Gravitation

• Given measurements:
  • $a_{apple}/a_{moon} = 9.8/0.0027 = 3600$
  • $d_{moon}/d_{apple} = 3.85 \times 10^5/6400 = 60$
• Relationship: $(d_{moon}/d_{apple})^2 = 60^2 = 3600$
• This matches the ratio of accelerations, showing:
  • $a \propto 1/r^2$ (acceleration is inversely proportional to square of distance)
• This fundamental insight led Newton to his universal law of gravitation

• Potential: $V = -\frac{GM}{\sqrt{a^2 + r^2}}$
• Significance:
  • Shows how extended mass creates different potential than point mass
  • Approaches point mass potential ($-\frac{GM}{r}$) when $r \gg a$
  • Near the ring ($r \ll a$), potential varies approximately linearly with $r$
  • Important building block for understanding more complex mass distributions
  • Demonstrates principle of superposition for gravitational potentials
  • Has cylindrical symmetry rather than spherical symmetry

The gravitational potential inside a uniform spherical shell remains constant because:

1. Mathematical explanation:

   • For an interior point at distance r from center, the potential equals -GM/a
   • This expression contains no r dependence, only the shell radius a
   • This results from the integration of the contributions from all mass elements over the shell

2. Physical principle:

   • This is a consequence of Newton's shell theorem
   • The net gravitational force at any point inside a spherical shell is exactly zero
   • Since force is the gradient of potential, zero force means constant potential

3. Conceptual understanding:

   • Mass elements on opposite sides of the shell exert forces that precisely cancel
   • The solid angle subtended by closer portions is smaller but their gravitational effect is stronger
   • The solid angle subtended by farther portions is larger but their gravitational effect is weaker
   • These effects exactly compensate throughout the interior

This principle is used in electrostatics (Faraday cage effect) and explains why the Earth's gravitational field would be zero if you could reach its center.

Outside the shell ($r > a$):

• Field is same as if entire mass were concentrated at center
• $E = \frac{GM}{r^2}$ directed toward center
• Follows inverse square law

Inside the shell ($r < a$):

• Field is exactly zero everywhere
• Gravitational pulls from all parts of shell cancel perfectly
• No net force regardless of position inside

Significance:

1. Newton's Shell Theorem allows treating spherically symmetric mass distributions as point masses for external points
2. Earth's gravity comes only from mass in shells with radius smaller than your position
3. Allows calculation of gravity within planets
4. Foundation for Gauss's law in gravitational theory
5. Inside a hollow planet, you would experience weightlessness

The gravitational field at point P is:

$E = \frac{GMr}{(a^2 + r^2)^{3/2}}$

Where:

• G is the gravitational constant
• M is the mass of the ring
• a is the radius of the ring
• r is the distance from the center of the ring to point P

The field is directed toward the center of the ring along the axis.

Note: This can be derived by integrating the contributions from all mass elements of the ring, using the fact that by symmetry, only the components along the axis add constructively.

Variation of gravitational field with axial distance:

1. Near field (r ≪ a):

   • Field increases approximately linearly with r
   • At r = 0 (center of ring), the field is zero due to symmetry

2. At r = a/√2:

   • Field reaches its maximum value

3. Far field (r ≫ a):

   • Field decreases proportionally to 1/r²
   • Formula approximates to: E ≈ GM/r²
   • Ring effectively behaves like a point mass

This behavior occurs because as r increases, the angle subtended by the ring decreases, and the geometry approaches that of a point mass at very large distances.

The complete formula showing this transition is: $E = \frac{GMr}{(a^2 + r^2)^{3/2}}$

Gravitational mass is defined by: $\frac{m_A}{m_B} = \frac{F_A}{F_B}$ or $m_A = \frac{F_A}{F_B}m_B$

Where $F_A$ and $F_B$ are the gravitational forces experienced by objects A and B due to the same massive body (like Earth).

Conceptual difference:

• Gravitational mass determines the strength of gravitational attraction
• Inertial mass measures resistance to acceleration

A spring balance measures gravitational mass, while acceleration measurements under non-gravitational forces reveal inertial mass.

Despite their different physical origins, experiments show these two masses are equivalent to high precision.

To find the force using the modified law: $F = \frac{G_{\infty}m_1m_2}{r^2}\left[1 + \left(1 + \frac{r}{\lambda}\right)\alpha e^{-\frac{r}{\lambda}}\right]$

Given:

• $r = 100$ m
• $m_1 = 1000$ kg
• $m_2 = \text{particle mass}$ (call it $m_p$)
• $\alpha = -0.007$
• $\lambda = 200$ m
• $G_{\infty} \approx 6.67 \times 10^{-11}$ N⋅m²/kg²

Calculation: $\frac{r}{\lambda} = \frac{100}{200} = 0.5$ $\left(1 + \frac{r}{\lambda}\right) = 1.5$ $e^{-\frac{r}{\lambda}} = e^{-0.5} \approx 0.607$ $\left[1 + \left(1 + \frac{r}{\lambda}\right)\alpha e^{-\frac{r}{\lambda}}\right] = [1 + (1.5)(-0.007)(0.607)] \approx 0.994$

Final force: $F \approx 0.994 \times \frac{6.67 \times 10^{-11} \times 1000 \times m_p}{(100)^2} = 6.63 \times 10^{-13} \times m_p$ N

This is approximately 0.6% less than the Newtonian prediction.

If the fifth force hypothesis were confirmed:

1. The simple equivalence principle would require modification, as objects would experience a combined force of standard gravity plus the fifth force component.

2. The observed "gravitational mass" would become distance-dependent:

   • At short distances: $m_{grav} = m(1 + \alpha) \approx 0.993m$
   • At large distances: $m_{grav} = m$

3. Free-fall acceleration would depend slightly on composition if different materials couple differently to the fifth force.

4. General relativity would need revision, as it's built on the exact equivalence of inertial and gravitational mass.

5. Experimental tests of the equivalence principle would show violations at specific distance scales (near λ ≈ 200m).

6. Precision tests using different materials at different distances could explicitly map the fifth force's strength and range.

This would represent a fundamental shift in our understanding of gravity and potentially open the door to new physics beyond the current standard model.

For the gravitational field outside a uniform solid sphere:

1. Shell Theorem:

   • The gravitational field at any external point (where r > radius of sphere) is identical to that of a point mass concentrated at the sphere's center
   • This fundamental principle allows us to simplify calculations for spherically symmetric mass distributions

2. Mathematical Expression:

   • E = GM/r²
   • Where:
     • E = gravitational field strength
     • G = gravitational constant
     • M = total mass of sphere
     • r = distance from center of sphere

3. Key Properties:

   • Field follows the inverse square law (1/r²)
   • Maximum field strength occurs at the surface (r = a): E = GM/a²
   • Field strength decreases as distance increases
   • Approaches zero as r approaches infinity

4. Practical Applications:

   • Calculating Earth's gravitational field for satellite orbits
   • Predicting gravitational effects of celestial bodies
   • Simplifying gravitational interactions between spherical astronomical objects

5. Important Limitation:

   • Only applies to spherically symmetric mass distributions