Class 12 Physics Unit 1 Gravitaions | FBISE Federal Board

15.1 Newton's Law of Universal Gravitation

Newton’s law of universal gravitation describes the force of attraction between objects in the universe. It served as the foundation for classical mechanics and explains natural phenomena such as tides and planetary orbits.

Key Definition

Every object in the universe attracts every other object with a force that is:
- Directly proportional to the product of their masses.
- Inversely proportional to the square of the distance between their centers.

Mathematical Derivation

Consider two spherical bodies with masses $m_1$ and $m_2$ separated by a distance $r$.

Mass Relationship: The gravitational force ($F_g$) is proportional to the product of the masses:$$F_g \propto m_1 \times m_2$$
Distance Relationship: The force is inversely proportional to the square of the distance:$$F_g \propto \frac{1}{r^2}$$
Combined Relation: Combining both factors gives:$$F_g \propto \frac{m_1 \times m_2}{r^2}$$
Final Equation: Replacing the proportionality with the gravitational constant ($G$):$$F_g = G \frac{m_1 \times m_2}{r^2}$$

The Gravitational Constant (G)

Definition: $G$ is the constant of proportionality known as the universal gravitational constant.
Value: $6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$

Core Concepts of Gravitational Force

Nature of Force: Gravitational force is always attractive and is independent of the medium between two masses.
Inverse-Square Law: The force decreases significantly as distance increases:
- Double the distance ($2r$) = Force decreases by a factor of 4 ($\frac{1}{4}F$).
- Triple the distance ($3r$) = Force decreases by a factor of 9 ($\frac{1}{9}F$).

Newton’s Law of Universal Gravitation & Motion

Interaction: Every object attracts every other object. If mass $m_1$ exerts a force ($F_{21}$) on mass $m_2$, then $m_2$ exerts an equal and opposite force ($F_{12}$) on $m_1$.
Consistency with Newton’s 3rd Law: Gravitational forces form an action-reaction pair, mathematically expressed as:
$F_{12} = -F_{21}$

Mathematical Calculation

The magnitude of gravitational force is calculated using the formula:

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

Key Constant ($G$): $6.673 \times 10^{-11}$
Observation: For everyday objects (like two people), the resulting force is "minuscule" (around $10^{-7}\text{ N}$), which is why it usually goes unnoticed without sensitive instruments.

Assignments

Problem Solving: Calculate the force exerted by the Earth on the Moon using:
- Mass of Earth: $6 \times 10^{24}\text{ kg}$
- Mass of Moon: $7.4 \times 10^{22}\text{ kg}$
- Distance: $3.84 \times 10^5\text{ km}$
Conceptual Question: Explain why the law of gravitation is referred to as a "universal law."

Study Guide: Gravitational Field Strength

1. The Gravitational Constant (G)

Significance: $G$ is crucial for accurate calculations of gravitational forces.
Historical Context: Measuring $G$ was difficult because it is a very small number. In 1798, Henry Cavendish used a torsion balance apparatus to measure the force between objects and determine the value of $G$.
Impact: Determining $G$ allowed scientists to calculate the precise mass of Earth and other astronomical bodies.

2. Gravitational Field and Strength

Definition: The area around a massive object where its gravitational force acts.
Field Lines: Used to represent the field.
- Direction: Points radially inward toward the mass.
- Strength: Closer lines indicate a stronger field; lines further apart indicate a weaker field.
Field Strength ($g$): Defined as the gravitational force per unit mass at a specific point ($g = \frac{F_g}{m}$). It is equivalent to the acceleration due to gravity.

3. Calculating 'g' on Earth's Surface

The value of $g$ is derived by comparing Newton’s Law of Universal Gravitation with Newton’s Second Law of Motion:

Force Equation: $F_g = G \frac{m_o M_E}{R_E^2}$ (where $m_o$ is the object mass, $M_E$ is Earth's mass, and $R_E$ is Earth's radius).
Weight Equation: $F_g = m_o g$.
Equating the two: $m_o g = G \frac{m_o M_E}{R_E^2}$.
Final Formula: $$g = G \frac{M_E}{R_E^2}$$

Earth's Value: Using $M_E = 6 \times 10^{24} \text{ kg}$ and $R_E = 6.4 \times 10^6 \text{ m}$, the value of $g$ is approximately $9.8 \text{ m s}^{-2}$.
Key Insight: The value of $g$ does not depend on the mass of the object ($m_o$); therefore, all bodies (light or heavy) fall at the same rate.

