Training Orbital Mechanics Newton's Law of Gravitation and Gravitational Fields
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Newton's Law of Gravitation and Gravitational Fields

25 min Orbital Mechanics

Newton's Law of Gravitation

In 1687, Isaac Newton published one of the most consequential equations in human history: the law of universal gravitation. Every object with mass attracts every other object with mass, and the force between them is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This single equation explains why apples fall, why the Moon orbits Earth, why planets orbit the Sun, and why galaxies hold together.

The gravitational force between two masses M and m separated by distance r is F equals G times M times m over r squared, where G is the universal gravitational constant, approximately 6.674 times 10 to the negative 11 in SI units. The force is always attractive and acts along the line connecting the two centers of mass.

Near Earth's surface, this simplifies to the familiar F equals mg, where g equals GM over R-squared is the local gravitational acceleration. Earth's g is approximately 9.81 meters per second squared at sea level, but it decreases with altitude because r increases. At the altitude of the International Space Station, about 400 kilometers up, g is still about 8.7 meters per second squared — the astronauts float not because gravity is absent, but because they are in continuous free fall around Earth.

The gravitational field strength at distance r from a mass M is simply g of r equals GM over r-squared. This field permeates all of space, weakening with the square of distance but never truly reaching zero. Understanding this inverse-square relationship is the key to all of orbital mechanics.

Newton's Law of Universal Gravitation

$$F = \frac{G M m}{r^2}$$

where $G = 6.674 \times 10^{-11}$ N·m²/kg², $M$ and $m$ are the masses, and $r$ is the center-to-center distance.

Gravitational Field Strength

$$g(r) = \frac{GM}{r^2}$$

At Earth's surface ($R_E = 6{,}371$ km): $g = \frac{GM_E}{R_E^2} \approx 9.81$ m/s².

Standard Gravitational Parameter

$$\mu = GM$$

For Earth: $\mu_E = 3.986 \times 10^{14}$ m³/s². This combined parameter appears in almost every orbital equation, eliminating the need to know $G$ and $M$ separately.

Example 1

Calculate the gravitational force between Earth ($M = 5.972 \times 10^{24}$ kg) and a 1000 kg satellite at altitude 400 km.

Step 1: Find the center-to-center distance.

$$r = R_E + h = 6{,}371{,}000 + 400{,}000 = 6{,}771{,}000 \text{ m}$$

Step 2: Apply Newton's law.

$$F = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})(1000)}{(6.771 \times 10^6)^2}$$

Step 3: Evaluate.

$$F = \frac{3.986 \times 10^{17}}{4.585 \times 10^{13}} = 8{,}694 \text{ N}$$

The satellite still weighs 8,694 N — about 89% of its surface weight (9,810 N). Gravity has only decreased by 11%.

Example 2

At what altitude above Earth is $g$ exactly half of its surface value?

Step 1: Set $g(r) = g_0/2$.

$$\frac{GM}{r^2} = \frac{1}{2} \cdot \frac{GM}{R_E^2}$$

Step 2: Solve for $r$.

$$r^2 = 2R_E^2 \implies r = R_E\sqrt{2} = 6371 \times 1.414 = 9{,}009 \text{ km}$$

Step 3: Find altitude.

$$h = r - R_E = 9009 - 6371 = 2{,}638 \text{ km}$$

Example 3

Find the gravitational acceleration on the Moon's surface. ($M_{\text{Moon}} = 7.342 \times 10^{22}$ kg, $R_{\text{Moon}} = 1{,}737$ km)

Step 1: Apply $g = GM/R^2$.

$$g = \frac{(6.674 \times 10^{-11})(7.342 \times 10^{22})}{(1.737 \times 10^6)^2} = \frac{4.899 \times 10^{12}}{3.017 \times 10^{12}} = 1.62 \text{ m/s}^2$$

Lunar gravity is about $1/6$ of Earth's, which is why Apollo astronauts could hop and bounce so easily.

Interactive Explorer: Gravitational Field Calculator
Distance from center (r) = 6771 km
g(r) = 8.69 m/s²
Fraction of surface g = 88.6 %
0g / g₀100%

Practice Problems

1. Calculate the gravitational force between Earth and the Moon. ($M_E = 5.972 \times 10^{24}$ kg, $M_M = 7.342 \times 10^{22}$ kg, $d = 384{,}400$ km)
2. At what altitude is $g = 1$ m/s²?
3. Find $g$ at the altitude of GPS satellites (20,200 km).
4. Two 100 kg spheres are 1 m apart. What is the gravitational force between them?
5. Mars has $M = 6.417 \times 10^{23}$ kg and $R = 3{,}390$ km. Find surface $g$.
6. Show that $g$ decreases by approximately $2h/R_E$ for small altitudes $h \ll R_E$.
7. A 70 kg person weighs 686 N on Earth's surface. What do they weigh at the ISS altitude (408 km)?
8. At what distance from Earth's center does the gravitational pull of the Moon exactly equal Earth's? ($d_{EM} = 384{,}400$ km)
9. If Earth's radius suddenly doubled but mass stayed the same, what would $g$ be at the new surface?
10. Calculate $\mu$ for the Sun ($M = 1.989 \times 10^{30}$ kg).
Show Answer Key

1. $F = G M_E M_M / d^2 = (6.674 \times 10^{-11})(5.972 \times 10^{24})(7.342 \times 10^{22})/(3.844 \times 10^8)^2 = 1.98 \times 10^{20}$ N.

2. $GM/r^2 = 1 \Rightarrow r = \sqrt{GM/1} = \sqrt{3.986 \times 10^{14}} = 1.997 \times 10^7$ m $= 19{,}965$ km. Altitude $= 19965 - 6371 = 13{,}594$ km.

3. $r = 6371 + 20200 = 26{,}571$ km. $g = 3.986 \times 10^{14}/(2.657 \times 10^7)^2 = 0.564$ m/s².

4. $F = (6.674 \times 10^{-11})(100)(100)/1^2 = 6.674 \times 10^{-7}$ N. Tiny!

5. $g = (6.674 \times 10^{-11})(6.417 \times 10^{23})/(3.39 \times 10^6)^2 = 3.72$ m/s².

6. $g(h) = GM/(R+h)^2 = g_0/(1+h/R)^2 \approx g_0(1-2h/R)$ for $h \ll R$.

7. $g(408) = 9.81 \times (6371/6779)^2 = 9.81 \times 0.882 = 8.66$ m/s². Weight $= 70 \times 8.66 = 606$ N.

8. $GM_E/r^2 = GM_M/(d-r)^2$. $r/(d-r) = \sqrt{M_E/M_M} = \sqrt{81.3} = 9.02$. $r = 9.02d/10.02 = 346{,}100$ km.

9. $g = GM/(2R)^2 = g_0/4 = 2.45$ m/s².

10. $\mu = (6.674 \times 10^{-11})(1.989 \times 10^{30}) = 1.327 \times 10^{20}$ m³/s².

Circular Orbits — Velocity, Period, and Altitude