e&m day three

The core idea is that gravity between two masses follows an inverse-square law and that the motion can be analyzed with force diagrams and Newton's laws.
Real-life systems may involve other forces beyond gravity (e.g., normal forces, friction, air resistance); the simple free-fall picture isolates gravity.

The classroom force diagram for two masses is not the whole story in real life because there are other forces acting (e.g., contact forces, friction).
In some situations (e.g., an object falling in vacuum, or a satellite orbiting with negligible air drag), gravity dominates and the inverse-square law governs the motion.
The problem often begins with identifying the force exerted by Earth on the object and the resulting acceleration of the object.

Center of mass: a concept used in multi-body systems; the gravitational force between two bodies acts along the line joining their centers of mass.
Mass, F, and acceleration: Newton's second law in its net form is
$F_{ ext{net}} = m a.$
For gravity, the force on an object of mass m due to Earth (mass M) at a distance r is
$F_g = rac{G M m}{r^2},$
where G is the gravitational constant.
The resulting acceleration is
$a = rac{F}{m} = rac{G M}{r^2}.$
This shows that, near Earth, the acceleration due to gravity does not depend on the object's mass (mass cancels in the equation). The acceleration is toward Earth's center.

The Earth is approximately a sphere with radius
$R_E \, ext{(Earth radius)} \, \approx 6.371 \times 10^6 \ \text{m} \approx 3.959 \text{ thousand miles}.$
A common classroom distance mentioned is about the Earth's radius (roughly 4000 miles).
Unit conversions (useful for quick checks):
- 1 mile ≈ 1.609 km; 1 km = 1000 m.
- Therefore, 4000 miles ≈ $4.0 \times 10^3 \text{ miles} \approx 6.44 \times 10^6 \text{ m}.$
A mental exercise often used: if you could tilt North America into the Earth, its width (~3000 miles) is on the order of the Earth’s radius, illustrating scale and distance relationships.

For a point mass (or a spherically symmetric mass distribution treated as a point at its center), the gravitational field magnitude is
$g(r) = \frac{G M}{r^2}.$
The graph of gravity versus distance r is an inverse-square curve: it decays as 1/r^2 and approaches zero as r → ∞, never quite reaching zero at finite r.
At the Earth’s surface (r ≈ RE), the acceleration is $g(R</em>E) = \frac{G M}{R_E^2} \approx 9.81\ \text{m s}^{-2}.$
This is the familiar “9.8 m/s^2” value used for near-surface problems.

Gravity inside a uniformly dense sphere changes with radius: the mass enclosed by a sphere of radius r is
$M(r) = \rho \frac{4}{3} \pi r^3,$
where ρ is the constant density.
The gravitational field inside, for r ≤ R_E, is
$g(r) = \frac{G M(r)}{r^2} = \frac{G \left( \rho \frac{4}{3} \pi r^3 \right)}{r^2} = \frac{4}{3} \pi G \rho \; r.$
This shows g(r) grows linearly with r as you go from the center outward.
At the very center (r = 0), g(0) = 0 due to symmetry: shells inside do not contribute any net gravitational force on the center.
Outside the Earth (r ≥ R_E), the full mass M contributes and the 1/r^2 law applies as usual.
The shell theorem helps explain why only the mass inside radius r contributes to g(r) inside the Earth (outer shells exert no net force on a mass located inside them).

The standard Newtonian gravity picture uses a point-m mass model for simplicity; however, real bodies are extended objects.
When you plot g(r) vs r for a point mass, you see an inverse-square curve with asymptotic approach toward zero as r grows.
Inside a planet, the gravity decreases linearly with r, reaching zero at the center.
This motivates the idea that the gravitational potential and field require calculus to fully describe, especially when dealing with variable r and internal structures.

Newton explained gravity as a central force with an inverse-square law and laid the groundwork for calculus-based descriptions of gravity.
The development of calculus by Newton (and independently by Leibniz) enabled the precise treatment of changing gravitational forces and potentials.
The Cavendish experiment (one of the earliest precise gravitational experiments) is historically notable for measuring the gravitational constant G, enabling quantitative predictions of F = GMm/r^2.
The gravitational constant G is a universal constant and, together with M and R, gives the exact value of g at any distance r from a spherical mass M.

Newton's second law (net force and acceleration):
$F_{ ext{net}} = m a.$
Gravitational force between Earth and a mass m:
$F_g = \frac{G M m}{r^2}.$
Resulting acceleration due to gravity (independent of m):
$a = \frac{F}{m} = \frac{G M}{r^2}.$
Gravity at the Earth's surface:
$g(R<em>E) = \frac{G M}{R</em>E^2} \approx 9.81 \ \text{m s}^{-2}.$
Radius of Earth (approximate values):
$R_E \approx 6.371 \times 10^6 \ \text{m} \approx 3.959 \text{ thousand miles}.$
Inside Earth, for constant density ρ:
$g(r) = \frac{4}{3} \pi G \rho \; r.$
Mass enclosed within radius r (uniform density):
$M(r) = \rho \frac{4}{3} \pi r^3.$
Outside Earth, the field follows the same inverse-square law as for any point mass:
$g(r) = \frac{G M}{r^2}.$

The key takeaway is that gravity near Earth acts as a central force toward the Earth's center and decreases with distance as 1/r^2 in the exterior, while inside a uniform Earth it increases linearly with r up to the surface.
Practice problems: distinguish when to treat Earth as a point mass vs. when to consider the Earth's extended structure (inside vs outside the radius).
When solving, start from the force picture (F = GMm/r^2) and then derive acceleration (a = F/m) to see how mass independence arises.
Remember to use the metric system for distances and speeds in gravity problems, and convert miles to meters when needed.

If you double your distance from Earth’s center, how does g change? Answer: it becomes four times smaller, since g ∝ 1/r^2.
Inside a uniform Earth, does gravity increase or decrease as you move toward the center? Answer: It decreases linearly with r, going to zero at the center.
Why does the mass of the falling object not affect its acceleration near the surface? Answer: Because a = F/m and F_g = GMm/r^2, so m cancels in the expression for a.
What is the shell theorem’s relevance to gravity inside the Earth? Answer: It implies that only the mass within radius r contributes to the gravitational force at distance r from the center; outer shells contribute no net force on a mass inside them.

One problem tomorrow night on circular motion (think in terms of centripetal acceleration and gravitational force).
A gravitation problem for Sunday (apply F = GMm/r^2 and a = F/m, with appropriate r for the scenario).
Sketch the qualitative shape of the gravity vs. distance graph (outside vs inside the Earth) and annotate different regions with the functional form of g(r).