Motion in Orbits (Senior 5 Physics): free lessons | Ganzaa

Motion in Orbits

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Senior 5 · Physics

Motion in Orbits

4 lessons

Lesson 1 of 4

Newton's Law of Gravitation and the Gravitational Field

25 min

The apple and the Moon

A mango falls from a tree in a garden in Huye. The Moon, high above the same garden, does not fall to the ground; it circles the Earth once every 27.3 days. Isaac Newton's great step, published in 1687, was to see that both are pulled by the same force: the Moon is falling too, but it is moving sideways so fast that it keeps missing the Earth. If the force weakens with the square of the distance, Newton argued, the numbers for the mango and for the Moon should agree. In this lesson you will check his calculation yourself.

Newton's law of universal gravitation

Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them:

F = \frac{Gm_1m_2}{r^2}

$G = 6.67 \times 10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}$ is the universal gravitational constant. Points to note:

The forces on the two bodies are equal in size and opposite in direction (Newton's third law), even if one mass is far larger.
The law is for point masses, but Newton proved that a uniform sphere attracts an outside body as if all its mass were at its centre, so $r$ is measured centre to centre.
Gravity is always attractive, and it is extremely weak: two 60 kg students 1.0 m apart attract each other with $F = \frac{6.67 \times 10^{-11} \times 60 \times 60}{1.0^2} = 2.4 \times 10^{-7}\,\mathrm{N}$ , far too small to feel. It dominates only when at least one mass is enormous.

Measuring G

Because gravity between ordinary objects is so weak, $G$ was first measured with a very delicate torsion balance, by Henry Cavendish in 1798. Two small spheres on a light rod hang from a fine fibre; two large lead spheres placed near them pull them sideways and twist the fibre by a tiny angle. Knowing the torque needed to twist the fibre, the masses and the distances, $G$ can be calculated.

Definition

Newton's law of gravitation: the gravitational force between two point masses $m_{1}$ and $m_{2}$ a distance $r$ apart is $F = \frac{Gm_1m_2}{r^2}$ , attractive, along the line joining them. $G = 6.67 \times 10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}$ . Example: the Earth ( $5.97 \times 10^{24}\,\mathrm{kg}$ ) and the Moon ( $7.35 \times 10^{22}\,\mathrm{kg}$ ), $3.84 \times 10^8\,\mathrm{m}$ apart, attract each other with $2.0 \times 10^{20}\,\mathrm{N}$ .

The gravitational field

A mass changes the space around it so that any other mass placed there feels a force. This region is a gravitational field. Its strength at a point is the force per unit mass on a small test mass placed there:

g = \frac{F}{m} \qquad\text{(unit } \mathrm{N\,kg^{-1}}\text{, which equals } \mathrm{m\,s^{-2}}\text{)}

For a spherical body of mass $M$ , the force on a test mass $m$ at distance $r$ from its centre is $\frac{GMm}{r^2}$ , so

g = \frac{GM}{r^2}

The field strength is a vector pointing towards the centre of the body. It equals the acceleration of free fall at that point, which is why we use the same symbol $g$ .

Worked example 1: weighing the Earth

At the Earth's surface $g = 9.81\,\mathrm{N\,kg^{-1}}$ and the Earth's mean radius is $6.37 \times 10^6\,\mathrm{m}$ . Find the mass of the Earth and its mean density.

Solution

$g = \frac{GM}{R^2}$ gives $M = \frac{gR^2}{G} = \frac{9.81 \times (6.37 \times 10^6)^2}{6.67 \times 10^{-11}} = 5.97 \times 10^{24}\,\mathrm{kg}$ .

Volume $= \frac{4}{3}\pi R^3 = 1.08 \times 10^{21}\,\mathrm{m^3}$ , so the mean density is $\frac{5.97 \times 10^{24}}{1.08 \times 10^{21}} = 5.5 \times 10^3\,\mathrm{kg\,m^{-3}}$ . Surface rocks are only about $2.7 \times 10^3\,\mathrm{kg\,m^{-3}}$ , so the Earth's interior must be much denser: evidence for its iron core. This is why Cavendish's experiment was called "weighing the Earth".

Worked example 2: Newton's Moon test

The Moon orbits at $r = 3.84 \times 10^8\,\mathrm{m}$ from the Earth's centre, with a period of 27.3 days. Compare (a) the Earth's gravitational field strength at the Moon's distance with (b) the Moon's centripetal acceleration.

Solution

(a) $g = \frac{GM}{r^2} = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(3.84 \times 10^8)^2} = 2.70 \times 10^{-3}\,\mathrm{m\,s^{-2}}$ .

(b) $T = 27.3 \times 86\,400 = 2.36 \times 10^6\,\mathrm{s}$ ; $a = \frac{4\pi^2r}{T^2} = \frac{4\pi^2 \times 3.84 \times 10^8}{(2.36 \times 10^6)^2} = 2.72 \times 10^{-3}\,\mathrm{m\,s^{-2}}$ .

They agree within 1%. The Moon is 60 Earth radii away, and its acceleration is about $\frac{1}{60^2} = \frac{1}{3600}$ of 9.81: the same force that drops the mango holds the Moon in orbit.

Worked example 3: where the pulls cancel

Where, on the line between the Earth and the Moon, is the resultant gravitational field zero? ( $M_E = 81.2\,M_M$ , separation $d = 3.84 \times 10^8\,\mathrm{m}$ .)

