AP Physics 1- 2.6 Gravitational Force- Study Notes- New Syllabus
AP Physics 1-2.6 Gravitational Force – Study Notes
AP Physics 1-2.6 Gravitational Force – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Gravitational Interaction Between Two Objects
- Field Concept
- Weight and Constant Gravitational Force
- Apparent Weight vs. Gravitational Force
- Inertial Mass vs. Gravitational Mass
Gravitational Interaction Between Two Objects
Any two objects with mass exert an attractive gravitational force on each other. This interaction is universal and depends only on the masses of the objects and the distance between them.
\( F = G \dfrac{m_1 m_2}{r^2} \)
- \( F \): Gravitational force between two objects
- \( G \): Universal gravitational constant \( (6.67 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2) \)
- \( m_1, m_2 \): Masses of the two objects
- \( r \): Distance between the centers of mass of the objects
Key Features:
- Gravitational force is always attractive.
- The force acts along the line joining the centers of the two masses.
- Both objects exert equal and opposite forces on each other (Newton’s Third Law).
- The gravitational interaction governs the motion of planets, satellites, and falling objects on Earth.
Example :
Find the gravitational force between Earth (\( m_E = 5.97 \times 10^{24} \, \text{kg} \)) and a 70 kg person standing on its surface. Earth’s radius = \( 6.37 \times 10^6 \, \text{m} \).
▶️ Answer/Explanation
\( F = G \dfrac{m_E m}{r^2} \)
\( F = (6.67 \times 10^{-11}) \dfrac{(5.97 \times 10^{24})(70)}{(6.37 \times 10^6)^2} \)
\( F \approx 686 \, \text{N} \)
Answer: Force = 686 N (equal to the person’s weight).
Example :
Two 10 kg masses are placed 2 m apart. Find the gravitational force between them.
▶️ Answer/Explanation
\( F = G \dfrac{m_1 m_2}{r^2} \)
\( F = (6.67 \times 10^{-11}) \dfrac{(10)(10)}{2^2} \)
\( F = (6.67 \times 10^{-11}) \dfrac{100}{4} \)
\( F = 1.67 \times 10^{-9} \, \text{N} \)
Answer: Force ≈ \( 1.67 \times 10^{-9} \, \text{N} \) (very small).
Field Concept
A field models the effects of a noncontact force exerted on an object at various positions in space. It assigns a value (such as force per unit mass or charge) to each point in space, representing the influence of the source object or system.
Key Ideas:
- A field describes how an object with certain properties (e.g., mass, charge) would experience a force at different locations.
- Fields are used to model noncontact forces like gravitational, electric, and magnetic forces.
- The strength and direction of the field may vary with position.
Examples of Fields:
- Gravitational Field: \( g = \dfrac{F}{m} = \dfrac{G M}{r^2} \) → represents the gravitational influence of a massive object.
- Electric Field: \( E = \dfrac{F}{q} = \dfrac{k Q}{r^2} \) → represents the electric influence of a charge.
- Magnetic Field: Represents the influence of moving charges or magnets on other charges in motion.
Example
Find the gravitational field strength at Earth’s surface.
▶️Answer/Explanation
\( g = \dfrac{G M_E}{R_E^2} \)
\( g = \dfrac{6.67 \times 10^{-11} (5.97 \times 10^{24})}{(6.37 \times 10^6)^2} \)
\( g \approx 9.8 \, \text{m/s}^2 \)
Answer: \( g = 9.8 \, \text{N/kg} \)
Example
Find the electric field 0.1 m away from a point charge of \( +2 \, \mu C \).
▶️Answer/Explanation
\( E = \dfrac{k Q}{r^2} \)
\( E = \dfrac{9 \times 10^9 (2 \times 10^{-6})}{(0.1)^2} \)
\( E = \dfrac{18 \times 10^3}{0.01} = 1.8 \times 10^6 \, \text{N/C} \)
Answer: \( E = 1.8 \times 10^6 \, \text{N/C} \)
Weight and Constant Gravitational Force
The gravitational force exerted by an astronomical body (e.g., Earth) on a relatively small nearby object is called its weight:
\( W = m g \)
where \( g \) is the gravitational field strength at that location.
Situations Where Gravitational Force Can Be Considered Constant:
- Near the surface of the Earth (or any large astronomical body), where the change in distance from the center is negligible compared to the body’s radius.
- For heights much smaller than the Earth’s radius (\( h \ll R_E \)), so that \( g \approx 10\, \text{m/s}^2 \) remains constant.
