Unit 3: Gravitation and Central Force Motion

Newton's Law of Gravitation

"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."

F_g = G (m₁m₂) / r²

• F_g is the magnitude of the gravitational force.
• G is the Universal Gravitational Constant (G ≈ 6.674 × 10⁻¹¹ N·m²/kg²).
• m₁ and m₂ are the two masses.
• r is the distance between their centers.

In vector form, the force on m₂ due to m₁ is:

F₂₁ = -G (m₁m₂) / r² r̂₁₂

Where r̂₁₂ is the unit vector pointing from m₁ to m₂. The negative sign indicates that the force is attractive.

---

Gravitational Potential and Potential Energy

Gravitational Potential Energy (U)

As defined in Unit 1, Potential Energy (U) is related to the work done by the conservative gravitational force. We set the reference point `U = 0` at `r = ∞` (infinitely far away).

The potential energy `U(r)` of a mass `m` at a distance `r` from a mass `M` is the work done by an external agent to bring `m` from `∞` to `r` *without acceleration*.

W_ext = -W_gravity = -∫_∞^r F_g · dr

F_g = -G(Mm)/r² r̂ and dr = dr r̂ (moving from infinity)

U(r) = -∫_∞^r [-G(Mm)/x²] dx = G(Mm) ∫_∞^r (1/x²) dx

U(r) = G(Mm) [-1/x] from ∞ to r = G(Mm) [-1/r - (-1/∞)] = -G(Mm)/r

U(r) = -G(Mm) / r

Note: Gravitational potential energy is always negative. This signifies a "bound system." You must *add* energy (do positive work) to separate the masses to infinity (where U = 0).

Gravitational Potential (V)

The Gravitational Potential `V` at a point in space is the potential energy per unit mass at that point. It's a property of the *field* itself, created by the source mass `M`.

V = U / m = -GM / r

• Units: Joules per kilogram (J/kg).
• Physical Meaning: The work done to bring a 1 kg mass from infinity to that point.

---

Potential and Field due to Spherical Shell and Solid Sphere

This is a standard application of Gauss's Law for gravity (which is analogous to Gauss's Law for electrostatics) or direct integration. The results are crucial.

1. Uniform Spherical Shell (Mass M, Radius R)

Case 1: Outside the shell (r > R)

• The shell behaves as if all its mass `M` is concentrated at its center.
• Field (g): g = -GM / r² (attractive)
• Potential (V): V = -GM / r

Case 2: On the surface (r = R)

• Field (g): g = -GM / R² (maximum magnitude)
• Potential (V): V = -GM / R

Case 3: Inside the shell (r < R)

• The gravitational forces from all parts of the shell cancel out perfectly.
• Field (g): g = 0
• Since the field `g = -dV/dr` is zero, the potential `V` must be constant. It is equal to the potential at the surface.
• Potential (V): V = -GM / R (constant)

2. Uniform Solid Sphere (Mass M, Radius R)

Case 1: Outside the sphere (r > R)

• Behaves as if all mass `M` is at the center.
• Field (g): g = -GM / r²
• Potential (V): V = -GM / r

Case 2: On the surface (r = R)

• Field (g): g = -GM / R²
• Potential (V): V = -GM / R

Case 3: Inside the sphere (r < R)

• Only the mass `m'` *inside* the radius `r` contributes to the field.
• `m' = M (Volume_r / Volume_R) = M ( (4/3)πr³ / (4/3)πR³ ) = M (r³/R³)`
• Field (g): g = -G(m') / r² = -G [M(r³/R³)] / r² = -(GM/R³) r
  The field is zero at the center (r=0) and increases linearly to the surface.
• Potential (V): Requires integration.
  V = -(GM/2R³) (3R² - r²)

---

Central Force: Definition and Characteristics

Definition

A central force is a force on a particle that is always directed along the line connecting the particle and a fixed center point.

Its magnitude depends only on the distance `r` from the center.

F(r) = f(r) r̂

Where `r̂` is the unit vector from the center to the particle. Gravity and the electrostatic Coulomb force are prime examples.