Solution

At a distance $x$ from the Earth's centre: $\frac{GM_E}{x^2} = \frac{GM_M}{(d - x)^2}$ , so $\frac{x}{d - x} = \sqrt{\frac{M_E}{M_M}} = \sqrt{81.2} = 9.01$ . Then $x = \frac{9.01}{10.01}d = 0.900d = 3.46 \times 10^8\,\mathrm{m}$ : 90% of the way to the Moon.

Warning

Common mistakes. (1) Measuring $r$ from the surface instead of from the centre of a planet. (2) Forgetting to square $r$ : doubling the distance divides the force by four. (3) Thinking the Earth pulls the Moon harder than the Moon pulls the Earth: the forces are equal; the Earth's acceleration is smaller because its mass is larger. (4) Confusing $g$ (field strength, $9.81\,\mathrm{N\,kg^{-1}}$ ) with $G$ (the universal constant).

Tip

Most gravitation calculations need only the product $G M$ . For the Earth, $GM_E = gR^2 = 3.99 \times 10^{14}\,\mathrm{m^3\,s^{-2}}$ : remember it, or rebuild it from $g$ and $R$ , and you avoid two error-prone powers of ten.

Watch gravity pull two metal balls together (Steve Mould). Watch a home-built torsion balance twist as large masses attract small ones, and how tiny the force is.

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Practice

State Newton's law of gravitation and give the unit of $G$ in base units.
Find the gravitational force between two 1.0 kg spheres whose centres are 10 cm apart.
Mars has mass $6.42 \times 10^{23}\,\mathrm{kg}$ and radius $3.39 \times 10^6\,\mathrm{m}$ . Find the gravitational field strength at its surface, and the weight there of a 70 kg astronaut.
By what factor does the gravitational force between two bodies change if both masses double and the distance between them also doubles?
The Moon has mass $7.35 \times 10^{22}\,\mathrm{kg}$ and radius $1.74 \times 10^6\,\mathrm{m}$ . Show that the field strength at its surface is about one sixth of the Earth's.
Draw the field lines near the Earth's surface over your school, and around the whole Earth. Explain why they differ.
Use $g = 9.81\,\mathrm{N\,kg^{-1}}$ and $R = 6.37 \times 10^6\,\mathrm{m}$ to find $G M_{E}$ , then the field strength at a height equal to one Earth radius.
Gatete says: "The Moon is not falling, because it is in orbit." Evaluate this statement.

Show answersHide answers

The force between two point masses is proportional to the product of the masses and inversely proportional to the square of their separation. $G$ in $\mathrm{N\,m^2\,kg^{-2}} = \mathrm{m^3\,kg^{-1}\,s^{-2}}$ .
$F = \frac{6.67 \times 10^{-11} \times 1.0 \times 1.0}{0.10^2} = 6.7 \times 10^{-9}\,\mathrm{N}$ .
$g = \frac{6.67 \times 10^{-11} \times 6.42 \times 10^{23}}{(3.39 \times 10^6)^2} = 3.73\,\mathrm{N\,kg^{-1}}$ ; weight $= 70 \times 3.73 = 261\,\mathrm{N}$ .
$\frac{2 \times 2}{2^2} = 1$ : unchanged.
$g = \frac{6.67 \times 10^{-11} \times 7.35 \times 10^{22}}{(1.74 \times 10^6)^2} = 1.62\,\mathrm{N\,kg^{-1}}$ ; $\frac{1.62}{9.81} = 0.165 \approx \frac{1}{6}$ .
Near the surface: parallel, equally spaced, vertical lines (uniform field over a small region). Around the Earth: radial lines pointing to the centre, spreading out with distance (field weakens as $\frac{1}{r^2}$ ).
$GM_E = 9.81 \times (6.37 \times 10^6)^2 = 3.98 \times 10^{14}\,\mathrm{m^3\,s^{-2}}$ . At $r = 2 R$ : $g = \frac{9.81}{4} = 2.45\,\mathrm{N\,kg^{-1}}$ .
Wrong. The Moon is accelerating towards the Earth all the time ( $2.7 \times 10^{-3}\,\mathrm{m\,s^{-2}}$ ), so it is falling; its sideways speed carries it past the Earth so it never gets closer. An orbit is continuous free fall.

Key points

$F = \frac{Gm_1m_2}{r^2}$ , with $G = 6.67 \times 10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}$ ; $r$ from centre to centre.
Field strength $g = \frac{F}{m}$ ; for a sphere $g = \frac{GM}{r^2}$ , pointing to its centre.
$GM_E = gR^2 \approx 3.99 \times 10^{14}\,\mathrm{m^3\,s^{-2}}$ ; from it, $M_E = 5.97 \times 10^{24}\,\mathrm{kg}$ .
The Moon's centripetal acceleration matches $\frac{GM_E}{r^2}$ : the same law governs falling bodies and orbits.

Sources and credits

13.1 Newton's Law of Universal Gravitation and 13.2 Gravitation Near Earth's Surface, from University Physics Volume 1 by William Moebs, Samuel J. Ling and Jeff Sanny, OpenStax, licensed under CC BY 4.0. Adapted by Ganzaa.
Video: Watch gravity pull two metal balls together, Steve Mould, embedded from YouTube.

The three diagrams, the worked examples and the practice exercises are written by Ganzaa.

1The apple and the Moon

2Newton's law of universal gravitation

Measuring G

3The gravitational field

The apple and the Moon

Newton's law of universal gravitation

The gravitational field