Term: 2^{nd} Term
Week: 3
Class: Senior Secondary School 1
Age: 15 years
Duration: 40 minutes of 5 periods each
Date:
Subject: Physics
Topic: Gravitational Field
SPECIFIC OBJECTIVES: At the end of the lesson, pupils should be able to
INSTRUCTIONAL TECHNIQUES: Identification, explanation, questions and answers, demonstration, videos from source
INSTRUCTIONAL MATERIALS: Videos, loud speaker, textbook, pictures
INSTRUCTIONAL PROCEDURES
PERIOD 12
PRESENTATION 
TEACHER’S ACTIVITY 
STUDENT’S ACTIVITY 
STEP 1 INTRODUCTION 
The teacher reviews the previous lesson on field 
Students pay attention 
STEP 2 EXPLANATION 
He explains the meaning of gravitational field

Students pay attention and participates 
STEP 3 DEMONSTRATION 
He states and explains Newton’s law of universal gravitation

Students pay attention and participate 
STEP 4 NOTE TAKING 
The teacher writes a summarized note on the board 
The students copy the note in their books 
NOTE
GRAVITATIONAL FIELD
ACCELERATION DUE TO GRAVITY (g)
The force of gravity is the pull of attraction between the earth and objects on the surface of the earth. The value of acceleration due to gravity is constant in the region of the laboratory.
g = GM/r^{2}
Where
G= Gravitational constant
M= Mass of the earth
R= Radius of the earth
g = Approximately 9.8ms^{2}
NEWTON’S LAW OF UNIVERSAL GRAVITATION
Newton’s law of universal gravitation states that every particle of matter 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 radius between their centers.
Mathematically,
F = Gm_{1}m_{2}/r^{2}
Where,
Solved Examples
Given:
Mass of Earth (m_{1}) = 5.98 × 10^{24}kg
Mass of object (m_{2}) = 2000kg
The radius of the Earth (r)= 6.38 × 10^{6}m
Acceleration due to gravity (g) = 9.8 m/s^{2}
Universal constant (G) = 6.67 x 10^{11} N m^{2} / kg^{2}
Solution:
F = Gm_{1}m_{2}/r^{2}
F = ( 6.67 x 10^{11}) (5.98 × 10^{24})(2 x 10^{3})/(6.38 × 10^{6})^{2}
F = (7.978 x 10^{17})/ (4.07044 × 10^{13})
F = 1.959 x 10^{4 }or F = 19.59 N
Solution
Given:
Mass of Earth (m_{1}) = 5.98 × 10^{24}kg
Mass of object (m_{2}) = 1000kg
The radius of the Earth (r)= 6.38 × 10^{6}m
Acceleration due to gravity (g) = 9.8 m/s^{2}
Universal constant (G) = 6.67 x 10^{11} N m^{2} / kg^{2}
h = 2 x 10^{4} m
F = Gm_{1}m_{2}/(r +h)^{2}
F = ( 6.67 x 10^{11}) (5.98 × 10^{24})(1 x 10^{3})/(6.38 × 10^{6} + 2 x 10^{4} )^{2}
F = (3.988 x 10^{17})/(4.058 x 10^{13})
F = 9,827.50
F = 0.9827 x 10^{4}
EVALUATION: 1. Define acceleration due to gravity and state its formula
CLASSWORK: As in evaluation
CONCLUSION: The teacher commends the students positively