2.6 Gravitational Force

Gravitational Interaction Between Objects

Newton's Law of Universal Gravitation

Newton's law of universal gravitation states that every particle in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This fundamental relationship is expressed mathematically as:

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

Where:

• $F_g$ is the gravitational force between objects
• $G$ is the universal gravitational constant ($6.67 \times 10^{-11} \text{ N} \cdot \text{m}^2/\text{kg}^2$)
• $m_1$ and $m_2$ are the masses of the two objects
• $r$ is the distance between the centers of mass of the objects

The gravitational force has several important characteristics:

• It always acts as an attractive force between objects with mass 🪐
• The force acts along the line connecting the centers of mass of the interacting objects
• For complex objects, the center of mass serves as the effective point where gravitational force acts

Gravitational Field Model

The gravitational field concept provides a way to visualize and calculate how gravity affects objects at different locations in space.

$\vec{g} = \frac{\vec{F}_g}{m}$

For a source mass $M$ at distance $r$, the magnitude of the gravitational field is:

$|\vec{g}| = \frac{|\vec{F}_g|}{m} = G\frac{M}{r^2}$

The field points toward the mass creating the field because gravity is attractive.

Where:

• $\vec{g}$ is the gravitational field vector at a point in space
• $\vec{F}_g$ is the gravitational force experienced by a test mass
• $m$ is the test mass placed at that point

This model helps us understand that:

• The gravitational field at any point equals the acceleration a test object would experience at that location
• The units of gravitational field strength are N/kg or m/s²
• The field extends throughout space, becoming weaker with distance but never reaching zero

Weight as Gravitational Force

Weight is the gravitational force exerted by a large astronomical body (like Earth) on a smaller object near its surface.

$W = F_g = mg$

This relationship shows that:

• Weight is proportional to mass
• Weight varies with location (due to variations in $g$)
• Weight is a force (measured in newtons), while mass is an intrinsic property (measured in kilograms)

Constant Gravitational Force

Negligible Change in Force

The gravitational force between two systems can be treated as constant when the change in distance between their centers of mass is negligible between the initial and final positions, so the value of $G\frac{m_1 m_2}{r^2}$ changes by only a very small amount.

For example, when analyzing the motion of objects near Earth's surface, we typically assume the gravitational force doesn't change with height for small vertical displacements. This simplification works because the change in distance from Earth's center is negligible compared to Earth's radius.

Earth's Gravitational Field Strength

Near Earth's surface, the gravitational field strength is approximately:

$g \approx 9.8 \text{ N/kg} \approx 10 \text{ N/kg}$ 🌍

This value varies slightly with:

• Latitude (due to Earth's rotation and non-spherical shape)
• Altitude (decreasing with height above sea level)
• Local geological features (affected by variations in density)

Apparent Weight vs Gravitational Force

Normal Force as Apparent Weight

The apparent weight of an object is what we perceive or measure, typically through the normal force exerted on a supporting surface.

When standing on a scale, the reading represents the normal force the scale exerts upward on you, which equals your apparent weight. This may or may not equal your true gravitational weight depending on your acceleration.

Acceleration Effects on Weight

When an object accelerates, its apparent weight differs from its true gravitational weight. Taking upward as positive, the vertical force equation is:

$F_N - mg = ma$

So $F_N = m(g + a)$ for upward acceleration and $F_N = m(g - a)$ for downward acceleration of magnitude $a$.

This explains why:

• You feel heavier in an elevator accelerating upward
• You feel lighter in an elevator accelerating downward
• You would feel weightless in free fall, even though gravity is still acting on you

Weightlessness Conditions

Objects appear weightless under specific conditions:

• When no forces are exerted on the object
• When gravity is the only force acting on the object, as in free fall 🪶
• In orbit, where the object is in continuous free fall around a planet

In these situations, there's no normal force to create the sensation of weight, even though gravitational force may continue to act.

Equivalence Principle

The equivalence principle states that an observer in a noninertial reference frame cannot distinguish between an object's apparent weight and the gravitational force exerted on the object by a gravitational field. This means:

• In an accelerating elevator, an observer may attribute the normal-force sensation to gravity-like effects of the noninertial frame.
• Without outside information, a person in a closed elevator cannot tell whether the apparent weight they feel is due to being at rest in a gravitational field or due to acceleration of the elevator.

Inertial vs Gravitational Mass

Inertia and Motion Resistance

Inertial mass measures an object's resistance to changes in motion, as described in Newton's Second Law:

$F = m_i a$

Where $m_i$ is the inertial mass. This property determines how much force is needed to accelerate an object.

Mass and Gravitational Attraction

Gravitational mass determines the strength of gravitational interaction between objects:

$F_g = G\frac{m_{g1} m_{g2}}{r^2}$

Where $m_{g1}$ and $m_{g2}$ are the gravitational masses of the two objects.

