PHYSICS SSS THREE WEEKS ONE AND TWO

PHYSICS SSS THREE

WEEK ONE

                                                        

                                                           FORCE FIELDS

TYPES OF FORCES: There are two types of forces namely:

1. Contact forces: Forces which are in contact with the body to which they are applied e.g push, pull, tension, reaction and frictional forces.

2. Force fields: These are forces whose sources do not require contact with the body to which they are applied e.g gravitational, electric (or electrostatic) and magnetic forces.

i. Gravitational force is the force with which the earth objects towards its centre.

ii. Electrostatic force is the force existing around charged bodies.

iii. Magnetic force is the one that exist around a bar magnet

The effects of these forces are felt by bodies which are not in contact with them.

FIELDS

            A field is a region of under the influence of some physical agencies such as gravitation, magnetism and electricity.

            The two types of force fields are the scalar field and vector field.

A scalar field is one that has only magnitude but no direction e.g temperature, density and energy,  while a vector field is the field that has both magnitude and direction e.g magnetic field , electric field and gravitational field.

GRAVITATIONAL FIELD

This is a region or space around a mass in which the gravitational force of the mass can be felt. It is a force field which acts over a distance and surrounds every object that has mass. Gravitational field permeates all of space.

            The earth attracts every object existing in its gravitational field. This attraction is gravitational attraction whose effect is to change the velocity of objects under its influence.

Acceleration due to gravity

            The acceleration of object due to the gravitational attraction of the earth is called the acceleration due to gravity represented by the symbol  ‘g’ with average value 9.81ms^-2.

            The value of g is the same for all bodies at a give place but varies from place to place, being minimum at the equator (9.78ms^-2) and maximum atbthe poles (9.83ms^-2).

The gravitational force of attraction is given by:

F = mg, when m = 1,  F = g.  This is known as the acceleration of free fall (g) due to gravity. It is the force of attraction on a unit mass.

            In the absence of friction and air resistance, all bodies fall with the same acceleration irrespective of their masses.

Gravitational Force between two masses

            Gravitation is a force of attraction exerted by a body on all other bodies in the universe. It acts between all masses and holds together; planets, stars and galaxies.

            Sir Isaac Newton proposed a famous law called Newton’s law of universal gravitation and it states that:

Every particle in the universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

Mathematically,

m₁  ʘ―――→F                                                   F←―――ʘ m₂

r

   F α m₁m₂ ,      F = Gm₁m₂  

             r²                         r²

‘G’   is the gravitational constant, m₁ and m₂ are the masses of the two particles, r is the distance between their centres. ‘G’ is measured in Nm²kg^-2 (6.67 x 10^-11Nm²kg^-2).

            The law shows that bodies which are closer together would attract each other more than those that are farther apart.

Gravitational attraction keeps the moon in its orbit around the earth and the earth in its orbit around the sun. Gravitational forces  are always forces of attraction.

Example:

Determine the force of attraction between the sun(mass 1.99 x 10³⁰kg) and the earth(mass 5.98 x 10²⁴kg) if the sun is 1.50 x 10⁸km from the earth

 (G = 6.67 x 10^-11Nm²kg^-2)

Solution:   F = Gm₁m₂ ,    where r= 1.50 x 10⁸ x 10³m

                               r² 

                     = 6.67 x 10^-11 x 1.99 x 10³⁰ x 5.98 x 10²⁴

                                           (1.50 x 10⁸ x 10³)²

                      =  3.52 x 10²²N

ASSIGNMENT

(1) Find the gravitational force of attraction between the electron of a hydrogen atom(mass 9.0 x 10^-31kg) and the proton(mass 1.67 x 10^-27kg) if the electron orbits the proton at its average distance of 5.3 x 10^-11m.

(2) Calculate the gravitational force of attraction between two protons each of mass 1.67 x 10 ^-27kg in a chemical compound whose distance apart is 10^-10m

WEEK  TWO

Relation between gravitational constant ‘G’ acceleration due to gravity ‘g’

Let r_e  be the radius of the earth and m_e be its mass. If m is the mass of a body on the earth’s surface, the force which tends to pull the body towards the centre of the earth i.e its weight (mg) has a magnitude W = F. The gravitational attraction of the earth on the body is  F = Gm_em ,  

                                                                               r_e²

The force of gravity on a body is expressed as  F = mg.

