Gravitational Field, Free Fall and Acceleration Due to Gravity for SS 1

Gravitational Field

Performance Objectives

By the end of the lesson, students should be able to:

1. explain gravitational field;
2. explain why two solid bodies of different masses released from rest at the same point simultaneously fall to the ground at the same time;
3. describe the shape of the Earth;
4. determine acceleration due to gravity from experiments.

What is a Gravitational Field?

A gravitational field is a region of space around a mass where another mass experiences a force of attraction.

Every object with mass creates a gravitational field around it. The Earth, being a very massive object, creates a significant gravitational field that extends far into space. This field is responsible for keeping objects on the ground, making things fall, and holding the Moon in orbit.

The strength of a gravitational field is represented by gravitational field lines. These lines show the direction of the gravitational force on a small test mass placed in the field. For the Earth, these lines are generally directed towards the centre of the Earth.

Acceleration Due to Gravity (g)

Acceleration due to gravity, denoted by ‘g’, is the acceleration experienced by an object solely due to the gravitational pull of a celestial body, such as the Earth.

When an object is dropped near the Earth’s surface, it falls towards the ground. This falling motion is an example of acceleration due to gravity. The value of ‘g’ is approximately 9.8 m/s² on the Earth’s surface. For simpler calculations, it is often approximated as 10 m/s².

Free Fall

Free fall is the motion of an object where gravity is the only force acting upon it. In a vacuum, all objects in free fall accelerate downwards at the same rate, regardless of their mass or size.

Imagine dropping a heavy stone and a light feather at the same time from the same height in a place with no air. Both objects would hit the ground at the exact same moment. This is because the acceleration due to gravity (‘g’) is constant for all objects near the Earth’s surface, assuming air resistance is negligible.

The gravitational force acting on an object is directly proportional to its mass (F = mg). However, according to Newton’s second law of motion (F = ma), the acceleration (a) is given by a = F/m. When we substitute F = mg into this, we get a = mg/m = g. This shows that the mass of the object cancels out, meaning that all objects experience the same acceleration ‘g’ in a gravitational field, irrespective of their individual masses.

Newton’s Law of Universal Gravitation

Newton’s Law of Universal Gravitation describes the attractive force between any two objects with mass.

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

This law can be expressed mathematically as:

F = G * (m₁m₂ / r²)

Where:

1. F is the gravitational force between the two objects.
2. G is the universal gravitational constant.
3. m₁ is the mass of the first object.
4. m₂ is the mass of the second object.
5. r is the distance between the centres of the two objects.

This law explains why the Earth pulls objects towards its centre and why planets orbit the Sun. The acceleration due to gravity ‘g’ on Earth’s surface is a direct consequence of this universal law, depending on the Earth’s mass and radius.

Shape and Dimension of the Earth

The Earth is not a perfect sphere; its shape is best described as an oblate spheroid.

An oblate spheroid is a sphere that is slightly flattened at the poles and bulges out at the equator. This shape is a result of the Earth’s rotation. The centrifugal force caused by rotation is greatest at the equator, causing the Earth to expand slightly there.

Due to this shape, the distance from the Earth’s centre to the surface is slightly greater at the equator than at the poles. This difference in distance, combined with the effect of the Earth’s rotation, causes the acceleration due to gravity (‘g’) to be slightly less at the equator (approximately 9.78 m/s²) and slightly greater at the poles (approximately 9.83 m/s²).

Motion Under Gravity (Free Fall Kinematics)

When objects move vertically under the influence of gravity, their motion can be described using the equations of motion for constant acceleration. In this case, the acceleration ‘a’ is replaced by ‘g’ (acceleration due to gravity).

The standard equations of motion are:

1. v = u + at
2. s = ut + ½at²
3. v² = u² + 2as

For motion under gravity, these equations become:

1. v = u + gt
2. s = ut + ½gt²
3. v² = u² + 2gs

Where:

1. u is the initial velocity.
2. v is the final velocity.
3. g is the acceleration due to gravity (approximately 9.8 m/s² or 10 m/s²).
4. t is the time taken.
5. s is the displacement (height or distance fallen).

It is important to choose a consistent direction for positive and negative values. Usually, upward motion is taken as positive and downward motion as negative, or vice versa. If downward is positive, then ‘g’ is positive. If upward is positive, then ‘g’ is negative.

Worked Examples

Example 1: A stone is dropped from a height of 45 m. Calculate the time it takes to reach the ground. (Take g = 10 m/s²)

Solution:

1. Given: Initial velocity (u) = 0 m/s (since it is dropped), displacement (s) = 45 m, acceleration (g) = 10 m/s².
2. We need to find time (t).
3. Using the equation: s = ut + ½gt²
4. 45 = (0 * t) + ½ * 10 * t²
5. 45 = 5t²
6. t² = 45 / 5
7. t² = 9
8. t = √9
9. t = 3 seconds

The stone takes 3 seconds to reach the ground.

Example 2: An object is thrown vertically upwards with an initial velocity of 20 m/s. Calculate the maximum height it reaches. (Take g = 10 m/s²)

Solution:

1. Given: Initial velocity (u) = 20 m/s, final velocity (v) = 0 m/s (at maximum height), acceleration (g) = -10 m/s² (since it’s moving against gravity).
2. We need to find the maximum height (s).
3. Using the equation: v² = u² + 2gs
4. 0² = 20² + 2 * (-10) * s
5. 0 = 400 – 20s
6. 20s = 400
7. s = 400 / 20
8. s = 20 metres

The maximum height reached by the object is 20 metres.

