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

where:

• $F_g$ = gravitational force (N)
• $G$ = universal gravitational constant = $6.67 \times 10^{-11}$ N·m²/kg²
• $m_1, m_2$ = masses of the two objects (kg)
• $r$ = distance between the centers of the masses (m)

💡 Key Insight: Gravity acts between ALL objects, but we only notice it for very massive objects (like planets) because $G$ is extremely small!

Characteristics of Gravitational Force

1. Always Attractive

Unlike electric force (which can attract or repel), gravitational force is always attractive.

2. Action-Reaction Pairs

By Newton's 3rd Law, the force Earth exerts on you equals the force you exert on Earth:

$F_{Earth \to you} = F_{you \to Earth}$

Both have magnitude $\frac{Gm_{Earth}m_{you}}{r^2}$

3. Inverse Square Law

Force is inversely proportional to $r^2$:

• If distance doubles, force becomes $\frac{1}{4}$ as strong
• If distance triples, force becomes $\frac{1}{9}$ as strong

4. Long Range

Gravity has infinite range (never becomes exactly zero), though it becomes negligibly small at large distances.

Gravitational Field

The gravitational field $g$ at a point is the gravitational force per unit mass:

$g = \frac{F_g}{m} = \frac{GM}{r^2}$

where $M$ is the mass creating the field and $r$ is the distance from its center.

At Earth's Surface

$g = \frac{GM_{Earth}}{R_{Earth}^2} = 9.8 \text{ m/s}^2$

where:

• $M_{Earth} = 5.97 \times 10^{24}$ kg
• $R_{Earth} = 6.37 \times 10^6$ m

Above Earth's Surface

At height $h$ above Earth's surface:

$g_h = \frac{GM_{Earth}}{(R_{Earth} + h)^2}$

As altitude increases, $g$ decreases.

Example: At the altitude of the International Space Station (~400 km):

$g_{ISS} \approx 8.7 \text{ m/s}^2$

(About 89% of surface gravity - astronauts aren't "weightless" due to zero gravity, but due to free fall!)

Weight vs. Mass

Mass

• Definition: Amount of matter in an object
• Property: Intrinsic to the object
• Units: kg
• Changes?: Never changes

Weight

• Definition: Gravitational force on an object: $W = mg$
• Property: Depends on location
• Units: N (newtons)
• Changes?: Yes! Different on Moon, Mars, in orbit, etc.

Example: A 70 kg person

• Mass: 70 kg (everywhere)
• Weight on Earth: $W = 70 \times 9.8 = 686$ N
• Weight on Moon: $W = 70 \times 1.6 = 112$ N (Moon's $g \approx 1.6$ m/s²)

Gravitational Force Inside vs. Outside

Outside a Uniform Sphere

Use the distance from the center of the sphere:

$F = \frac{GMm}{r^2}$

This applies to any point outside the sphere (including at the surface).

Inside a Uniform Sphere

Only the mass inside radius $r$ contributes to the force. The shell of mass outside $r$ has zero net effect!

$F = \frac{GM_{inside}m}{r^2}$

For a uniform sphere: $M_{inside} = M\frac{r^3}{R^3}$

So: $F = \frac{GMm}{R^3}r$

The force is proportional to $r$ inside a uniform sphere (increases linearly from zero at center).

Orbital Motion

When gravitational force provides the centripetal force, objects orbit!

Circular Orbits

Set gravitational force equal to required centripetal force:

$\frac{GMm}{r^2} = \frac{mv^2}{r}$

Simplify (mass $m$ cancels):

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

Orbital Speed

The orbital speed is independent of the satellite's mass and depends only on:

• The mass of the central body ($M$)
• The orbital radius ($r$)

Important: Closer orbits are faster! (Smaller $r$ means larger $v$)

Orbital Period

Using $v = \frac{2\pi r}{T}$:

$\frac{2\pi r}{T} = \sqrt{\frac{GM}{r}}$

$T = 2\pi\sqrt{\frac{r^3}{GM}}$

Kepler's 3rd Law: $T^2 \propto r^3$

Escape Velocity

The escape velocity is the minimum speed needed to escape a planet's gravitational pull:

$v_{esc} = \sqrt{\frac{2GM}{R}}$

For Earth: $v_{esc} \approx 11,200$ m/s (about 25,000 mph!)

