what is the gravitational force?

The gravitational force is an attractive force that a mass exerts on another mass when the former is placed in the latter’s gravitational field.

 

Fig 1

In fig 1 above we have two masses m₁ and m₂ that are situated at a distance r apart. As you know these two masses will create gravitational fields around themselves such that mass M₁ will exert a gravitational force F₂ on mass M₂ while the mass M₂ will exert a gravitational force F₁ on mass M₁. That is each mass will exert a gravitational force on the other mass. 

And from the third law of motion we know that “For every action there is an equal and opposite reaction.

Hence the gravitational force F₂ will be equal to the gravitational force F₁ .

F₁  = F₂

If the two forces are equal then they have the same magnitude and as a result

|F₁ | = |F₂ | = F

We can thus replace F₁ and  F₂  by F as shown in fig 2 below.

Fig 2

Law of gravitation

In order to calculate the magnitude of the force that each mass will exert on the other the law of gravitation must be used.

It merely states that the gravitational force that the two objects will exert on each other is directly proportional to the product of the two masses and inversely proportional to the square of their distance of separation.

Simply written in an equation it will be as shown below:

Removing the proportionality sign you will obtain the equation below:

Where G is the universal gravitational constant

G = 6.67 x 10^-11N m² kg^-2

Example 1

The moon and the earth are separated by a distance of 3.8x 108 m. The mass of the moon is 6.4 x 1022 kg while that of the earth is 6.0x1024 kg. Calculate the gravitational force between the moon.

Conservation of kinetic energy in collisions

As we have seen in this post during a collision between two object the velocity of the two colliding objects changes. Hence you would agree that he kinetic energy of the objects would change.

We have also seen in this post that in all collisions the sum of linear momentum is a constant. This is the principle of conservation of linear momentum.  However as we are going to see in some collisions, elastic collisions, the sum of kinetic energy is also constant. That is the sum of the kinetic energies of the objects before the collision is the same as the sum of the kinetic energies after the collision.

We are now going to see an example of how to use this “principle” which i am going to call the “principle of conservation of kinetic energy”.

Fig 1 below shows two objects travelling towards each other and fig 2 shows the two objects separating after the two objects have separated.

So if the principle of conservation of kinetic energy applies then it means that Sum of kinetic energy before collision = sum of kinetic energy after collision

Fig 1 Before collision

Before the collision

Sum of kinetic energy before collision = kinetic energy of object 1 + kinetic energy of object 2

= ¹/₂m₁u₁² + ¹/₂m₂v₁²

Fig 2 After collision

 

After the collision

Sum of kinetic energy after collision = kinetic energy of object 1 + kinetic energy of object 2

= ¹/₂m₁u₂² + ¹/₂m₂v₂²

If you refer back to the principle then

Sum of kinetic energy before collision = sum of kinetic energy after collision

 ¹/₂m₁u₁² + ¹/₂m₂v₁²   =   ¹/₂m₁u₂² + ¹/₂m₂v₂²

Example

Two balls of mass 1.0 kg are travelling towards each other at a speed of 5.0 m/s towards the right (ball 1)  and 5.0 m/s towards the left (ball 2) respectively. If after separation one of them moves moves off a speed of 5 m/s towards the right (ball 2 ) calculate the speed of the other ball.

Sum of kinetic energy before collision = sum of kinetic energy after collision

 ¹/₂m₁u₁² + ¹/₂m₂v₁²   =   ¹/₂m₁u₂² + ¹/₂m₂v₂²

 ¹/₂*1.0*5.0² + ¹/₂*1.0*(-5.0)²   =   ¹/₂*1.0*u₂² + ¹/₂*1.0*5.0²

Solving for u₂ the velocity of object 1 after the collision

12.5+ 12.5 = 0.5*u₂²+12.5

12.5= 0.5*u₂²

12.5 /0.5 = 25 =u₂²

u₂² = 25

There can be two solution to u₂

First u₂=-5 m/s and secondly u₂= 5 m/s

Since the second ball is already travelling towards the right and according to the principle of conservation of linear momentum then the only solution is

u₂=-5 m/s

Conservation of linear momentum in collisions

As we have seen in this previous post, in a close system the sum of linear momentum is a constant. Hence in any collision if the sum of linear momentum of the objects before the collision would be equal to the sum of linear momentum after the collision.

