Acceleration Due to Gravity (Earth’s Gravitational Acceleration)

• The Earth exerts a gravitational force on all objects near its surface.
• According to Newton's Second Law of Motion (F = ma), this force causes a body to accelerate.
• This specific acceleration caused by the Earth's gravity is called acceleration due to gravity.
• It is a vector quantity, denoted by the letter 'g', and is always directed vertically downwards towards the Earth's center.

The gravitational force due to the earth on a body results in its acceleration. This is called acceleration due to gravity and is denoted by ‘g’.

OR

When a body falls towards the Earth under gravity, then the acceleration produced in the body due to gravity is called acceleration due to gravity, which is denoted by g.

The value of the acceleration due to gravity (g) on the surface of the Earth is given by the formula:

Where:

• g = Acceleration due to gravity (in m/s²).
• G = Newton's Universal Gravitational Constant (≈ 6.67 × 10⁻¹¹ N · m² / kg²).
• M = Mass of the Earth (≈ 6 × 10²⁴ kg).
• R = Radius of the Earth (≈ 6.4 × 10⁶ m).

Direction: Always directed towards the center of the Earth (vertically downwards).

Independence: The value of g at a given point does not depend on the properties or mass (m) of the object being attracted.

Variability: The value of g is not constant and varies with:

• Location (latitude) on the Earth's surface.
• Height (altitude) above the surface.
• Depth below the surface.

Highest/Lowest Value: Highest at the poles (≈ 9.832 m /s²) and lowest at the equator (≈ 9.78 m/s²).

The acceleration due to gravity (g) is derived by equating the general gravitational force and the force from Newton's Second Law.

1. Gravitational Force (F): According to Newton's Universal Law of Gravitation, the force between the Earth (Mass M) and an object (Mass m) at a distance r from the Earth's center is:
   \[F = \frac{G M m}{r^2} \quad \text{.........(1)}\]
2. Newton's Second Law Force (F): The force causing the object's acceleration (g) is:
   \[F = m g \quad \text{.........(2)}\]
3. Equating (1) and (2):
   \[m g = \frac{G M m}{r^2}\]
4. Solving for g: The mass of the object (m) cancels out:
   \[g = \frac{G M}{r^2} \quad \text{.........(3)}\]
5. On the Surface: If the object is on the Earth's surface, the distance r equals the Earth's radius R, leading to the final surface value formula:
   \[g = \frac{G M}{R^2} \quad \text{.........(4)}\]

• The Earth is flatter at the poles and bulges at the equator.
• The radius (R) is largest at the equator and smallest at the poles.
• Since g is inversely proportional to R², a smaller radius at the poles means a higher value of g (highest at the poles), and a larger radius at the equator means a lower value of g (lowest at the equator). 

Aim: To calculate the mass of the Earth (M_E) using the acceleration due to gravity (g).

Given Data:

• Acceleration due to gravity, g = 9.81 m/s²
• Radius of the Earth, R_E = 6.37 × 10⁶ m
• Universal Gravitational Constant, G = 6.67 × 10^⁻¹¹ N · m² /kg²

Steps:

1. Start with the formula for g on the surface:
   \[g = \frac{G M_E}{R_E^2}\]
2. Rearrange the formula to solve for M_E:
   \[M_E = \frac{g R_E^2}{G}\]
3. Substitute the given values into the equation:
   \[M_E = \frac{9.81 \times (6.37 \times 10^6)^2}{6.67 \times 10^{-11}}\]
4. Calculate the result:
   \[M_E = 5.97 \times 10^{24} \text{ kg}\]

Result: The calculated mass of the Earth is 5.97 × 10²⁴ kg.

Aim: To calculate the acceleration due to gravity on the Moon's surface (g_m).

Given Data (relative to Earth):

• Mass of the Moon (M_m) = M/80 (where M is Earth's mass).
• Radius of the Moon (R_m) = R/4 (where R is Earth's radius).
• Acceleration due to gravity on Earth, g = 9.8 m/s².

Steps:

1. Write the formulas for g on Earth and g_m on the Moon:
   \[g = \frac{G M}{R^2} \quad \text{...(1)}\]
   \[g_m = \frac{G M_m}{R_m^2} \quad \text{...(2)}\]
2. Find the ratio \[\frac {g_m}{g}\]:
   \[\frac{g_m}{g} = \frac{M_m}{M} \times \left(\frac{R}{R_m}\right)^2\]
3. Substitute the given ratios:
   \[\frac{g_m}{g} = \frac{1}{80} \times \left(\frac{4}{1}\right)^2\]
4. Simplify:
   \[\frac{g_m}{g} = \frac{1}{80} \times 16 = \frac{16}{80} = \frac{1}{5}\]
5. Solve for g_m:
   \[g_m = \frac{g}{5} = \frac{9.8}{5}\]
6. Calculate the result:
   \[g_m = 1.96 \text{ m/s}^2\]

Result: The acceleration due to gravity on the Moon's surface is 1.96 m/s² (about 1/5th of Earth's value).

• Weight on the Moon: You weigh 1/6 as much on the Moon because gravity there is weaker than on Earth.
• Dropping Objects: A small rock and a heavy ball fall at the same speed (without air) — gravity doesn’t depend on mass.
• Tidal Forces: Tides happen because the Moon pulls harder on the side of Earth closest to it, causing water to bulge.
• Satellite Orbits: Satellites need to be at the right height where gravity keeps them in orbit — not falling or flying away.

• Due to Altitude: The acceleration due to gravity decreases with altitude as we move away from the surface of the Earth. As h↑, g↓.
• Due to Depth: The acceleration due to gravity decreases as we move into the Earth's interior, i.e., with increasing depth. As d↑, g↓.
• Due to Latitude: The acceleration due to gravity increases with latitude. At the poles (θ = 90°), g = g_max⁡​; at the equator (θ = 0°), g = g_min​.
• Due to Shape of Earth: The equatorial radius of the Earth is greater than the polar radius. Since g ∝ \[\frac {1}{R^2}\], the acceleration due to gravity is greater at the poles compared to the equator, i.e., R_equator > R_pole​ and g_equator < g_pole​.