Note: This is $\sqrt{2}$ times the orbital velocity at the surface.

⚠️ Common Mistakes

Mistake 1: Forgetting r²

❌ Wrong: $F_g = \frac{Gm_1m_2}{r}$ ✅ Right: $F_g = \frac{Gm_1m_2}{r^2}$

Mistake 2: Using Surface Distance for Satellites

❌ Wrong: Using Earth's radius for a satellite 1000 km up ✅ Right: Use $r = R_{Earth} + h = 6370 + 1000 = 7370$ km

Mistake 3: Confusing Weight and Mass

❌ Wrong: "I'm weightless in space, so my mass is zero" ✅ Right: Mass is constant; weight is $mg$ where $g$ varies with location

Mistake 4: Thinking Gravity Stops

❌ Wrong: "There's no gravity in space" ✅ Right: Gravity extends infinitely (though it weakens with distance). Astronauts feel "weightless" because they're in free fall, not because gravity is absent.

Problem-Solving Strategy

1. Identify the masses and the distance between their centers
2. Use $F_g = \frac{Gm_1m_2}{r^2}$ for gravitational force
3. For orbits, set $F_g = F_c$: $\frac{GMm}{r^2} = \frac{mv^2}{r}$
4. Remember the satellite's mass cancels in orbital calculations
5. Check if distance is from center or from surface (add radius if needed)

Real-World Applications

GPS Satellites

• Orbit at ~20,200 km altitude
• Must account for weaker gravity (slower time due to general relativity)
• Orbital period ~12 hours

Geostationary Satellites

• Orbit at ~35,800 km altitude
• Period = 24 hours (matches Earth's rotation)
• Appear stationary above a point on equator

Tides

• Moon's gravity creates tidal bulges
• Differential force (near side vs. far side) causes tides
• Sun also contributes (spring and neap tides)

Key Formulas Summary

Constants:

• $G = 6.67 \times 10^{-11}$ N·m²/kg²
• $M_{Earth} = 5.97 \times 10^{24}$ kg
• $R_{Earth} = 6.37 \times 10^6$ m

Find: Gravitational force $F_g$

Solution:

Use Newton's Law of Universal Gravitation:

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

Substitute values:

$F_g = (6.67 \times 10^{-11})\frac{(5.97 \times 10^{24})(7.35 \times 10^{22})}{(3.84 \times 10^8)^2}$

Calculate the numerator: $6.67 \times 5.97 \times 7.35 = 292.6$ $10^{-11} \times 10^{24} \times 10^{22} = 10^{35}$ $\text{Numerator} = 292.6 \times 10^{35} = 2.926 \times 10^{37}$

Calculate the denominator: $(3.84)^2 = 14.75$ $(10^8)^2 = 10^{16}$ $\text{Denominator} = 14.75 \times 10^{16} = 1.475 \times 10^{17}$

Divide: $F_g = \frac{2.926 \times 10^{37}}{1.475 \times 10^{17}}$

$F_g = 1.98 \times 10^{20} \text{ N}$

Answer: The gravitational force is approximately $1.98 \times 10^{20}$ N (or 198 billion billion newtons).

Note: By Newton's 3rd Law, this is both the force Earth exerts on the Moon AND the force the Moon exerts on Earth!