Characteristics

1. Torque is always Zero: The force vector F is parallel to the position vector r (or anti-parallel, like gravity).
   τ = r × F = r × (f(r) r̂) = f(r) (r × r̂) = 0.
2. Angular Momentum is Conserved: Since τ_net = 0, we know from τ = dL/dt that L = constant.
   This is the *most important* characteristic of central force motion.
3. Motion is in a Plane: The angular momentum vector L = r × p is constant. This means L always points in the same direction. Since both r (position) and p (momentum) must be perpendicular to L (by definition of the cross product), the motion is confined to a plane that is perpendicular to the constant vector L.

---

Kepler’s Laws with Derivation

Kepler's laws describe the motion of planets around the Sun.

Kepler's First Law (Law of Orbits)

Law: "All planets move in elliptical orbits with the Sun at one of the two foci."

Derivation: This is the most complex derivation. It involves solving the differential equation of motion for an inverse-square central force (F ∝ 1/r²). The general solution for the path `r(θ)` is the equation of a conic section:

1/r = C [1 + e cos(θ)]

Here, `e` is the eccentricity of the orbit.

• For a bound orbit (negative total energy), 0 ≤ e < 1, which is the equation of an ellipse.
• `e = 0` is a circle (a special case of an ellipse).
• `e = 1` is a parabola (unbound orbit).
• `e > 1` is a hyperbola (unbound orbit).

Since planets are in bound orbits, their paths must be ellipses.

Kepler's Second Law (Law of Areas)

Law: "A line segment joining a planet and the Sun sweeps out equal areas in equal intervals of time."

Derivation: This is a direct consequence of the conservation of angular momentum (which is true for *all* central forces).

1. Consider the small area `dA` swept out by the position vector `r` in a time `dt`.
2. This area is a triangle with base `r` and height `v_perp * dt = (v sinφ) dt`.
   Or, more formally, `dA = (1/2) |r × dr|`.
3. Since `dr = v dt`, we have `dA = (1/2) |r × v dt| = (1/2) |r × v| dt`.
4. We know `L = r × p = m(r × v)`. So, `|r × v| = L/m`.
5. `dA = (1/2) (L/m) dt`
6. Rearranging gives the areal velocity:

   dA/dt = L / (2m)

7. Since the force is central, L is constant. Mass `m` is also constant.
8. Therefore, dA/dt = constant. This is Kepler's Second Law.

Kepler's Third Law (Law of Periods)

Law: "The square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (a) of its orbit."

T² ∝ a³

Derivation (for a simple circular orbit of radius r):

1. A planet (mass m) orbits the Sun (mass M) in a circle of radius `r`.
2. The gravitational force provides the necessary centripetal force.
3. F_gravity = F_centripetal
4. G(Mm) / r² = m (v²) / r
5. G M / r = v²
6. The orbital speed `v` is the circumference (2πr) divided by the period (T): `v = 2πr / T`.
7. Substitute `v`: G M / r = (2πr / T)² = 4π²r² / T²
8. Rearrange to solve for T²:
   T² G M = 4π²r³

   T² = [ 4π² / (GM) ] r³

9. Since `(4π² / GM)` is a constant, we have T² ∝ r³. (For an ellipse, `r` is replaced by the semi-major axis `a`).

---

Deduction of Newton’s Law from Kepler’s Law

This shows how Newton "reverse-engineered" his law of gravitation.

1. From Kepler's 1st & 2nd Laws: The elliptical orbit and the equal areas law (L=const) imply that the force must be central and conservative.
2. From Kepler's 3rd Law (T² ∝ r³): We can deduce the inverse-square nature.
   Start from `F = mv²/r` (for a circular orbit).
3. From T² = Kr³, we have T = K^1/2 r^3/2.
4. Speed `v = 2πr / T = 2πr / (K^1/2 r^3/2) = (2π / K^1/2) * r^-1/2.
5. This means v ∝ 1 / √r, or v² ∝ 1/r.
6. Substitute this into the force equation:
   F = m (v²) / r ∝ m (1/r) / r = m/r²
7. So, the force must be F ∝ 1/r².
8. We also know the force is proportional to the planet's mass `m` (from F=ma). By Newton's 3rd Law, it must also be proportional to the Sun's mass `M`.
9. Combining these: F ∝ Mm/r². Introducing the constant G gives Newton's Law.

