circular motion lec3

Introduction to the Concept of Gravitation

• Context and Importance: Gravitation is an integral part of circular and rotational motion. While it contains only two to three primary topics, it is exceptionally important for competitive exams like MDCAT and ECAT. In MDCAT, the examiner frequently asks questions regarding the variation in the value of $g$ to connect this topic with projectile motion or the time period of a simple pendulum in Simple Harmonic Motion (SHM).
• Definition: Gravitation refers to the force of attraction that exists between any two objects in the universe due to their masses.

Newton’s Law of Universal Gravitation

• The Principle: Newton stated that every object in the universe attracts every other object 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.

• Mathematical Expression: If two objects have masses $m_1$ and $m_2$ and the distance between their centers is $r$, the force of attraction $F$ is given by:          $F = \frac{Gm_1m_2}{r^2}$

• Action and Reaction Pairs:

  • $F_{12}$ represents the force applied by object 1 on object 2.
  • $F_{21}$ represents the force applied by object 2 on object 1.
  • Relationship: These forces are equal in magnitude but opposite in direction. Mathematically: $F_{12} = -F_{21}$. Even if one mass is significantly larger than the other (e.g., Earth vs. a person), the magnitude of the force exerted by each on the other is identical.

• Inverse Square Law: The force has an inverse square relationship with the distance ($F \propto \frac{1}{r^2}$). If the distance between two objects is increased by a factor of 3 ($3r$), the new force ($F^{\prime}$) becomes nine times smaller than the original force ($F^{\prime} = \frac{F}{9}$).

The Universal Gravitational Constant ($G$)

• Definition: The constant $G$ is known as the Universal Gravitational Constant and remains the same throughout the universe.
• Numerical Value: The value was determined by Henry Cavendish using the Cavendish apparatus.          $G = 6.673 \times 10^{-11} \, \text{N}\cdot\text{m}^2/\text{kg}^2$

Gravitational Acceleration ($g$) and the "Golden Equation"

• Earth's Physical Data:

  • Mass of Earth ($M_e$) \approx $6 \times 10^{24} \, \text{kg}$.
  • Radius of Earth ($R_e$) \approx $6400 \, \text{km}$ = $6400 \times 10^3 \, \text{m}$.

• Derivation: The force between an object of mass $m$ on the surface of the Earth and the Earth itself is its weight ($w = mg$). Equating this to the gravitational force formula:          $mg = \frac{GM_e m}{R_e^2}$          Canceling $m$ from both sides results in the Golden Equation for this topic:          $g = \frac{GM_e}{R_e^2}$

• General Application: This equation can be used to calculate the gravitational acceleration on the surface of any planet ($X$):          $g_x = \frac{GM_x}{R_x^2}$

• Example Calculation (Planet Y): If a planet $Y$ has a mass 9 times that of Earth ($M_y = 9M_e$) and a radius 2 times that of Earth ($R_y = 2R_e$), the gravitational acceleration on $Y$ ($g_y$) is:          $g_y = \frac{G(9M_e)}{(2R_e)^2} = \frac{9}{4} \times \frac{GM_e}{R_{e}^2} = 2.25g$          Using $g = 10 \, \text{m/s}^2$, the value would be $22.5 \, \text{m/s}^2$.

Variation in the Value of $g$ with Altitude (Height)

• Mechanism: As an object moves to a height $h$ above the Earth's surface, the distance from the center of the Earth becomes $R_e + h$. Consequently, the value of $g$ decreases.

• Equation for Acceleration at Height $h$ ($g_h$):          $g_h = \frac{GM_e}{(R_e+h)^2}$

• Comparison Formula: To find the ratio of $g_h$ to the surface gravity $g$:          $\frac{g_h}{g} = \left(\frac{R_e}{R_e+h}\right)^2$

• Approximation for Small Altitudes: If $h$ is much smaller than $R_e$ ($h \ll R_e$), the formula simplifies via binomial expansion to:          $g_h = g(1 - \frac{2h}{R_e})$Note: This approximation is valid only for very small heights (meters). For significant heights (multiples of $R_e$), the original square ratio formula must be used.

• Specific Altitude Examples:

  • At $h = R_e$: $\frac{g_h}{g} = (\frac{R_e}{2R_e})^2 = \frac{1}{4}$. Thus, $g_h = \frac{g}{4} = 2.5 \, \text{m/s}^2$.
  • At $h = 2R_e$: $\frac{g_h}{g} = (\frac{R_e}{3R_e})^2 = \frac{1}{9}$. Thus, $g_h = \frac{g}{9} \approx 1.11 \, \text{m/s}^2$.

Variation in the Value of $g$ with Depth

• Mechanism: Depth ($d$) is the distance measured into the Earth from the surface. Inward movement reduces the distance to the center to $(R_e - d)$. Unlike altitude, variation with depth involves two factors: the decreasing distance (which would increase $g$) and the decreasing effective mass of the Earth attracting the object.

• Effective Mass Concept: Only the sphere of mass inside the radius $(R_e - d)$ exerts a net gravitational pull on the object. This is analogous to atmospheric pressure: higher up, fewer air molecules exist above you to press down. As you go deeper into Earth, the outer shells of mass effectively "cancel out," decreasing the effective mass.

