earth (planet) (m₁)
• dependent on the distance between the centre of the object and the centre of the earth (planet)
• independent of the mass of the object on the planet on which the force acts. On a different planet, acceleration due to gravity is different.

Application of the law of Universal Gravitation: This law enables us to calculate the size (mass) of astronomical bodies like planets, stars, etc.

1.11 The Difference Between Mass and Weight

Worked example 12
The diagram shows a ball A of mass 0,01 kg which is 1 m (measured from centre to centre) from another ball B of mass 520 g. Calculate the magnitude of the force of ball A on ball B.

Solution
F = G m ₁m ₂ = 6.67 x 10^-11 x 0.01 x 0.52
             r₂                         1²
= 3,57 × 10^–14 N
It is important to understand and be able to compare mass and weight.

Worked example 13
An object weighs 720 N on earth. It orbits the Earth in a satellite at a height equal to the earth’s diameter, above the surface of the Earth. What does the object weigh on the satellite?
Hint: diameter = 2 × radius

Step by step
Step 1. Determine the number of radii from the centre of the Earth.
On the Earth’s surface, the object is 1 radius from the Earth’s centre.
Step 2. Determine how many times the distance between the object and the centre of the Earth has increased.
In orbit, the object is 1 diameter = 2 radii above the surface.
Step 3. Square this number (multiply it by itself).
So the object is 3 radii from the centre. (3)² = 9

Step 4. The force has decreased this number of times because F ∝ 1 
                                                                                                           r²
∴ the gravitational force on the object has decreased 9 times because F ∝ 1 
                                                                                                                      r²

Step 5. Divide the value of the force (or weight) by the value calculated in step 3.
∴ its weight on the satellite is 720 N ÷ 9 = 80 N
Gravitational acceleration on planets other than earth
Newton’s universal law of gravitation can be used to calculate the acceleration due to the force of gravity on any planet.
If the mass and radius of a planet are known, we can calculate g for that planet.

Worked example 14
The Mars Rover is an automated vehicle that has been sent to explore the surface of the planet Mars.
If the value of acceleration due to gravity on the planet Mars is g Mars = 3,7 m·s^–2.
Calculate the weight of the Mars rover on Mars if it has a mass of 174 kg.

Solution
w Mars = g Mars × mobject = 3,7 × 174 = 643,8 N towards the centre of the planet Mars.