Fig 1 below shows two spherical objects travelling toward each other. The mass of the two objects are 1.0 kg each. Object 1 is travelling at a velocity of 5.0 m/s  and object 2 is travelling at a velocity of –5.0 m/s.

Fig 1

If  p_1b is the linear momentum of object 1 before the collision and p_2b

is the linear momentum of  object 2 before the collision then

the sum of linear momentum = p_1b +p_2b

= m₁u₁ + m₂v₁

=1.0*5.0 + 1.0*(-5.0)

= 5.0 - 5.0

=0 Kg m/s

As you can see before the collision the sum of the linear momentum is 0 kg m/s. According to the principle of conservation of linear momentum the sum of linear momentum at all time will thus be 0 kg m/s before and after the collision.

Fig 2 below shows the spherical objects after the collision while they are separating. According to the principle of conservation of linear momentum after the collision the sum of linear will thus be 0 kg m/s.

Question

If after the collision the two objects separate such that object 2 moves with a velocity of  5.0 m/s, calculate the velocity of object 1.

Fig 2

If  p_1a is the linear momentum of object 1 after the collision and p_2a

is the linear momentum of  object 2 after the collision then

the sum of linear momentum = p_1a +p_2a

= m₁u₂ + m₂v₂

=1.0*u₂  + 1.0 *5.0

= u₂  +5.0

Now if we use the principle of  of conservation of linear momentum

sum of linear momentum before the collision = sum of linear momentum before the collision

0 =   u₂  +5.0

u₂  = - 5.0 m/s

Principle of conservation of linear momentum

The linear momentum is the product of the mass of an object and the velocity of an object.

Now if you have two objects that are moving toward each other or one object explodes into two different objects at all time the sum of the linear momentum of the objects will give a constant number.

Hence the principle of conservation of linear momentum states that in a close system the sum of linear momentum is constant.

The equation will be that at all time

p1 +p2+p3+p4+p5+ ………= constant

Consider the diagram below:

Fig 1 Before firing ball

Fig 2 After firing ball

As you can see from the diagram above, you have a close system that is made up of the the cannon and the cannon ball. Hence at all time the sum of the linear momentum will be a constant. Before the firing of the ball and after the firing of the ball.

Example

A canon ball is fired such that the ball leaves the cannon at a velocity of  150 m/s. If the cannon has a mass of 1000 kg and the cannon ball has a mass of 15 kg, determine the velocity of recoil of the cannon.

Hence if you apply the principle of conservation of linear momentum.

Before firing cannon ball

Sum of linear momentum before firing = p_ic  +p_ib

where p_ic is the initial momentum of the cannon and p_ib is the initial momentum of the ball.

After firing the cannon ball

sum of linear momentum after firing= p_fc  +p_fb

where p_fc is the final momentum of the cannon and p_fb is the final momentum of the cannon ball.

We are now going to determine the speed of recoil of the cannon after recoil.

Sum of linear momentum before firing = p_ic  +p_ib

= 1000*0 + 15*0

= 0 kg m/s

sum of linear momentum after firing= p_fc  +p_fb

=1000*v_rc +15*150

=1000*v_rc +2250

where is the v_rc velocity or recoil of the cannon.

Hence if the sum of linear momentum is a constant

then the sum of linear momentum before firing = sum of linear momentum after firing

0 = 1000*v_rc +2250

0 – 2250 = 1000*v_rc

-2250 = 1000*v_rc

v_rc = –2.25 m/s

This principles is used also when collisions are considered. We shall see collisions in a future post.

What is the weight of an object?

Every object has a quantity known as the mass. The mass is simply the quantity of matter that it contains.

When according to the law of gravitation, a mass that is situated inside the gravitational field of the earth would experience an attractive force, a pull, as shown in fig 1 below.

Fig 1

Now this pull on the object is called the weight.

Hence the weight can be defined as the gravitational pull that the earth exerts on an object. 

Since it is gravitational pull it is a  thus a force and its unit is the Newton (N).

