NEET UG - Universal law of gravitation, acceleration due to gravity and its variation with altitude and depth E-Book Study Material

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Universal Law of Gravitation

The universe is a vast expanse filled with celestial bodies, all of which exert a force on one another. This fundamental interaction is described by Newton's Universal Law of Gravitation. This 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 centers.

Mathematically, this can be expressed as:

$\mathbf{F} = G \frac{m_1 m_2}{r^2}$

Where:

F is the magnitude of the gravitational force between the two particles.
G is the universal gravitational constant. This constant is the same for all pairs of particles, regardless of their composition or the medium between them. Its value is approximately 6.674 × 10^-11 N m²/kg².
m₁ and m₂ are the masses of the two particles.
r is the distance between the centers of the two particles.

The force is always attractive and acts along the line joining the centers of the two particles.

Key Takeaway: The gravitational force weakens rapidly with distance (proportional to 1/r²). This is why we feel the Earth's gravity strongly but not the Sun's, even though the Sun is much more massive. The Sun's distance is immense.

The Gravitational Constant (G)

The value of G was first accurately measured by Henry Cavendish in 1798 using a torsion balance experiment. This experiment was crucial because it allowed scientists to determine the mass of the Earth by measuring the gravitational attraction between known masses. The value of G is very small, which explains why gravitational forces between everyday objects are negligible.

Properties of Gravitational Force

Attractive: It always pulls objects towards each other.
Universal: It acts between any two objects with mass.
Long-range: Its influence extends over vast distances, though it diminishes rapidly.
Central Force: It acts along the line joining the centers of the two masses.
Conservative Force: The work done by gravity depends only on the initial and final positions, not the path taken.
Proportional to the product of masses: More massive objects exert a stronger gravitational pull.
Inversely proportional to the square of the distance: Doubling the distance reduces the force to one-fourth.

Acceleration Due to Gravity (g)

When an object is near the Earth's surface, it experiences a gravitational force exerted by the Earth. According to Newton's second law of motion (F = ma), this force causes the object to accelerate. This acceleration is known as the acceleration due to gravity, denoted by 'g'.

We can derive the expression for 'g' by equating the gravitational force exerted by the Earth on an object of mass 'm' to the force causing acceleration 'g'. Let 'M' be the mass of the Earth and 'R' be its radius. The distance 'r' in the universal law of gravitation is approximately equal to 'R' for objects near the Earth's surface.

Gravitational Force, F = G * (M * m) / R²

According to Newton's second law, F = m * g

Equating the two expressions for F:

$m \mathbf{g} = G \frac{M m}{R^2}$

Canceling out the mass 'm' of the object from both sides, we get:

$\mathbf{g} = G \frac{M}{R^2}$

This formula shows that the acceleration due to gravity 'g' depends on the mass of the planet (M) and its radius (R), but not on the mass of the object itself.

The average value of 'g' on the Earth's surface is approximately 9.8 m/s². This means that, neglecting air resistance, any object falling freely near the Earth's surface will accelerate downwards at this rate.

Mnemonic for G: Think of "G" as the "Gravitational Glue" that holds the universe together. Its small value means this glue is weak for ordinary objects.

Mnemonic for g: "g" is for "gravity's pull" on objects near a planet.

Example: Calculating 'g' for the Moon

Mass of the Moon (M_Moon) ≈ 7.34 × 10²² kg

Radius of the Moon (R_Moon) ≈ 1.74 × 10⁶ m

Using G = 6.674 × 10^-11 N m²/kg²:

$\mathbf{g}_{\text{Moon}} = G \frac{M_{\text{Moon}}}{R_{\text{Moon}}^2} = (6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2) \frac{7.34 \times 10^{22} \text{ kg}}{(1.74 \times 10^6 \text{ m})^2}$

$\mathbf{g}_{\text{Moon}} \approx 1.62 \text{ m/s}^2$

As you can see, the acceleration due to gravity on the Moon is much less than on Earth, about one-sixth.

Variation of Acceleration Due to Gravity

The formula g = GM/R² applies to a uniform spherical body. However, in reality, the Earth is not a perfect sphere, and we often consider variations in 'g' with altitude and depth.

