6.5 Newton’s Universal Law of Gravitation

Newton's Universal Law of Gravitation

Every object in the universe attracts every other object through gravity. Newton's Universal Law of Gravitation quantifies this attraction, showing exactly how mass and distance determine the gravitational force between any two objects. This single law explains everything from why you have weight on Earth to why the Moon causes tides to why planets follow their orbits.

Newton's Universal Law of Gravitation

Newton's law states that every particle attracts every other particle with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. That inverse-square relationship is key: double the distance, and the force drops to one-quarter.

The mathematical form is:

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

• $F$ = gravitational force between the two objects (N)
• $G$ = gravitational constant = $6.674 \times 10^{-11} \text{ N} \cdot \text{m}^2 / \text{kg}^2$
• $m_1$ and $m_2$ = masses of the two objects (kg)
• $r$ = distance between the centers of the two objects (m)

Notice that $r$ is measured center-to-center, not surface-to-surface. For a person standing on Earth, $r$ is Earth's radius (~6,371 km), not zero.

Connection to $F_g = mg$

Near Earth's surface, the universal law simplifies to the familiar weight equation:

$F_g = mg$

• $m$ = mass of the object (kg)
• $g$ = acceleration due to gravity near Earth's surface ($9.8 \text{ m/s}^2$)

Where does $g = 9.8 \text{ m/s}^2$ come from? Set the two expressions equal for an object of mass $m$ near Earth's surface:

$mg = G \frac{M_E \, m}{R_E^2}$

The object's mass $m$ cancels, giving $g = G \frac{M_E}{R_E^2}$. Plug in Earth's mass ($5.97 \times 10^{24}$ kg) and radius ($6.37 \times 10^6$ m), and you get $9.8 \text{ m/s}^2$. So $g$ isn't a separate constant; it comes directly from Newton's universal law.

Calculation of Gravitational Forces

To find the gravitational force between any two objects:

1. Identify the two masses $m_1$ and $m_2$ (in kg).
2. Determine the center-to-center distance $r$ (in m).
3. Substitute into $F = G \frac{m_1 m_2}{r^2}$ and solve.

Example: Earth–Moon Gravitational Force

• $m_1$ (Earth) = $5.97 \times 10^{24}$ kg
• $m_2$ (Moon) = $7.34 \times 10^{22}$ kg
• $r$ = $3.84 \times 10^8$ m

$F = (6.674 \times 10^{-11}) \frac{(5.97 \times 10^{24})(7.34 \times 10^{22})}{(3.84 \times 10^8)^2} \approx 1.98 \times 10^{20} \text{ N}$

That's roughly $2 \times 10^{20}$ N, an enormous force that keeps the Moon in orbit. By Newton's Third Law, the Moon pulls on Earth with the exact same force.

Moon's Gravity and Earth's Tides

The Moon's gravitational pull isn't uniform across Earth. The side of Earth closest to the Moon feels a stronger pull than the far side. This difference in force across Earth's diameter creates tidal bulges on both the near and far sides, which is why most coastlines experience two high tides per day as Earth rotates.

• Spring tides occur during new and full moons, when the Sun, Earth, and Moon are roughly aligned. The gravitational effects of the Sun and Moon reinforce each other, producing higher high tides and lower low tides.
• Neap tides occur during first-quarter and third-quarter moons, when the Sun and Moon are at right angles relative to Earth. Their gravitational effects partially cancel, producing a smaller tidal range.

Tidal range varies dramatically by location due to coastline shape and ocean basin geometry. The Bay of Fundy in Canada sees tidal ranges up to 16 m, while parts of the Gulf of Mexico experience less than 1 m.

Apparent Weightlessness in Orbit

Astronauts on the International Space Station (orbiting at ~400 km altitude) appear to float, but gravity hasn't disappeared. At that altitude, $g$ is still about $8.7 \text{ m/s}^2$, roughly 89% of its surface value.

The reason astronauts feel weightless is that both the spacecraft and everything inside it are in free fall together. The station has enough forward velocity (~7,660 m/s) that as it falls toward Earth, it continually "misses" the surface. Since the astronauts and the station accelerate at the same rate, there's no normal force pushing on the astronauts. With no normal force, you feel no weight.

This is apparent weightlessness, not zero gravity. Gravity is very much still there; it's what keeps the station in orbit.

Celestial Mechanics and Orbital Motion

Celestial mechanics is the study of how gravitational forces govern the motion of planets, moons, and satellites. A few core ideas tie this together:

• Orbital paths are elliptical, as described by Kepler's First Law. Circular orbits are a special case of an ellipse. Both planets orbiting the Sun and satellites orbiting Earth follow these paths.
• Gravitational field strength decreases with distance from a massive object, following the same inverse-square relationship. Farther from Earth, the gravitational field is weaker, which is why orbits at higher altitudes have longer periods.
• Escape velocity is the minimum speed an object needs to completely break free from a body's gravitational pull without any further propulsion. For Earth's surface, escape velocity is about 11,200 m/s (roughly 40,000 km/h). It depends on the mass of the body and the distance from its center: $v_{escape} = \sqrt{\frac{2GM}{r}}$.