4. Variation of 'g' with Altitude

The value of $g$ is not constant; it changes as you move away from the center of the Earth.

General Formula: For any altitude $h$, the formula becomes: $$g_h = \frac{g R_E^2}{(R_E + h)^2}$$
Inverse Relationship: As the distance from the center of the Earth increases (higher altitude), the value of $g$ decreases.

Altitude (km)	$g_h$ ($m s^{-2}$)	Example Location
0	9.8	Average Earth radius
8.8	9.8	Mount Everest
36.6	9.7	Highest manned balloon
400	8.7	Space shuttle orbit
35,700	0.2	Communication satellites

Gravitational Field Strength

Definition: The gravitational field strength ($g$) represents the pull of gravity at a specific location.
Behavior of Earth's Field:
- Outside the Sphere: Earth behaves as a "point mass" where all mass is concentrated at its center. The field strength reduces as distance increases according to the inverse-square law.
- Inside the Sphere: The gravitational field strength increases steadily toward the surface, as only the mass within the radius affects the point.
Formula for Altitude: $$g_h = \frac{g R_E^2}{(R_E + h)^2}$$
- $R_E$ = Radius of Earth ($6.4 \times 10^6$ m)
- $h$ = Height above the earth's surface.

Satellites and Orbits

Natural Satellites: A natural body (like a moon) orbiting a planet where the larger body's gravity dominates.
Artificial Satellites: Objects intentionally placed into orbit for specific missions such as communication, weather monitoring, or research.
Types of Orbits:
1. Equatorial: Orbiting along the plane of the equator.
2. Inclined: Orbiting at an angle to the equator.
3. Polar: Orbiting over the North and South poles.
4. Geostationary: Orbits at 35,700 km, appearing stationary over a fixed point on Earth.

The Launching Process

Launching involves moving a satellite to a high altitude and accelerating it to a specific tangential speed:

Low Speed: The satellite falls back to Earth.
Adjusted Speed: The satellite maintains a stable circular orbit.
High Speed: The satellite moves into an elliptical orbit.
Escape Speed: The satellite overcomes Earth's gravity and never returns.

Orbital Velocity

Definition: The minimum velocity required for a satellite to orbit a celestial body at a specific altitude.
Physics Principle: The centripetal acceleration ($a_c$) is supplied by gravity ($g$). Since $a_c = \frac{v_o^2}{r}$, then $v_o = \sqrt{gr}$.
General Formula: $$v_o = \sqrt{\frac{G M_E}{R_E + h}}$$
Key Relationship: A satellite moves faster when it is close to the gravitating body and slower when it is further away.

Core Examples & Values

International Space Station (ISS): Orbits at an altitude of ~400 km with a velocity of approximately $7.66$ km s⁻¹. It completes 16 orbits around Earth in 24 hours.
Jupiter: Has a surface gravity of $24.77$ m s⁻². An astronaut weighing 588.6 N on Earth would weigh 1486.2 N on Jupiter.
Earth's Orbital Speed: Earth moves around the Sun at a speed of approximately $30$ km s⁻¹.

Study Guide: Geostationary Satellites

1. Definitions and Key Concepts

Geostationary Satellite: A satellite that circles the Earth at a specific distance and takes exactly 24 hours to complete an orbit. It remains stationary above a fixed point on Earth's equator.
Geostationary Orbit: The specific circular path followed by a geostationary (or geosynchronous) satellite.
Hill Sphere: The region around a celestial body where its gravitational pull is stronger than that of a larger, nearby body.

