## Key Concepts:

- Gravitational acceleration
- Variation in gravitational acceleration
- Weight

### Introduction:

Suppose an object of mass m is kept on the surface of the Earth, then according to the universal law of gravitation, the force of gravity acting on the body F_{F}_{ }can be given as follows.

Mass of the earth = m_{E}

The radius of the earth = r_{E}

Mass of the object = m

According to the universal law of gravitation the gravitational force on the mass m due to the gravitational pull of the Earth = F_{G}

Gravitational force on the object = F_{G} = G mm_{E} / r^{2}_{E}

**We know from Newton’s second law = Force = mass x acceleration**

**We can write**

**Gravitational force = mass x gravitational acceleration**

We denote gravitational acceleration = g

F_{G} = mg

mg = G mm_{E} / r^{2}_{E}

g = GmE / r^{2}_{E}

**Gravitational acceleration = g = ** **GmE / r ^{2}_{E}**

Mass of the earth = m_{E}_{ }= 6 x 10^{24 }kg

Radius of the earth = r_{E}_{ }= 6.4 x 10^{3} km

Gravitational constant = G = 6.7 x 10^{-11} Nm^{2}/kg^{2 }

g = **GmE / r ^{2}_{E}**

g = 6.7 ×10^{−11}× 6 × 10^{24} = 9.8 m/s^{2}

Gravitational acceleration on the earth =** g = 9.8 m/s**^{2}

## Explanation:

- The below image is taken from a multi-flash photograph showing the pattern of movement when an object falls on the surface of the earth.

- There are seven images of the ball, taken at equal intervals of time. The ball falls further in each successive time interval. This shows that its speed is increasing.
- This ball falling under gravity is executing uniformly accelerated motion.

**Example – An apple falling under gravity **

**Acceleration due to gravity does not depend upon mass.**

The acceleration due to gravity is independent of the mass of the object falling towards the center of the earth. If air resistance and friction are negligible (or in vacuum) both the object fell at the same rate.

We are so accustomed to the effect of air resistance and friction that we think that the light object (feather) falls slower than the heavy (hammer) one. If light and a heavy object dropped from the same height simultaneously, they will reach the ground at the same time.

The earth applies gravitational force on all objects and objects also apply the same gravitational force on the earth then why do we not see the Earth be moving towards the mango or the moon?

From Newton’s 2^{nd} law for a given force, the acceleration is proportional to the mass of the object.

I.e., a α 1/m

m_{mango} <<< m_{earth}

Thus, a_{mango} >>> a_{earth}

The Earth also moves towards the mango as a result of the gravitational pull of the mango on the Earth. However, the acceleration of the earth towards the mango is negligibly small and hence, not apparent.

We do not see the Earth moving towards the mango or the ball similarly, the mass of the Moon is very small as compared to that of the Earth.

m_{moon} << m_{earth}

Thus, a_{moon} >> a_{earth}

**Weight of a body:**

The earth attracts everybody towards its centre with a certain force that depends on the mass of the body and the gravitational acceleration at that place. Weight is determined by the force it is attracted towards the centre of the earth.

Weight of a body on earth = Gravitational force exerted by the earth on the body

W = mass of the body x gravitational acceleration

W = mg

Weight is measured using a spring balance.

**Variation in g due to shape of the earth:**

Earth is elliptical in shape. It is flattened at the poles and bulged out at the equator. The equatorial radius is about 21 km longer than the polar radius.

Mass of the earth = M

Gravitational acceleration = g = GM/R^{2}

Gravitational acceleration at equator = g_{e} = GM**/R ^{2}**

_{ e}

Gravitational acceleration at pole = g_{p} = GM/**R ^{2}**

_{p}

Since R_{e} > R_{p} therefore g_{equator}_{ }< g_{pole} and approximately g_{e} + 0.018 m/s^{2 }= g_{p}

Therefore, the weight of a body increases as it is taken from equator to the pole.

It has been experienced when astronauts land on the surface of the moon they feel less weight.

**Weight on the surface of the Moon:**

If we take an object to the moon, it will be weightless, because the Moon’s gravity is weaker than the Earth’s as comparatively its mass is small. However, the mass of the object will remain the same.

Size of Earth = 4 x Size of Moon

Gravitational Acceleration on Earth = g_{e}

Gravitational Acceleration on Moon = gm=1/6g_{e}

gm = 1/6g_{e}= 9.8/6 = 1.66 m/s^{2}

Weight on Earth = mg_{e}

Weight on Moon = mg_{m}

Weight on Moon = (1/6) × Weight on Earth

**Question:**

What would be the weight of a person on the Moon and Earth whose mass is 12 kg?

**Answer:**

Weight of the person on earth = mg_{E}_{ } = 12 kg × 9.8 m/ s^{2}

= approximately 120 N

Weight of the person on moon = mg_{m} = 120 N/6 = 20 N

## Summary:

- According to the universal law of gravitation the gravitational force on the mass m due to the gravitational pull of the Earth =
- According to Newton’s second law-: Force = mass x acceleration
- Gravitational force = mass x gravitational acceleration
- We denote gravitational acceleration = g

F_{g} = mg

mg = G mm_{E} / **r ^{2}**

_{E}

Gravitational acceleration = g = Gm_{E}**/r ^{2}**

_{E}

- Gravitational acceleration on the earth = g = 9.8 m/s?
- The acceleration due to gravity is independent of the mass of the object falling towards

the canter of the earth. If air resistance and friction are negligible (or in a vacuum) all

objects fall down at the same rate. - Weight of a body on earth = Gravitational force exerted by the earth on the body
- W= mass of the body x gravitational acceleration
- W=mg
- Earth is elliptical in shape. It is flattened at the poles and bulged out at the equator. The

equatorial radius is about 21 km longer than the polar radius; therefore, the weight of a

body increases as it is taken from the equator to the pole. - If we take an object to the moon, it will weightless, because the Moon’s gravity is weaker

than the Earth’s as comparatively its mass is small. However, the mass of the object will

remain the same. - Weight on Moon = (1/6) X Weight on Earth

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