Therefore, it follows that:

               F = mg =  Gm_em

                                    r_e²

The force per unit mass is given by:

               F = g = Gm_e

              m            r_e² 

Hence   g = Gm_e

                      r_e²

It means that the acceleration due to gravity  ‘g’  can be referred to as the force per unit mass on the surface of the earth.

Example:     

Determine the mass of the earth whose radius is approximately 6.38 x 10⁶m,

 G = 6.67 x 10-11Nm²kg^-2 and g = 9.8ms^-2

Solution:

From g = Gm_e

                  r_e²

          m_e = gr_e²        =    9.8 x (6.38 x 10⁶)²

                   G                    6.67 x 10^-11

                               =  5.98 x 10²⁴kg

Gravitational Potential

Gravitational potential at a point is defined as the work done in taking a unit mass from infinity to that point. The unit is Jkg^-1.

The gravitational potential V is given by:   

                         V = -Gm  

                                   r

‘m’ is the mass producing the gravitational field, r is the  distance of the point to the mass. As r  increases, the gravitational potential decreases, and becomes zero when r is infinitely large. The negative sign in the equation shows that the potential at infinity (i.e zero) is higher than the potential close to the mass.

Example:  Find the gravitational potential at a point on the earth’s surface. Take the mass of the earth as 5.98 x 10²⁴kg, its radius as 6.38 x 10⁶m and

 G = 6.67 x 10^-11Nm²kg^-2.

Solution:

   V = Gm

            r

      =  6.67 x 10^-11 x 5.98 x 10²⁴

                       6.38 x 10⁶

          

           =   6.25 x 107Jkg^-1                 

Escape Velocity

            One common feature of man-made satellites that circle round the earth is that they are held in an approximately circular path by the earth’s gravitational pull. This force provides the needed centripetal force required to keep the satellites in their orbits.

            The centripetal force which is equal to the gravitational force is given by:

mv_s² =  F =  Gm_em   ( where  v_s  is the velocity of the satellite as it

  r_e                   r_e²         moves round the earth)

 Hence   V_s = √(Gm_e/r_e).

For a satellite to escape from the earth and never return, it must be launched with a velocity greater than that required to make it orbit.

Escape Velocity  V_e  is the minimum velocity required for an object (satellite or rocket) to just leave or escape the gravitational influence(or field) of an astronomical body(e.g earth) permanently.

As the mass moves away from the field at the surface, its kinetic energy decreases because its gravitational potential increases.

An object will escape if the K.E is just enough to provide the needed potential energy.

The work done to carry a mass m away from the centre of the earth is

force x distance.

 W  =  Gm_em  x  r   =   Gm_em

               r_e²                      r_e

This work done equals the K.E of the body of mass m given by  K.E = ½mv_e²

Thus   ½mv_e²  =  Gm_em

                               r_e

              V_e  =  √[2Gm_e/r_e]     r_e is the radius of the earth and m_e = gr_e²/G

Then     V_e = √[2G x gr_e²/r_e x G]   = √(2gr_e), r_e is the radius of the earth denoted R

Hence   V_e = √(2gR)

Example: Calculate the escape velocity of a satellite from the earth’s gravitational field if g = 9.8ms^-2 and earth’s radius is 6.4 x 10⁶m.

Solution:

V_e  =  √(2gR)

      =  √(2 x 9.8 x 6.4 x 10⁶)

      =  11 x 10³ms^-1

The satellite must be launched with a velocity greater than 11kms^-1 to escape the gravitational influence of the earth

ASSIGNMENT

1. The average radius of Jupiter’s orbit round the sun of mass 2 x 10³⁰kg is 7.8 x 10¹¹m. If Jupiter is 1.9 x 10⁷kg, find the gravitational force the sun exerts on Jupiter. (G = 6.67 x 10^-11Nm2kg^-2)

2. The escape velocity of a rocket from the earth is 11kms^-1. Calculate the radius of the earth if the acceleration due to gravity is 9.8ms^-2

3. The radius of the moon is one-fourth, and its mass is one eighty-first that of the earth. If the acceleration due to gravity on earth is 9.8ms-2; what is its value on the surface of the moon?

4. On a mission to the planet Mars, of radius r_m = 3.4 x 10⁶m  and mass

m_m =   6.42 x 10²³kg. The earth weight of Mars lander is 39200N, calculate its weight F_g and the acceleration g_m due to gravity:

i. 6 x 10⁶m above the surface of Mars

ii. at the surface of Mars (neglect gravitational effects of other planetary bodies around Mars