2. Mathematical Derivation of Orbital Radius

To find the orbital radius (r) of a synchronous satellite, we use the following process:

Orbital Velocity Formula: v o = r GM E
Circumference Velocity Formula: Since a satellite travels the circumference (2πr) in a time period (T), v o = T 2πr
Comparison: Equate the two formulas: T 2πr = r GM E
Rearrangement: Square both sides and solve for r to get:
r=( 4π 2 GM E T 2 ) 1/3

3. Earth Calculation Results

Using Earth’s mass (M E =6×10 24 kg) and a period of 24 hours (T=86400 s):

Orbital Radius (r): 4.23×10 7 m (or 4.23×10 4 km) measured from the center of Earth.
Orbital Speed (v o ): 3.07×10 3 m s −1 .

4. Practical Limitations: Moon and Venus

Even if the math provides a radius, a stable stationary orbit may be impossible due to the Hill Sphere.

The Moon:

Required radius for a "lunar-stationary" orbit: 88,417 km.
Moon's Hill Sphere limit: ≈60,000 km.
Result: Earth’s gravity interferes and destabilizes the orbit because the required radius is outside the Hill Sphere.

Venus:

Required radius for a synchronous orbit: 1.53×10 6 km.
Result: Due to Venus's slow rotation and low mass, this distance is likely outside its Hill Sphere. The Sun's gravity would make the orbit unstable.

5. Practice Assignments

Assignment 15.4: Calculate the orbital speed of a satellite orbiting Earth at an altitude equal to Earth's radius.
Assignment 15.5: Calculate the height (from the surface of Earth) of a satellite in geostationary orbit.

Gravitational Potential and Potential Energy

The standard formula $\Delta P.E = mgh$ is only accurate near Earth's surface where gravity is constant. For larger distances, gravity weakens, necessitating an expression based on Newton's law of universal gravitation.

1. Core Definitions

Absolute Potential Energy ($U$): The work done in moving a body from Earth's surface to a point at infinite distance where $g$ is negligible.
Zero-Reference Point: Potential energy is set to zero at infinity. This results in negative potential energy for any finite distance $r$.
Gravitational Potential ($V$): The work done per unit mass in bringing a small test mass from infinity to a specific point.
- Formula: $V = \frac{U}{m}$

2. Deriving Gravitational Potential Energy ($U$)

Because gravitational force is variable over large distances, the total work is calculated by dividing the distance into small intervals ($\Delta r$) where force is considered constant.

Calculate Average Force ($F_{av}$): For an interval between $R_E$ and $r_1$, the average force is:$$F_{av} = \frac{GM_Em}{R_Er_1}$$
Work Done for One Interval ($W_1$):$$W_1 = -GM_Em \left( \frac{1}{R_E} - \frac{1}{r_1} \right)$$
Total Net Work ($W_{net}$): Summing all intervals from Earth's surface ($R_E$) to a point ($r_n$) leads to the cancellation of intermediate terms (telescoping sum):$$W_{net} = -GM_Em \left( \frac{1}{R_E} - \frac{1}{r_n} \right)$$
Final Expression for $U$: When the final point is at infinity ($r_n = \infty$), then $\frac{1}{\infty} = 0$:$$U = -\frac{GM_Em}{r}$$

3. Gravitational Potential ($V$) at Earth's Surface

General Formula: $V = -\frac{GM_E}{r}$
Calculation at Surface: Using $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$, $M_E = 6 \times 10^{24} \text{ kg}$, and $R_E = 6.4 \times 10^6 \text{ m}$:$V = -62.3 \text{ MJ kg}^{-1}$

4. Key Takeaways from Example 15.7 (Sun-Earth System)

These formulas apply to any gravitating bodies, such as the Sun ($M_s$) and Earth ($M_E$) at a distance $r$:

Potential Energy: $U = -\frac{GM_sM_E}{r}$ (calculated as $-5.29 \times 10^{33} \text{ J}$)
Gravitational Potential: $V = -\frac{GM_s}{r}$ (calculated as $-885 \text{ MJ kg}^{-1}$)

Constants Used

Gravitational Constant (G): 6.674 × 10^-11 N m²/kg²
Mass of Earth (M_E): 6 × 10²⁴ kg
Radius of Earth (R_E): 6.4 × 10⁶ m