1. Variation with Altitude

As we move away from the Earth's surface to a higher altitude, the distance from the Earth's center increases. Let the altitude be 'h'. The new distance from the center will be (R + h).

The acceleration due to gravity at altitude 'h' is given by:

$\mathbf{g}_{\text{h}} = G \frac{M}{(R+h)^2}$

We can rewrite this in terms of g (acceleration at the surface):

$\mathbf{g}_{\text{h}} = G \frac{M}{R^2 \left(1 + \frac{h}{R}\right)^2} = \mathbf{g} \left(1 + \frac{h}{R}\right)^{-2}$

Using the binomial approximation (1 + x)ⁿ ≈ 1 + nx for small x (i.e., when h << R):

$\mathbf{g}_{\text{h}} \approx \mathbf{g} \left(1 - \frac{2h}{R}\right)$

This approximation shows that as altitude 'h' increases, the value of 'g' decreases. The decrease is approximately proportional to the altitude for small heights.

Altitude Shortcut: For heights much smaller than the Earth's radius (h << R), the change in 'g' is approximately Δg ≈ - (2h/R) * g. The negative sign indicates a decrease.

2. Variation with Depth

When an object is at a depth 'd' below the Earth's surface, its distance from the Earth's center is (R - d). However, the gravitational force on the object is only due to the mass of the Earth within the sphere of radius (R - d). The mass of the shell outside this radius exerts no net force on the object.

Let the density of the Earth be ρ. Assuming the Earth has uniform density (a simplification), the mass of the Earth (M) can be written as:

$M = \rho \times \frac{4}{3} \pi R^3}$

The mass of the inner sphere (of radius r = R - d) is:

$M' = \rho \times \frac{4}{3} \pi r^3 = \rho \times \frac{4}{3} \pi (R-d)^3}$

The acceleration due to gravity at depth 'd' (at a distance r = R - d from the center) is:

$\mathbf{g}_{\text{d}} = G \frac{M'}{r^2} = G \frac{\rho \times \frac{4}{3} \pi (R-d)^3}{(R-d)^2}$

$\mathbf{g}_{\text{d}} = G \times \rho \times \frac{4}{3} \pi (R-d)$

We know that at the surface, g = G * M / R² = G * (ρ * 4/3 * π R³) / R² = G * ρ * 4/3 * π R.

So, we can write:

$\mathbf{g}_{\text{d}} = \left( G \times \rho \times \frac{4}{3} \pi \right) (R-d) = \frac{\mathbf{g}}{R} (R-d)$

$\mathbf{g}_{\text{d}} = \mathbf{g} \left(1 - \frac{d}{R}\right)$

This formula shows that as depth 'd' increases, the value of 'g' decreases. At the center of the Earth (d = R), g_d becomes zero.

Depth Shortcut: For any depth 'd', the acceleration due to gravity is g_d = g * (1 - d/R). This means 'g' decreases linearly with depth.

Other Factors Affecting 'g'

While altitude and depth are the primary factors considered, 'g' also varies slightly due to:

Earth's Rotation: The Earth's rotation causes a centrifugal effect, which slightly reduces the effective acceleration due to gravity, especially at the equator. This effect is maximum at the equator and zero at the poles.
Earth's Shape: The Earth is not a perfect sphere; it is an oblate spheroid, bulging at the equator and flattened at the poles. This means the radius at the equator is larger than at the poles. Since g is inversely proportional to R², 'g' is slightly smaller at the equator than at the poles.
Local Density Variations: Differences in the density of the Earth's crust beneath a location can cause minor variations in 'g'.

Summary Table of Variations

Condition	Distance from Center	Expression for g	Effect on g
Surface	R	$g = G \frac{M}{R^2}$	Reference value
Altitude 'h' (h << R)	R + h	$g_h \approx g (1 - \frac{2h}{R})$	Decreases
Depth 'd'	R - d	$g_d = g (1 - \frac{d}{R})$	Decreases

Understanding the universal law of gravitation and the factors affecting acceleration due to gravity is crucial for comprehending the motion of celestial bodies, satellites, and objects on Earth.

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