Equivalence of Mass Types

Numerous experiments have confirmed that inertial mass and gravitational mass are equivalent to extremely high precision. This equivalence is:

• A fundamental assumption in Newtonian mechanics
• A cornerstone of Einstein's general relativity
• Verified to approximately 1 part in 10¹³ by modern experiments

Gravitational Force of Spherical Mass

Net Force from Mass Distribution

The net gravitational force on an object from a distributed mass equals the vector sum of the gravitational forces from all the individual mass elements:

$\vec{F}_{net} = \sum \vec{F}_i = G\sum\frac{m_i M}{r_i^2}\hat{r}_i$

For complex shapes, this calculation often requires calculus and integration.

Newton's Shell Theorem

Newton's shell theorem provides powerful insights about gravitational interactions with spherical objects:

• For a point outside a uniform spherical shell: The gravitational force is identical to that of a point mass at the center with the shell's total mass
• For a point inside a uniform spherical shell: The net gravitational force is zero (forces from opposite sides cancel)
• For a point inside a solid sphere with uniform density: Only the mass closer to the center than the point contributes to the gravitational force

These principles allow us to simplify many gravitational calculations involving spherical bodies.

Force Inside Uniform Sphere

For a sphere of uniform density, the mass enclosed within radius $r$ is:

$m_{\text{partial}} = \rho \frac{4}{3}\pi r^3$

Only this enclosed mass contributes to the net gravitational force on an object located a distance $r$ from the center.

For a point at distance $r$ from the center of a uniform sphere of radius $R$ and mass $M$, when $r < R$:

$F_g = G\frac{M_r m}{r^2} = G\frac{m}{r^2}\left(\frac{r^3}{R^3}M\right) = G\frac{Mm}{R^3}r$

In restoring-force form, this can be written as $F_{g,\partial} = -kr$, where the negative sign indicates that the gravitational force points toward the center, opposite the displacement from the center.

This shows that inside a uniform sphere:

• The gravitational force is directly proportional to distance from the center
• The force decreases linearly to zero as you approach the center
• This creates a harmonic oscillator effect for objects moving through the sphere

Practice Problem 1: Gravitational Force Calculation

Solution

To find the gravitational force (weight) on Mars, we'll use Newton's law of universal gravitation:

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

Where:

• $m_1$ = mass of Mars = 6.42 × 10²³ kg
• $m_2$ = mass of astronaut = 70 kg
• $r$ = radius of Mars = 3.39 × 10⁶ m
• $G$ = 6.67 × 10⁻¹¹ N·m²/kg²

Substituting these values:

$F_g = (6.67 \times 10^{-11})\frac{(6.42 \times 10^{23})(70)}{(3.39 \times 10^6)^2}$

$F_g = (6.67 \times 10^{-11})\frac{4.494 \times 10^{25}}{1.149 \times 10^{13}}$

$F_g = (6.67 \times 10^{-11})(3.91 \times 10^{12})$

$F_g = 261 \text{ N}$

Therefore, the astronaut would weigh approximately 261 N on Mars, which is about 38% of their weight on Earth.

Practice Problem 2: Apparent Weight in an Elevator

Solution

When the elevator accelerates upward, the apparent weight increases. The scale reading equals the normal force, which we can find using Newton's Second Law.

First, identify the forces acting on the person:

• Weight (downward): $F_g = mg = (60 \text{ kg})(9.8 \text{ m/s}^2) = 588 \text{ N}$
• Normal force from scale (upward): $F_N$ (this is what we're solving for)

Since the person is accelerating upward at 2.0 m/s², we can write: $F_{net} = ma$ $F_N - F_g = ma$

$F_N - 588 \text{ N} = (60 \text{ kg})(2.0 \text{ m/s}^2)$

$F_N - 588 \text{ N} = 120 \text{ N}$

$F_N = 708 \text{ N}$

Therefore, the scale would read 708 N, which is greater than the person's actual weight of 588 N. This demonstrates how acceleration affects apparent weight.

Practice Problem 3: Gravitational Force Inside Earth

Solution

For an object inside a uniform sphere, we need to use the result from Newton's shell theorem that states the gravitational force is proportional to the distance from the center.

The gravitational force at a distance $r$ from the center is: $F_g = G\frac{Mm}{R^3}r$ (for $r < R$)

Where:

• $M$ = mass of Earth = 5.97 × 10²⁴ kg
• $m$ = mass of object = 1 kg
• $R$ = radius of Earth = 6370 km
• $r$ = distance from center = 6370 km - 2000 km = 4370 km

To find the fraction of surface gravity, we need to calculate $\frac{F_g \text{ at depth}}{F_g \text{ at surface}}$:

$\frac{F_g \text{ at depth}}{F_g \text{ at surface}} = \frac{G\frac{Mm}{R^3}r}{G\frac{Mm}{R^2}} = \frac{r}{R}$

$\frac{F_g \text{ at depth}}{F_g \text{ at surface}} = \frac{4370 \text{ km}}{6370 \text{ km}} = 0.686$

Therefore, at a depth of 2000 km, the object would experience approximately 68.6% of the gravitational force it would feel at Earth's surface.