9th Class Physics Chapter 5 Notes are the perfect starting point if Unit 5, Gravitation, seems difficult to understand. This chapter explains why objects fall toward the Earth, how the Earth’s mass is calculated, and how artificial satellites stay in orbit.
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These 9th Class Physics Chapter 5 Notes cover every definition, formula, and numerical from Unit 5 in easy language. Whether you are revising for a test or preparing for the final exam, this guide gives you the complete chapter in one place.
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What Chapter 5 Gravitation Covers
Chapter 5 explains the force of attraction between objects and how this force governs the motion of planets and satellites. The main topics included in these 9th Class Physics Chapter 5 Notes are:
- Newton’s law of universal gravitation
- Gravitational field and field strength
- Mass of the Earth and how it is calculated
- Variation of g with altitude
- Artificial satellites and their orbital speed
Newton’s Law of Universal Gravitation
Newton’s law of universal gravitation states that every body in the universe attracts every other body with a force. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
Mathematically, this relationship is written as:
F = G (m1 m2) / d²
Here, G is the universal gravitational constant, with a value of 6.673 × 10⁻¹¹ Nm²kg⁻². Because G is so small, the gravitational force between everyday objects is too weak to notice, while the Earth’s large mass produces a significant pull on nearby objects.
Law of Gravitation and Newton’s Third Law
The gravitational force between two masses always acts as an action-reaction pair. If mass m1 attracts m2 with a force F, then m2 also attracts m1 with an equal and opposite force.
This behavior matches Newton’s third law of motion, which states that every action has an equal but opposite reaction, confirming that gravitational attraction always acts in pairs.
Gravitational Field and Field Strength
A gravitational field is the region around the Earth where its gravitational pull acts on any object, whether or not that object is in direct contact with the Earth. This is why a ball thrown upward continues to slow down even while in the air.
The gravitational force per unit mass at a point is called the gravitational field strength. Near the surface of the Earth, this value is 10 N kg⁻¹, which is the same as the standard value of g.
How the Mass of the Earth Is Determined
These 9th Class Physics Chapter 5 Notes explain how scientists calculate the mass of the Earth using the relationship between weight and gravitational force.
Since the weight of a body equals both mg and GmMe/R², the two expressions can be set equal to find:
Me = R²g / G
Using known values of the Earth’s radius, gravitational acceleration, and the gravitational constant, this formula gives a mass of approximately 6.0 × 10²⁴ kg for the Earth.
Variation of g With Altitude
The value of g, the acceleration due to gravity, does not stay constant everywhere. It decreases as altitude, the height above sea level, increases, because g is inversely proportional to the square of the distance from the Earth’s centre.
Key Points on Altitude and g
According to these 9th Class Physics Chapter 5 Notes, some important results include:
- At a height equal to one Earth radius above the surface, g becomes one-fourth of its surface value
- At a height equal to two Earth radii, g becomes one-ninth of its surface value
- g is greater at sea level than on hills or mountains
The formula used to calculate g at any altitude h is:
gh = G Me / (R + h)²
Artificial Satellites Explained
A satellite is any object that revolves around a planet. The moon is a natural satellite of the Earth, while objects launched into space by scientists for communication or research purposes are called artificial satellites.
How Gravitation Explains Satellite Motion
A satellite needs a centripetal force to stay in its orbit around the Earth, and this centripetal force is provided entirely by gravitational attraction. Setting the centripetal force equal to the satellite’s weight gives the orbital velocity formula:
vo = √(gh (R + h))
For a satellite orbiting very close to the Earth, where R is much greater than h, this simplifies to vo = √(gR), giving an orbital speed of nearly 8 km/s or 29,000 km/h.
Geostationary Satellites
Communication satellites are placed in a special orbit called a geostationary orbit, positioned at an altitude where they take exactly 24 hours to complete one revolution. Since the Earth also completes one rotation in 24 hours, these satellites appear stationary relative to a fixed point on Earth, allowing dish antennas to remain fixed in one direction.
Important Numerical Problems from Chapter 5
Practicing numericals is essential for exam preparation, and these 9th Class Physics Chapter 5 Notes summarize the key solved examples from the exercise:
- The gravitational force between two 1000 kg spheres placed 0.5 m apart is 2.67 × 10⁻⁴ N.
- The value of g at an altitude of 1000 km, using Earth’s mass of 6.0 × 10²⁴ kg and radius 6400 km, comes out to 7.3 ms⁻².
- The acceleration due to gravity on the surface of Mars, with mass 6.42 × 10²³ kg and radius 3370 km, is approximately 3.77 ms⁻².
- A polar satellite launched at 850 km above the Earth has an orbital speed of about 7431 ms⁻¹.
Why These Notes Matter for Exam Preparation
Chapter 5 combines conceptual questions about gravitational force with numerical problems involving g, altitude, and satellite motion. Short questions often ask students to define gravitational force or field strength, while long questions require deriving the mass of the Earth or the orbital velocity formula.
Using these 9th Class Physics Chapter 5 Notes alongside your textbook makes revision faster since every law, formula, and solved numerical is organized in one place. Students preparing for board exams should also check [https://taleemworld.com/] for a complete chapter-wise list of 9th class physics notes.
FAQs
Q1: What is covered in 9th Class Physics Chapter 5 Notes? These notes cover Newton’s law of gravitation, gravitational field strength, mass of the Earth, variation of g with altitude, and artificial satellites, along with solved numerical problems from the chapter.
Q2: What is Newton’s law of gravitation in simple words? Newton’s law of gravitation states that every body attracts every other body with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.
Q3: What is the value of the universal gravitational constant G? The value of G is 6.673 × 10⁻¹¹ Nm²kg⁻². It is the same everywhere in the universe, but its small value means the gravitational pull between everyday objects is too weak to feel.
Q4: Why does the value of g decrease with altitude? The value of g decreases with altitude because it is inversely proportional to the square of the distance from the Earth’s centre. As altitude increases, this distance increases, so g becomes smaller.
Q5: What is a geostationary satellite? A geostationary satellite orbits the Earth once every 24 hours, matching the Earth’s own rotation. This makes the satellite appear stationary from a fixed point on Earth, which is ideal for communication purposes.
Q6: How is the mass of the Earth calculated? The mass of the Earth is calculated using the formula Me = R²g / G, where R is the Earth’s radius, g is the acceleration due to gravity, and G is the universal gravitational constant.