Determining Acceleration Due to Gravity Experimentally

The acceleration due to gravity (‘g’) can be determined through various experiments. Two common methods involve using a ticker-timer and a simple pendulum.

Method 1: Determining ‘g’ Using a Ticker-Timer

A ticker-timer is a device that makes dots on a paper tape at regular, short intervals. By analysing the spacing of these dots when a mass falls, the acceleration due to gravity can be calculated.

The required materials are:

1. Ticker-timer
2. Power supply (12 V AC)
3. Ticker tape
4. G-clamp or retort stand
5. Heavy mass (e.g., a 50 g or 100 g slotted mass)
6. Metre rule

The procedure involves the following steps:

1. Set up the ticker-timer and connect it to a 12 V AC power supply.
2. Thread a long piece of ticker tape through the timer.
3. Attach a heavy mass to one end of the ticker tape.
4. Hold the ticker tape vertically so that the mass is just below the timer and the tape is straight.
5. Switch on the ticker-timer and immediately release the mass, allowing it to fall freely.
6. After the mass has fallen a sufficient distance, switch off the timer and detach the tape.
7. Examine the ticker tape. The first few dots may be too close together due to initial slow motion or uneven release. Discard these and choose a clear starting point (mark it as ‘0’).
8. Count a convenient number of spaces (e.g., 10 spaces) from the starting point and mark the end of this interval. This represents a time interval (t₁). Measure the length of this interval (s₁).
9. Count another equal number of spaces (e.g., the next 10 spaces) and mark the end of this second interval. This represents a second time interval (t₂). Measure the length of this interval (s₂).
10. Calculate the time for one space. If the ticker-timer makes 50 dots per second, then the time for one space (period) is 1/50 = 0.02 seconds. If you counted 10 spaces, then t₁ = t₂ = 10 * 0.02 = 0.2 seconds.
11. Calculate the average velocity for the first interval (v₁) using v₁ = s₁ / t₁.
12. Calculate the average velocity for the second interval (v₂) using v₂ = s₂ / t₂.
13. The acceleration ‘g’ can then be calculated using the formula: g = (v₂ – v₁) / (t₂ – t₁). Since t₁ and t₂ are consecutive equal time intervals, the total time difference between the midpoints of these intervals is simply t₁ (or t₂). So, g = (v₂ – v₁) / t.

Safety Precautions:

1. Ensure the ticker-timer is connected to an AC power supply, not DC.
2. Handle the heavy mass carefully to avoid dropping it on feet.
3. Keep the work area clear to prevent tripping hazards.

Method 2: Determining ‘g’ Using a Simple Pendulum

A simple pendulum consists of a small, heavy mass (pendulum bob) suspended from a fixed point by a light, inextensible string. The period of oscillation of a simple pendulum depends on its length and the acceleration due to gravity.

The required materials are:

1. Pendulum bob
2. String (about 1 metre long)
3. Retort stand with a clamp
4. Stopwatch
5. Metre rule

The procedure involves the following steps:

1. Set up the retort stand and clamp. Suspend the pendulum bob from the clamp using the string. Ensure the string is long enough, about 80-100 cm.
2. Measure the length (L) of the pendulum from the point of suspension to the centre of the pendulum bob.
3. Displace the pendulum bob slightly from its equilibrium position (about 5-10 degrees) and release it to oscillate freely. Ensure it swings in a single plane and does not wobble.
4. Allow the pendulum to make a few oscillations to settle into a steady swing.
5. Start the stopwatch when the pendulum bob passes a fixed reference point (e.g., its lowest point) in a particular direction. Count this as ‘0’ oscillation.
6. Count 20 complete oscillations and stop the stopwatch precisely when the pendulum bob passes the reference point for the 20th time in the same direction.
7. Record the time (t) for 20 oscillations.
8. Repeat steps 5-7 at least two more times to get an average time for 20 oscillations.
9. Calculate the period (T) of one oscillation by dividing the average time for 20 oscillations by 20 (T = t / 20).
10. Repeat the entire experiment for at least five different lengths of the pendulum (L), varying the length by about 10 cm each time.
11. For each length, calculate T and then T².
12. Plot a graph of T² (y-axis) against L (x-axis). The graph should be a straight line passing through the origin.
13. Determine the gradient (slope) of the graph. The gradient (m) = T²/L.
14. The formula relating the period of a simple pendulum to its length and ‘g’ is T = 2π√(L/g).
15. Squaring both sides gives T² = 4π²L/g.
16. Rearranging for ‘g’, we get g = 4π²L/T².
17. Since the gradient of the T² vs L graph is T²/L, then g = 4π² / gradient.
18. Substitute the value of the gradient into this formula to calculate ‘g’.

Safety Precautions:

1. Ensure the retort stand is stable and placed on a firm surface to prevent it from toppling over.
2. Do not displace the pendulum bob too much, as large angles can affect the accuracy of the simple pendulum formula.
3. Ensure the string is securely tied to both the clamp and the bob to prevent it from falling during oscillations.