Newton's Law of Gravitation

Gravity is an invisible force shared between any two objects. Often we think of gravity as a force that simply pulls things down toward Earth, but gravity acts between any two objects that have mass.

For example, two asteroids with similar velocities in relatively close proximity will eventually gravitate toward each other and collide in orbit. The two objects attract one another simply because they possess mass. The magnitude and direction of this force are described by Newton's Law of Universal Gravitation:

Newton's Law of Universal Gravitation

The gravitational force $\vec{F}_{ab}$ that object $b$ exerts onto object $a$ can be expressed as:

\vec{F}_{ab} = G \frac{m_a m_b}{r_{ab}^2} \hat{r}_{ab}

Where $G$ is the universal gravitational constant, $m_a$ and $m_b$ are the masses of objects $a$ and $b$ , $r_{ab}$ is the distance between their centers of mass, and $\hat{r}_{ab}$ is a unit vector pointing from object $a$ toward object $b$ .

The universal gravitational constant $G$ is approximately defined as:

G \approx 6.674 \times 10^{-11} \text{ } \frac{\text{N} \cdot \text{m}^2}{\text{kg}^2}

When direction is already known or accounted for, the more frequently used scalar version of this law expresses just the magnitude of the force:

F_{ab} = G \frac{m_a m_b}{r_{ab}^2}

Example 1

Three masses are placed on a coordinate grid: Mass $A$ ( $100 \text{ kg}$ ) is at the origin $(0,0)$ , Mass $B$ ( $200 \text{ kg}$ ) is at $(4,0) \text{ m}$ , and Mass $C$ ( $300 \text{ kg}$ ) is at $(0,3) \text{ m}$ .

Determine the net gravitational force vector $\vec{F}_{\text{net}}$ acting on Mass $A$ in terms of unit vectors $\hat{i}$ and $\hat{j}$ .

Hint Click to unblur

\vec{F}_{g,net} = \vec{F}_{g,AB} + \vec{F}_{g,AC}

Solution Click to unblur

First, calculate the individual attractive force magnitudes along each axis using the scalar form, keeping units attached to each term:

F_{AB} = G \frac{m_A m_B}{r_{AB}^2} = G \frac{(100\text{ kg})(200\text{ kg})}{(4\text{ m})^2} = 1250\,G\,\frac{\text{kg}^2}{\text{m}^2}

F_{AC} = G \frac{m_A m_C}{r_{AC}^2} = G \frac{(100\text{ kg})(300\text{ kg})}{(3\text{ m})^2} \approx 3333.3\,G\,\frac{\text{kg}^2}{\text{m}^2}

Since Mass $B$ pulls along the positive x-axis and Mass $C$ pulls along the positive y-axis, we apply unit vectors $\hat{i}$ and $\hat{j}$ to construct the net force vector:

\vec{F}_{\text{net}} = G \left( 1250\,\hat{i} + 3333.3\,\hat{j} \right) \frac{\text{kg}^2}{\text{m}^2}

Finally, substituting the gravitational constant $G \approx 6.674 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2$ , we evaluate the actual force in Newtons:

\vec{F}_{\text{net}} \approx \left( 8.34 \times 10^{-8}\,\hat{i} + 2.22 \times 10^{-7}\,\hat{j} \right) \text{ N}

Example 2

A spacecraft of mass $m$ travels along the straight line connecting the center of the Earth (mass $M_E$ ) and the center of the Moon (mass $M_M$ ). Let $d$ be the total center-to-center distance between Earth and the Moon.

At what distance $x$ measured from the center of the Earth does the net gravitational force on the spacecraft equal zero? Express your answer symbolically in terms of $M_E$ , $M_M$ , and $d$ .

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Set the magnitude of the pull from Earth equal to the pull from Moon: $F_{E} = F_{M}$ .

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At the neutral point, the gravitational pull toward Earth must exactly balance the pull toward the Moon:

G \frac{m M_E}{x^2} = G \frac{m M_M}{(d - x)^2}

Notice that the spacecraft's mass $m$ and the universal gravitational constant $G$ appear on both sides and cancel out immediately:

\frac{M_E}{x^2} = \frac{M_M}{(d - x)^2}

Taking the square root of both sides allows us to solve directly for position $x$ :

\frac{\sqrt{M_E}}{x} = \frac{\sqrt{M_M}}{d - x}

Cross-multiplying and isolating $x$ yields the final symbolic formula:

x = \frac{d}{1 + \sqrt{\frac{M_M}{M_E}}}

Example 3: Orbital Acceleration at the ISS

The International Space Station (ISS) orbits Earth at an average altitude of $400\text{ km}$ above the surface. Earth's mass is approximately $M_E = 5.972 \times 10^{24}\text{ kg}$ and its mean radius is $R_E = 6371\text{ km}$ .

Calculate the instantaneous gravitational acceleration $a_g$ experienced by the ISS in its orbit, and compare it to standard surface gravity ( $g = 9.81\text{ m/s}^2$ ).

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Combine Newton's Law of Universal Gravitation with Newton's Second Law ( $F = ma$ ). What happens to the spacecraft's mass $m$ ?

Solution Click to unblur

First, find the total orbital radius $r$ from the center of the Earth:

r = R_E + \text{altitude} = 6371\text{ km} + 400\text{ km} = 6771\text{ km} = 6.771 \times 10^6\text{ m}

Using Newton's Second Law ( $F = m_{\text{ISS}} a_g$ ) set equal to the gravitational force:

m_{\text{ISS}} a_g = G \frac{M_E m_{\text{ISS}}}{r^2}

The spacecraft mass cancels out, leaving the formula for gravitational acceleration at any altitude:

a_g = G \frac{M_E}{r^2}

(1)

Substituting the known values:

a_g = (6.674 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2) \frac{5.972 \times 10^{24}\text{ kg}}{(6.771 \times 10^6\text{ m})^2}

a_g \approx 8.68\text{ m/s}^2

(2)

Compared to surface gravity ( $9.81\text{ m/s}^2$ ), gravity at ISS altitude is still about $88.6\%$ as strong. Astronauts float not because gravity is absent, but because they are in continuous freefall around the planet.