• Dominant Factor: Even though the object is closer to the center, the decrease in effective mass is the dominant factor, resulting in an overall decrease in $g$ as you go deeper.

• Equation for Acceleration at Depth $d$ ($g_d$):          $g_d = g(1 - \frac{d}{R_e})$

• Specific Depth Examples:

  • At $d = \frac{R_e}{2}$: $g_d = g(1 - \frac{0.5R_e}{R_e}) = \frac{g}{2} = 5 \, \text{m/s}^2$.
  • At the Center of the Earth ($d = R_e$): $g_d = g(1 - 1) = 0 \, \text{m/s}^2$. This is because at the center, the surrounding mass pulls the object equally in all directions, and the effective mass is zero.

Comprehensive Problem: Finding Points of Equivalent Gravity

• Scenario: Find the distance from the center of the Earth to points where gravity is half of the surface value ($g/2$).
• Case 1 (Above the Surface):
  • Using $\frac{g_h}{g} = (\frac{R_e}{R_e + h})^2$, set $\frac{g_h}{g} = 0.5$.
  • $1/2 = (\frac{R_e}{R_e+h})^2 \rightarrow \frac{1}{\sqrt{2}} = \frac{R_e}{R_e+h}$.
  • $R_e + h = \sqrt{2}R_e = 1.414R_e$.
  • The total distance from the center is $1.414R_e$.
• Case 2 (Below the Surface):
  • Using $\frac{g_d}{g} = (1 - \frac{d}{R_e})$, set $\frac{g_d}{g} = 0.5$.
  • $0.5 = 1 - \frac{d}{R_e} \rightarrow \frac{d}{R_e} = 0.5 \rightarrow d = 0.5R_e$.
  • Distance from the center = $R_e - d = R_e - 0.5R_e = 0.5R_e$.

Concepts of Real and Apparent Weight

• Real Weight ($w$): The actual force with which Earth attracts an object. It is a constant pull intended to accelerate a body toward the center at $g \approx 9.8 \, \text{m/s}^2$.
• Apparent Weight ($T$): The reading shown by a scale or the tension in a support string. It represents the force required to keep an object within its frame of reference.

Apparent Weight in an Elevator

• Case A: Static or Constant Velocity ($a = 0$):
  • There is no unbalance in the system. The tension in the string equals the weight: $T = w$. Apparent weight equals real weight.
• Case B: Accelerating Upward ($a \neq 0$):
  • To accelerate upward, the upward force must overcome the downward pull. Using Newton's Second Law ($T - w = ma$):
  • $T = w + ma$.
  • Implication: The body feels heavier. The apparent weight increases by the factor of the inertial force $ma$.
• Case C: Accelerating Downward ($a \neq 0$):
  • Using Newton's Second Law ($w - T = ma$):
  • $T = w - ma$.
  • Implication: The body feels lighter. The apparent weight decreases.
• Case D: Free Fall ($a = g$):
  • If the cable breaks and the lift falls freely:
  • $T = w - mg = mg - mg = 0$.
  • Implication: This is the state of Weightlessness. Objects inside the elevator become detached from the floor and float.

Practical Elevator Example

• Details: A woman with mass $m = 45 \, \text{kg}$ is in an elevator accelerating upward at $a = 1.2 \, \text{m/s}^2$.
• Calculations (using $g = 10 \, \text{m/s}^2$):
  • Real weight ($w$) = $45 \times 10 = 450 \, \text{N}$.
  • Upward inertial force needed ($ma$) = $45 \times 1.2 = 54 \, \text{N}$.
  • Apparent weight ($T$) = $w + ma = 450 + 54 = 504 \, \text{N}$.
• Alternative Explanation: If the woman were static, Earth pulls with $450 \, \text{N}$. Because she is moving up, Earth must apply an extra force to overcome the upward acceleration, making the perceived force $504 \, \text{N}$.

Weightlessness and Artificial Gravity in Satellites

• Free Fall in Orbit: Artificial satellites are in a state of continuous free fall toward Earth. Their forward velocity combined with the curvature of Earth prevents them from hitting the surface. Because they are in free fall, objects inside experience zero apparent weight.
• Artificial Gravity Necessity: To allow astronauts to work safely without floating, artificial gravity must be generated by rotating the spacecraft.
• Mechanism: Rotation generates centripetal acceleration. By setting the frequency of rotation correctly, one can simulate the gravitational pull of Earth.
• Calculation Formula: The frequency ($f$) of rotation to simulate Earth's gravity ($g$) is given by:          $f = \frac{1}{2\pi}\sqrt{\frac{g}{r}}$          Where $r$ is the radius of the spacecraft. If a spacecraft has a tunnel length $L$, then $r = \frac{L}{2}$.
• Numerical Example: If the tunnel length $L = 30 \, \text{m}$, then $r = 15 \, \text{m}$. To simulate $g = 10 \, \text{m/s}^2$:
  • $f = \frac{1}{2\pi}\sqrt{\frac{10}{15}} \approx 0.13 \, \text{Hz}$ (Cycles per second).
  • In minutes: $0.13 \times 60 = 7.8$ cycles per minute.
  • Rotating the spaceship at this speed creates an environment that feels exactly like Earth's surface gravity.