The weight is calculated according to the equation below:

weight = mass * acceleration due to gravity

w  = mg 

Example

A stone has a mass of 5.0 kg. Calculate it weight if the acceleration due to gravity is 9.81 m s^-2 .

Weight w = mg

               = 5.0 * 9.81

                = 49.05 N

                = 49 N

What is linear momentum?

Linear momentum is a physical quantity that depends on the mass and the velocity of the object.

Fig 1 below shows an object that has mass m and is moving with a velocity v.

 

The linear momentum which is denoted by the symbol p and is calculated using the equation below

 

linear momentum  = mass of object  x velocity of object

p = mv

Hence the unit for linear momentum is kg m s^-1

The linear momentum is thus a vector quantity since it is the product of a scalar quantity (the mass) and the vector quantity (the velocity).

We can thus say that the linear momentum of a body is defined as the product of the mass of that body and its velocity

Gravitational potential energy

It is simply the energy that a body possesses due to its position above the surface of the earth. There is a second form of potential energy called elastic potential energy that we are going to see later.

The equation to calculate the gravitational energy is

Gravitational potential energy Ep = mgh

Where m is the mass of the object
g is the acceleration due to gravity
h is the height of the object above the ground


Let us have a look at example where the gravitational potential energy of an object.

Example 1

A high jumper has a mass of 62 kg. He jumps to a maximum height of 2.3 m. If the acceleration due to gravity is 9.81 ms^-2, calculate the gravitational potential energy of the object.

Gravitational potential energy Ep = mgh
= 62*9.81*2.3
= 1398 kg
= 1400 kg


You should also know how to obtain the mass or the height of an object if the gravitational potential energy is known.

Let us see an example.

Example 2

A person has a potential energy of 11000 J. If the mass of the person is 85 kg and the acceleration due to gravity is 9.81ms^-2, determine his height above the ground.

Gravitational potential energy Ep = mgh
11000 = 85*9.81*h
h = 11000/ (85*9.81)
` = 13.2 m
= 13 m

I hope you have understood. If you have any questions, leave them in the comment section below. See you later my students.

If you want to test your knowledge of gravitational potential energy, do the following questions leave the answer in the comment section?

Question

A bird of mass 0.12 kg is flying at a height of 12 m above the ground. If the acceleration due to gravity is 9.81 m s^-2, what would be the gravitational potential energy of the bird?

If the bird has a gravitational potential energy of 110 J, at what height would it be flying?

Kinetic energy

Kinetic energy is one if the eight forms of energy. It is simply the energy that a body possesses due to its motion.

The kinetic energy of a body is simply calculated using the equation

Kinetic energy Ek = ½ mv²

Where m is the mass of the mass of the object in kg
v is the velocity of the object in ms^-1

As you can see the kinetic energy of the object depends on both the velocity and the mass of the object. So it is possible for an elephant with a large mass to have the same kinetic energy as a small object moving at a high velocity.

Now let us see an example where the kinetic energy of an object is calculated.

Example 1

An elephant of mass _{3.00 x 10}³ kg is moving at a velocity of 3.00 ms^-1. What is the kinetic energy of the elephant?

So let us recall that the kinetic energy of the elephant is calculated using the equation

Ek = ½ mv²

Hence the kinetic energy Ek = ½ mv²
= ½(_{3.00* 10}³)*(3.00)²
= 13500 J
= _{1.35 *10}⁴ J

Now you can be given the kinetic energy and be asked to calculate the mass or the velocity. Let us look at a second example.

Example 2

An object has a mass of 4.0 kg and a kinetic energy of 16 J. Determine the velocity of the object.

Ek = ½ mv²
16 = ½ (4.0)*v²

Making v the subject of formula

v² = (16 *2)/4.0
v = (16*2/4.0)^½
v = 2.8 ms^-1

Now to test your newly acquired you can do the following questions

1. Calculate the kinetic energy possessed by an aeroplane of mass _{3.4 * 10}⁵ kg flying at 110 ms.
2. If a body has _{3.4 x 10}⁵ J of kinetic energy and mass _{3.2* 10}² kg, what is its velocity?