Numerical Problems Solutions

1) Mass of Earth

Given: R = 6.4 × 10⁶ m, g = 9.8 m/s²

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

Calculation:
$M = \frac{9.8 \times (6.4 \times 10^6)^2}{6.67 \times 10^{-11}} \approx 6.0 \times 10^{24} \text{ kg}$

2) Distance between Alpha Centauri A and B

Given: M₁ = 2.19 × 10³⁰ kg, M₂ = 1.80 × 10³⁰ kg, F = 2.24 × 10²⁵ N

Formula: $r = \sqrt{\frac{G M_1 M_2}{F}}$

Calculation:
$r = \sqrt{\frac{6.674 \times 10^{-11} \times 2.19 \times 10^{30} \times 1.80 \times 10^{30}}{2.24 \times 10^{25}}} \approx 3.43 \times 10^{12} \text{ m}$

3) 'g' on an Exoplanet

Given: M_p = 5M_E, R_p = 2R_E

Formula: $g_p = \frac{G (5 M_E)}{(2 R_E)^2} = 1.25 \times g$

Calculation:
$g_p = 1.25 \times 9.8 = 12.25 \text{ m/s}^2$

4) Hubble Space Telescope

Given: h = 6.13 × 10⁵ m, R_E = 6.4 × 10⁶ m

Total Radius (r): 7.013 × 10⁶ m

(a) Speed (v): $v = \sqrt{\frac{G M_E}{r}} \approx 7550 \text{ m/s}$
(b) Period (T): $T = \frac{2 \pi r}{v} \approx 5836 \text{ s}$

5) Mars Satellite Orbital Velocity

Given: h = 3 × 10⁵ m, M = 6.42 × 10²³ kg, R = 3.39 × 10⁶ m

Total Radius (r): 3.69 × 10⁶ m

Formula: $v = \sqrt{\frac{G M}{r}} \approx 3406 \text{ m/s} = 3.4 \text{ km/s}$

Assignment 15.6: Gravitational Potential

Formula used: $V = -\frac{GM}{r}$, where $r = R_E + h$

Altitude (h)	Total Radius (r)	Gravitational Potential (V)
1,000 km	7.4 × 10⁶ m	-5.41 × 10⁷ J/kg
50,000 km	5.64 × 10⁷ m	-7.10 × 10⁶ J/kg
100,000 km	1.064 × 10⁸ m	-3.76 × 10⁶ J/kg

Comparison with E_p = mgh

The formula E_p = mgh is a linear approximation intended for use near the Earth's surface. At the altitudes calculated above (1,000 km to 100,000 km), the gravitational field strength g is no longer constant, making the simple mgh formula inaccurate for space-scale calculations.

Short Questions Solutions

1) Why can gravitation not account for the formation of molecules?

Gravitation is the weakest of the four fundamental forces. At the atomic and molecular scale, the masses of atoms are so incredibly small that the gravitational pull between them is negligible. Instead, electromagnetic forces (the attraction between protons and electrons) are responsible for holding atoms together to form molecules.

2) Why don't two books on your desk attract each other gravitationally, despite Newton's law of gravitation?

They actually do attract each other, but the force is extremely weak due to the small value of the gravitational constant $G$ ($6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2$). This tiny force is easily overcome by the static friction between the books and the desk, preventing any visible movement.

3) Why does an apple fall towards the Earth due to gravity, while the Earth doesn't move towards the apple?

According to Newton's Third Law, the force exerted by the Earth on the apple is equal and opposite to the force exerted by the apple on the Earth. However, Newton's Second Law ($F = ma$) states that acceleration is inversely proportional to mass. Because the Earth's mass is so vast, its acceleration is infinitesimally small and undetectable, whereas the apple's small mass allows for a large, visible acceleration.

4) Can gravitational field strength be negative? Explain.

Gravitational field strength ($g$) is a vector quantity. In a radial coordinate system, it is often assigned a negative sign to indicate that the direction of the force is attractive—pointing inward toward the center of the mass, opposite to the outward radial direction.