(a) Find gravitational field strength $g$ at altitude $h$

Step 1: Find the distance from Earth's center

$r = R_E + h = 6.37 \times 10^6 + 4.0 \times 10^5$

$r = 6.77 \times 10^6 \text{ m}$

Step 2: Calculate gravitational field

$g = \frac{GM_E}{r^2}$

$g = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(6.77 \times 10^6)^2}$

Numerator: $6.67 \times 5.97 \times 10^{13} = 3.98 \times 10^{14}$

Denominator: $(6.77)^2 \times 10^{12} = 45.8 \times 10^{12} = 4.58 \times 10^{13}$

$g = \frac{3.98 \times 10^{14}}{4.58 \times 10^{13}} = 8.69 \text{ m/s}^2$

(b) Find orbital speed $v$

Step 3: Set gravitational force = centripetal force

$\frac{GM_E m}{r^2} = \frac{mv^2}{r}$

The mass $m$ cancels:

$\frac{GM_E}{r} = v^2$

$v = \sqrt{\frac{GM_E}{r}}$

Step 4: Calculate

$v = \sqrt{\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{6.77 \times 10^6}}$

$v = \sqrt{\frac{3.98 \times 10^{14}}{6.77 \times 10^6}}$

$v = \sqrt{5.88 \times 10^7}$

$v = 7.67 \times 10^3 \text{ m/s} = 7670 \text{ m/s}$

Alternative for part (b):

Since $g = \frac{GM_E}{r^2}$, we have $GM_E = gr^2$

$v = \sqrt{\frac{GM_E}{r}} = \sqrt{gr}$

$v = \sqrt{(8.69)(6.77 \times 10^6)}$

$v = \sqrt{5.88 \times 10^7} = 7670 \text{ m/s}$

Answers:

• (a) $g \approx$ 8.69 m/s² (about 89% of surface gravity)
• (b) Orbital speed $\approx$ 7670 m/s (about 17,000 mph!)

Note: Despite $g$ being nearly the same as on Earth's surface, astronauts feel "weightless" because both they and the ISS are in free fall toward Earth!

(a) Find orbital radius $r$

Step 1: Use the period formula

$T = 2\pi\sqrt{\frac{r^3}{GM_E}}$

Step 2: Solve for $r$

Square both sides: $T^2 = 4\pi^2\frac{r^3}{GM_E}$

$r^3 = \frac{GM_E T^2}{4\pi^2}$

$r = \left(\frac{GM_E T^2}{4\pi^2}\right)^{1/3}$

Step 3: Calculate

Numerator: $GM_E T^2 = (6.67 \times 10^{-11})(5.97 \times 10^{24})(86400)^2$

$= (6.67)(5.97)(86400)^2 \times 10^{13}$

$= (6.67)(5.97)(7.464 \times 10^9) \times 10^{13}$

$= 297.3 \times 10^{22} = 2.973 \times 10^{24}$

Denominator: $4\pi^2 = 39.48$

$r^3 = \frac{2.973 \times 10^{24}}{39.48} = 7.53 \times 10^{22}$

$r = (7.53 \times 10^{22})^{1/3}$

Using calculator or estimation: $r \approx 4.22 \times 10^7 \text{ m}$

(b) Find altitude $h$ above Earth's surface

$h = r - R_E$

$h = 4.22 \times 10^7 - 6.37 \times 10^6$

$h = 4.22 \times 10^7 - 0.637 \times 10^7$

$h = 3.58 \times 10^7 \text{ m} = 35,800 \text{ km}$

(c) Find orbital speed $v$

Method 1: Using $v = \frac{2\pi r}{T}$

$v = \frac{2\pi (4.22 \times 10^7)}{86400}$

$v = \frac{2.65 \times 10^8}{8.64 \times 10^4}$

$v = 3.07 \times 10^3 \text{ m/s} = 3070 \text{ m/s}$

Method 2: Using $v = \sqrt{\frac{GM_E}{r}}$

$v = \sqrt{\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{4.22 \times 10^7}}$

$v = \sqrt{\frac{3.98 \times 10^{14}}{4.22 \times 10^7}}$

$v = \sqrt{9.43 \times 10^6}$

$v \approx 3070 \text{ m/s}$

Answers:

• (a) Orbital radius: 42,200 km (from Earth's center)
• (b) Altitude: 35,800 km (above Earth's surface)
• (c) Orbital speed: 3070 m/s (about 6,900 mph)

Application: Geostationary satellites at this altitude complete one orbit in exactly 24 hours, so they remain above the same point on Earth's equator - perfect for communication satellites!