Theory

Gravitation

The law of universal gravitation, the gravitational field and potential energy, the variation of g with altitude, circular orbits of satellites, escape velocity and Kepler's laws derived from dynamics.

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Law of universal gravitation
Gravitational field

Law of universal gravitation

Two point masses $m_{1}$ and $m_{2}$ , at distance $r$ from each other, attract each other with forces directed along the line joining them, of magnitude

F = G \frac{m _{1} m _{2}}{r ^{2}},

where $G = 6.67 \cdot 1 0^{- 11} N \cdot m^{2} /k g^{2}$ is the universal gravitational constant. In vector form, with $\overset{u}{^}_{r}$ the unit vector pointing from $m_{1}$ toward $m_{2}$ , the force that $m_{1}$ exerts on $m_{2}$ is

F_{12} = - G \frac{m _{1} m _{2}}{r ^{2}} \overset{u}{^}_{r},

and the force that $m_{2}$ exerts on $m_{1}$ is $F_{21} = - F_{12}$ , in agreement with Newton’s third law.

It is an experimental law, formulated by Newton and verified in the laboratory by Cavendish with the torsion balance, with which the value of $G$ was measured for the first time. It is used to compute the force between two bodies when their sizes are small compared with their distance, and, by superposition, the force between extended bodies. Two people of $70 kg$ and $60 kg$ at a distance of $1.0 m$ attract each other with a force of $6.67 \cdot 1 0^{- 11} \cdot \frac{70 \cdot 60}{1. 0 ^{2}} N = 2.8 \cdot 1 0^{- 7} N$ , less than a billionth of the weight of each; gravitation becomes relevant only when at least one of the masses is astronomical.

Acceleration of the Moon

The $\frac{1}{r ^{2}}$ dependence can be checked with the motion of the Moon. The Moon travels around the Earth on a nearly circular orbit of radius $d = 3.84 \cdot 1 0^{8} m$ in $T = 27.3$ days, so its centripetal acceleration is

a_{L} = \frac{4 π ^{2} d}{T ^{2}} = 2.72 \cdot 1 0^{- 3} m/ s^{2} .

The Earth-Moon distance is about $60$ times the Earth’s radius $R_{T} = 6.37 \cdot 1 0^{6} m$ ; if the Earth’s attraction decreases as $\frac{1}{r ^{2}}$ , at the distance of the Moon the gravitational acceleration is $g_{0} (\frac{R _{T}}{d})^{2} = 2.70 \cdot 1 0^{- 3} m/ s^{2}$ , where $g_{0} = 9.81 m/ s^{2}$ is the value at the Earth’s surface. The two values differ by about $1%$ , therefore the same force that makes bodies fall on the Earth keeps the Moon in orbit.

Superposition principle

If a point mass $m$ interacts with the point masses $m_{1}, \dots, m_{n}$ , the total gravitational force on $m$ is the vector sum of the forces that each mass would exert alone:

F = F_{1} + F_{2} + \dots + F_{n} .

It is an experimental fact: the presence of the other masses does not change the force exerted by each one. The principle makes it possible to compute the force due to several bodies and, by dividing an extended body into small parts, the force due to an extended body. Forces add as vectors, so in a planar problem the components $F_{x}$ and $F_{y}$ are added separately.

The total force on

m

is the vector sum of the forces

F_{1}

and

F_{2}

, directed toward

m_{1}

and toward

m_{2}

Forces that cancel

A body of mass $m = 1.0 kg$ is at the origin of an $x$ axis; a mass of $3.0 kg$ is at $x = 0.30 m$ and one of $12 kg$ at $x = - 0.60 m$ . The two forces have magnitudes

F_{1} = G \frac{3.0 \cdot 1.0}{0.3 0 ^{2}} N = 2.2 \cdot 1 0^{- 9} N, F_{2} = G \frac{12 \cdot 1.0}{0.6 0 ^{2}} N = 2.2 \cdot 1 0^{- 9} N,

since $\frac{3.0}{0.09} = \frac{12}{0.36}$ , and opposite directions, so the total force on the body is zero. The larger mass, being at twice the distance, exerts the same force as the smaller mass.

Gravitational field

Let $F$ be the gravitational force that a system of masses exerts on a point mass $m$ placed at $P$ . The gravitational field of the system at the point $P$ is

g (P) = \frac{F}{m} .

The field does not depend on $m$ , because the force is proportional to $m$ ; it depends only on the masses that generate it and on the position of $P$ . It is measured in $N/kg$ , a unit that coincides with $m/ s^{2}$ . Once the field is known, the force on any mass placed at $P$ is $F = m g (P)$ , and if it is the only force acting, the acceleration of the mass is exactly $g (P)$ .

Field of a point mass

From the law of gravitation, the field generated by a point mass $M$ at distance $r$ is

g = - \frac{GM}{r ^{2}} \overset{u}{^}_{r}, g = \frac{GM}{r ^{2}},

where $\overset{u}{^}_{r}$ is the unit vector pointing from $M$ toward the point considered. The field is directed toward the mass that generates it and its magnitude is reduced to one quarter when the distance doubles. By the superposition principle the field of several masses is the vector sum of the fields of each, $g = g_{1} + \dots + g_{n}$ .

The field of a mass

M

points toward

M

; at twice the distance the magnitude is

\frac{1}{4}

, as the length of the arrows shows

Point of zero field between the Earth and the Moon

With $M_{T} = 5.97 \cdot 1 0^{24} kg$ , $M_{L} = 7.35 \cdot 1 0^{22} kg$ and distance $d = 3.84 \cdot 1 0^{8} m$ between the centers, we look for the point of the Earth-Moon segment, at distance $x$ from the Earth, where the two fields have the same magnitude. Since the two fields have opposite directions, the total field is zero if

\frac{G M _{T}}{x ^{2}} = \frac{G M _{L}}{( d - x ) ^{2}} ⟺ \frac{x}{d - x} = \frac{M _{T}}{M _{L}} = 9.01,

where the square root of two positive quantities has been taken. We obtain $x = \frac{9.01}{10.01} d = 3.46 \cdot 1 0^{8} m$ , that is, the point lies at about $90%$ of the distance, close to the Moon.

The theory continues with 6 more sections (gravitational potential energy, spherically symmetric bodies, gravitational acceleration and altitude…), plus all worked examples, the solutions and the PDF.

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