5) What factors determine the strength of the gravitational field around a planet?

Based on the formula $g = \frac{GM}{R^2}$, the two primary factors are:

Mass of the planet ($M$): Strength is directly proportional to mass.
Radius of the planet ($R$): Strength is inversely proportional to the square of the distance from the center.

6) If two planets have the same mass but different radii, how would their gravitational field strengths compare?

The planet with the smaller radius will have a higher gravitational field strength at its surface. This is because the surface is closer to the center of mass, and the field strength increases as distance ($R$) decreases.

7) Why do satellites in higher orbits have lower orbital velocities?

To maintain a stable orbit, the gravitational pull must equal the required centripetal force. The relationship is defined by $v = \sqrt{\frac{GM}{r}}$. As the orbital radius ($r$) increases, the gravitational pull weakens, requiring a lower orbital velocity to balance the forces and stay in orbit.

8) How does the mass of the Earth affect the orbital velocity required for a satellite to stay in orbit?

The orbital velocity is directly proportional to the square root of the Earth's mass ($v \propto \sqrt{M}$). If the Earth were more massive, a satellite would require a higher velocity to generate enough centripetal force to keep from being pulled down by the stronger gravity.

9) Is it possible for an object’s gravitational potential energy to become negative? If so, what does this mean for the object's motion?

Yes, gravitational potential energy is typically negative because the reference point of zero energy is set at infinity. A negative value means the object is "bound" by the gravitational field; it lacks the kinetic energy to escape and will remain in motion (like an orbit) or fall toward the mass unless external work is done on it.

10) How the gravitational potential energy of two point masses is related to concept of gravitational potential?

Gravitational potential ($V$) is defined as the gravitational potential energy per unit mass ($V = \frac{U}{m}$). If the potential energy of two masses is $U = -\frac{GMm}{r}$, then the potential created by the source mass $M$ is $V = -\frac{GM}{r}$. Essentially, the potential energy of a system is the product of the potential at a point and the mass placed there.

No.	Correct Option	Explanation/Formula
1	B. $6.67 \times 10^{-11} \text{ N}$	Using $F = G \frac{m_1 m_2}{r^2}$, where $G = 6.67 \times 10^{-11}$. Since $m_1, m_2$, and $r$ are all 1, $F = G$.
2	D. $16 \text{ F}$	$F \propto \frac{m_1 m_2}{r^2}$. Doubling both masses $(2 \times 2 = 4)$ and halving distance $(\frac{1}{2})^2 = \frac{1}{4}$ in denominator results in $4 \times 4 = 16$ times the force.
3	B. $6.4$	$g_h = g (\frac{R_E}{R_E + h})^2$. With $R_E \approx 6400 \text{ km}$ and $h = 1500 \text{ km}$, the value drops from $9.8$ to approx $6.4 \text{ m/s}^2$.
4	D. $7.9 \text{ km s}^{-1}$	Orbital velocity $v = \sqrt{gR}$. Using $g = 9.8$ and $R = 6.4 \times 10^6 \text{ m}$, we get $\approx 7900 \text{ m/s}$.
5	B. decreases	As orbit radius increases, orbital velocity decreases ($v \propto \frac{1}{\sqrt{r}}$), therefore Kinetic Energy decreases.
6	C. $3.07 \text{ km s}^{-1}$	A geostationary satellite orbits at a much higher altitude $(\approx 35,786 \text{ km})$, resulting in a slower orbital speed compared to surface level.
7	C. $3$	Three geostationary satellites placed $120^\circ$ apart can cover the entire Earth's surface (except the poles).
8	A. increases	Gravitational potential energy $U = -G\frac{Mm}{r}$. As $r$ increases, the value becomes "less negative," which is an increase.
9	A. $-94.7 \text{ M J kg}^{-1}$	Gravitational potential $V = -\frac{GM}{r}$. At geostationary altitude, the calculated value is approximately $-9.47 \times 10^7 \text{ J/kg}$.

Class 12 Physics Unit 1 Gravitaions | FBISE Federal Board | Read and Download