# Weight

Show/Hide Sub-topics (Mass, Weight & Density | O Level Physics)

## Weight

All bodies of matter near the surface of Earth experience gravitational force due to Earth’s gravitational field. This gravitational pull is commonly referred to as the weight.

The weight (W) of a body is the gravitational force exerted on it by Earth.

$$W = mg$$

, where $m$ is mass (in kg), $g$ is gravitational field strength (in $\text{N kg}^{-1}$)

• SI unit of weight is newton (N). It is a vector quantity.
• Its direction is towards the centre of the Earth or commonly referred to as vertically downwards.
• Weight is measured using a newton-meter or by hanging the object on a spring balance (force meter) marked in newtons and letting the pull of the gravity stretch the spring in the balance. The greater the pull, the more the spring stretches.

## Gravitational Field & Weight Discrepancies On Earth

Measurement indicates that a mass of 1 kg, when placed on the Earth’s surface, encounters a force of 9.8 N, signifying its weight. The intensity of the Earth’s gravitational field is consequently 9.8 N/kg (approximately 10 N/kg), represented by the symbol g, which is also employed to signify the acceleration due to free fall. Therefore, $g = 9.8 \, \text{N kg}^{-1} = 9.8 \, \text{m s}^{-2}$.

• We currently have two perspectives on g.
• In the context of freely falling objects, we can perceive it as an acceleration of 9.8 m/s².
• Alternatively, when an object with a known mass is stationary and we want to determine the force of gravity (in N) acting on it, we view g as the Earth’s gravitational field strength, which is 9.8 N/kg. The gravitational field strength is synonymous with the acceleration experienced during free fall.

For an object above or on the Earth’s surface, the nearer it is to the centre of the Earth, the more the
Earth attracts it. Since the Earth is not a perfect sphere but is flatter at the poles, the weight of a
body varies over the Earth’s surface – at the north pole, one would weigh about 3 N heavier than one near the equator. This is due to the gravitational field strength being slightly DIFFERENT at different places on Earth. Normally, you neglect this difference in your calculations.

## Comparison Between Mass & Weight

The relationship between an object’s weight and its mass is directly proportional, elucidating the consistency of the gravitational field strength (g) for all objects. When an object possesses greater mass, it experiences a stronger gravitational force; however, this does not result in a faster acceleration during free fall. This is due to the object’s greater inertia, indicating its heightened resistance to acceleration. While the mass of an object remains constant, its weight fluctuates depending on the value of g. For instance, on the Moon, where the acceleration due to gravity is only about $1.6 \text{m s}^{2}$, a 1 kg mass would have a weight of merely 1.6 N.

### Weight Is An External Force

When a body is placed in a region of free space far away from any massive bodies, it experiences no gravitational pull and thus are considered weightless. When this same body is now placed near the surface of the Earth, the body experiences the pull of gravity. This shows that the weight must come from an external source. Hence, it is an external force.

## Worked Examples

### Example 1

The gravitational field strength on the surface of the Moon is one-sixth that of Earth’s. For a body of 60 kg, deduce its mass and weight on the Moon?

Mass of body on Moon = 60 kg. (Mass don’t change)

Earth’s gravitational field strength, $g_{Earth} = 10 \, \text{m s}^{-2}$

\begin{aligned}\text{Weight of body on moon} &= m \times g_{Moon} \\ &= (60)(\frac{10}{6}) \\ &= 100 \, \text{N}\end{aligned}

### Example 2

An object with a mass of 1 kg experiences a weight of 10 N at a specific location. Determine the weights for the following masses:

• $100 \, \text{g}$
• $5 \, \text{kg}$
• $50 \, \text{g}$

– $\text{Weight}_{100\,g} = 1 \, \text{N}$ (since $1 \, \text{kg} = 1000 \, \text{g}$)

– $\text{Weight}_{5\,kg} = 50 \, \text{N}$

– $\text{Weight}_{50\,g} = 0.5 \, \text{N}$

### Example 3

The gravitational force on the Moon is reported to be one-sixth of that on Earth. Calculate the weight of a mass of 12 kg:

• on Earth
• on the Moon

– $\text{Weight}_{\text{Earth}} = 120 \, \text{N}$ (since $\text{Weight} = \text{mass} \times \text{gravity}$ and $\text{gravity}_{\text{Earth}} = 10 \, \text{m/s}^2$)

– $\text{Weight}_{\text{Moon}} = 20 \, \text{N}$ (since $\text{gravity}_{\text{Moon}} = \frac{1}{6} \times 10 \, \text{m/s}^2$)

### Example 4

An astronaut possesses a mass of 80 kg.
a) Determine the astronaut’s weight on the Moon, where the gravitational field strength is 1.6 N/kg.
b) During the return journey to Earth, the astronaut reaches point X where the gravitational field strengths due to the Earth and the Moon are equal in magnitude but opposite in direction. State:
i) The resultant value of the gravitational field strength at X.
ii) The weight of the astronaut at X.

a) To calculate the weight $W$ on the Moon, we use the formula $W = m \times g$, where $m$ is the mass of the astronaut and $g$ is the gravitational field strength on the Moon.
\begin{aligned} W_{\text{Moon}} &= 80 \, \text{kg} \times 1.6 \, \text{N/kg} \\ &= 128 \, \text{N} \end{aligned}

b) At point X, where the gravitational field strengths due to the Earth and the Moon are equal in magnitude but opposite in direction:
i) The resultant gravitational field strength $g_{\text{X}}$ is zero since the Earth’s and Moon’s gravitational effects cancel each other.
ii) The weight of the astronaut at X, $W_{\text{X}}$ is zero, as there is no net gravitational force acting on the astronaut at this point.

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