---

Satellites and Orbits

Satellite in Circular Orbit

This is the same as the derivation for Kepler's 3rd Law. A satellite of mass `m` orbits a planet of mass `M` (e.g., Earth) at a radius `r` from the *center* of the planet.

F_gravity = F_centripetal

G(Mm) / r² = m (v²) / r

Orbital Velocity (v)

v = √(GM / r)

Note: This speed is *independent* of the satellite's mass `m`.

Time Period (T)

T = 2πr / v = 2πr / √(GM/r) = 2πr * √(r/GM) = 2π √(r³/GM)

T = 2π √(r³ / GM)

This is just Kepler's 3rd Law again: T² = (4π²/GM) r³.

---

Geosynchronous Orbits and Weightlessness

Geosynchronous (or Geostationary) Orbits

A Geosynchronous satellite is a satellite that has an orbital period `T` of exactly 24 hours (the same as Earth's rotation).

A Geostationary satellite is a special case: it is a geosynchronous satellite that is also in a circular orbit and travels directly above the equator in the same direction as Earth's rotation.

• Result: From the ground, a geostationary satellite appears to be fixed at a single point in the sky.
• Application: This is extremely useful for communications (TV broadcasting, satellite phones) and weather monitoring, as the ground-based antennas don't need to track it.
• Calculation: We can find its required altitude `r` by setting T = 24 hours in Kepler's 3rd Law.
  T = 24 hr ≈ 86400 s
  T² = (4π²/GM) r³
  r³ = (GM T²) / (4π²)
  Plugging in values for G, M_earth, and T gives:
  r ≈ 4.22 × 10⁷ m (This is the radius from the *center* of Earth)
  Altitude (h) = r - R_earth ≈ (4.22 × 10⁷) - (6.37 × 10⁶) ≈ 3.58 × 10⁷ m, or ~35,800 km.

Weightlessness

Weightlessness is the sensation (or state) of having no apparent weight. It is *not* the absence of gravity.

• Your sensation of "weight" is the normal force (F_N) exerted on you by the floor or a scale.
• Weightlessness occurs when this normal force is zero.
• Example 1: Freefall. In a falling elevator (with cut cables), both you and the elevator accelerate downwards at `g`. Your feet don't press against the floor, so F_N = 0. You feel "weightless".
• Example 2: Orbit. An astronaut in a satellite is in a constant state of freefall *around* the Earth. The satellite (and everything in it) is accelerating towards the Earth at `g = v²/r`. Since the astronaut and the satellite "fall" together, there is no normal force. They are weightless.

---

Global Positioning System (GPS)

GPS is a satellite-based navigation system that provides location and time information.

• Constellation: It consists of a network of ~30 satellites (a "constellation") orbiting the Earth.
• Orbit: They are *not* in geostationary orbits. They are in Medium Earth Orbits (MEO) at an altitude of ~20,200 km, with orbital periods of ~12 hours.
• How it works (Trilateration):
  1. Each satellite continuously broadcasts a signal containing its exact location and the precise time the signal was sent.
  2. Your GPS receiver (in your phone) receives these signals.
  3. By measuring the *time delay* (t) for the signal to arrive, it calculates its distance (d) from that satellite (d = c*t, where c is the speed of light).
  4. Receiving a signal from one satellite tells you you are on the surface of a "sphere" of radius `d` centered on that satellite.
  5. A signal from a second satellite narrows your location to the circle where the two spheres intersect.
  6. A signal from a third satellite narrows your location to just two points (where the circle and the third sphere intersect). One point is usually impossible (e.g., in space) so your 2D location (latitude, longitude) is found.
  7. A signal from a fourth satellite is needed to solve for the fourth unknown, which is the precise time synchronization of your receiver's clock, thus giving a highly accurate 3D position and time.
• Key Point: The system's accuracy *depends* on applying corrections from Einstein's Special and General Relativity. (Special relativity for the high speed of the satellites, and General Relativity for the weaker gravity at their altitude, both of which affect the passage of time on their atomic clocks).