- In laboratory or classroom experiments on Earth, where variations in \( g \) due to altitude, latitude, or local density are extremely small.
Mathematical Reason:
The exact expression for gravitational acceleration is:
\( g = \dfrac{G M}{(R_E + h)^2} \)
If \( h \ll R_E \), then:
\( g \approx \dfrac{G M}{R_E^2} \) → nearly constant.
Example :
Find the percentage change in gravitational acceleration at a height of 1 km above Earth’s surface compared to sea level. (Take \( R_E = 6.37 \times 10^6 \, \text{m} \))
▶️Answer/Explanation
\( g_h = g \left(\dfrac{R_E}{R_E + h}\right)^2 \)
\( g_h = 9.8 \left(\dfrac{6.37 \times 10^6}{6.37 \times 10^6 + 1 \times 10^3}\right)^2 \)
\( g_h \approx 9.7997 \, \text{m/s}^2 \)
Percentage change = \( \dfrac{9.8 – 9.7997}{9.8} \times 100 \approx 0.003\% \)
Answer: Practically negligible → \( g \) can be considered constant.
Example :
A 5 kg object is lifted to a height of 500 m. Approximate its weight using constant \( g \).
▶️Answer/Explanation
Weight: \( W = m g = 5 \times 10 = 50 \, \text{N} \)
Exact change in \( g \) at \( h = 500 \, \text{m} \) is extremely small (~0.0015%).
Answer: Weight ≈ 50 N (constant \( g \) approximation is valid).
Apparent Weight vs. Gravitational Force
The apparent weight of a system is the normal (contact) force exerted on it by a surface (or tension in a supporting cable). It is what is “felt” as weight.
The gravitational force is the true force of gravity acting on the system: \( F_g = m g \).
Key Idea:
The apparent weight differs from the gravitational force when the system experiences acceleration, because the normal force must balance both gravity and the net acceleration.
Conditions:
| Situation | Acceleration | Apparent Weight |
|---|---|---|
| At rest or constant velocity | \( a = 0 \) | \( W_{app} = m g \) (equal to true weight) |
| Accelerating upward | \( a > 0 \) | \( W_{app} = m(g + a) \) (feels heavier) |
| Accelerating downward | \( a > 0 \) | \( W_{app} = m(g – a) \) (feels lighter) |
| Free fall | \( a = g \) | \( W_{app} = 0 \) (weightless) |
Example :
A 60 kg person stands in an elevator accelerating upward at 2 m/s². Find their apparent weight.
▶️Answer/Explanation
True weight: \( W = m g = 60 \times 9.8 = 588 \, \text{N} \)
Apparent weight: \( W_{app} = m(g + a) = 60(9.8 + 2) = 708 \, \text{N} \)
Answer: Apparent weight = 708 N (feels heavier)
Example :
A 75 kg astronaut inside a spacecraft orbiting Earth is in free fall. What is their apparent weight?
▶️Answer/Explanation
True weight: \( W = m g = 75 \times 9.8 = 735 \, \text{N} \)
In free fall: \( a = g \), so \( W_{app} = 0 \)
Answer: The astronaut is weightless (apparent weight = 0).
Inertial Mass vs. Gravitational Mass
Inertial Mass:
A measure of how much an object resists changes in its state of motion when a net force is applied.
- Appears in Newton’s Second Law: \( F = m_i a \)
- Large inertial mass → object is harder to accelerate.
Gravitational Mass:
A measure of the strength of the gravitational interaction between an object and a gravitational field.
- Appears in Newton’s Law of Gravitation: \( F = G \dfrac{m_g M}{r^2} \)
- Large gravitational mass → stronger gravitational force.
Equivalence Principle:
- Experiments show that inertial mass and gravitational mass are equivalent in value (\( m_i = m_g \)).
- This principle is fundamental to Einstein’s General Theory of Relativity.
Example:
Discuss when a 2 kg ball is dropped near Earth’s surface.
▶️Answer/Explanation
Gravitational force: \( F = m_g g = 2 \times 9.8 = 19.6 \, \text{N} \)
Acceleration: From Newton’s 2nd law, \( a = \dfrac{F}{m_i} = \dfrac{19.6}{2} = 9.8 \, \text{m/s}^2 \)
This works only if \( m_i = m_g \).
Answer: The ball accelerates at 9.8 m/s², confirming equivalence of inertial and gravitational mass.