﻿ Newton's Law of Universal Gravitation - College Physics

# Newton's Law of Universal Gravitation

## Introduction

In 1686, Isaac Newton first formulated his law of universal gravitation in his work Philosophiae Naturalis Principia Mathematica. The law states that every point mass attracts every other point mass by a force directed along the connecting line.

$$F = G \cdot \dfrac{m_1 \cdot m_2}{r^2}$$ $$G$$ = Gravitational constant,
$$m_1, m_2$$ = Masses of the two bodies, $$r$$ = Distance between the centers of the masses

### Gravitational constant

The value of the gravitational constant is: $$G = \SI{6.673e-11}{ \dfrac{m^3}{kg \cdot s^2} }$$

### Relations

$$F \sim m_1$$ $$F \sim m_2$$

The force is proportional to the amount of the two masses.

$$F \sim \dfrac{1}{r^2}$$

The force is inversely proportional to the square of the distance between the two masses.

### Examples

Die Kraft mit der sich zwei $$1 \,\, \rm kg$$ schwere Körper in $$5 \,\, \rm m$$ Entfernung anziehen, lässt sich über die obige Formel berechnen: The force with which two $$1 \,\, \rm kg$$ heavy bodies in $$5 \,\, \rm m$$ distance attract each other can be calculated using the above formula:

$$F = G \cdot \dfrac{m_1 \cdot m_2}{r^2} = \num{6.673e-11} \rm \dfrac{m^3}{kg \cdot s^2} \cdot \dfrac{1 \,\, kg \cdot 1 \,\, kg}{(5 \,\, m)^2} = \SI{2.67e-12}{N}$$

A $$80 \,\, \rm kg$$ heavy man is attracted to the ground (Earth) with the following force:

$$F = G \cdot \dfrac{m_\rm{Erde} \cdot m_\rm{Mann}}{(r_\rm{Erde})^2} = \num{6.673e-11} \rm \dfrac{m^3}{kg \cdot s^2} \cdot \dfrac{ \SI{5.976e24}{kg} \cdot 80 \,\, kg }{( \SI{6378e3}{m} )^2} = \SI{784.25}{N}$$

As can be seen from the above example, the gravitational force between two bodies on the Earth is very small. The attraction of the Earth, however, is clearly noticeable for us due to its large mass and it ensures that we stay on the ground and the apples fall from the trees in autumn.

## Comparison with Coulomb's law

However, the force of gravity is the weakest of the four fundamental forces, which can be seen when it is compared with Coulomb's law. Two electrons repel each other much more strongly than they are attracted by gravity:

\begin{aligned} F_\rm{el} &= \dfrac{1}{4 \pi \cdot \epsilon_0 \cdot \epsilon_r} \cdot \dfrac{q_\rm{e} \cdot q_\rm{e}}{r^2} \\ &= \rm \dfrac{1}{4 \pi \cdot \num{8.854e-12} \frac{A \cdot s}{V \cdot m} \cdot 1} \cdot \dfrac{ \SI{1.602e-19}{C} \cdot \SI{1.602e-19}{C} }{(1 \,\, m)^2} \\ \\ &= \SI{2.31e-28}{N} \\ \\ \\ \\ F_\rm{G} &= G \cdot \dfrac{m_\rm{e} \cdot m_\rm{e}}{r^2} \\ &= \num{6.673e-11} \rm \dfrac{m^3}{kg \cdot s^2} \cdot \dfrac{ \SI{9.11e-31}{kg} \cdot \SI{9.11e-31}{kg} }{r^2} \\ \\ &= \SI{5.54e-71}{N} \end{aligned}

You can now calculate the ratio between the two forces:

$$\dfrac{F_\rm{el}}{F_\rm{G}} = \dfrac{ \SI{2.31e-28}{N} }{ \SI{5.54e-71}{N} } \approx \num{4.2e42}$$

The repulsion of the two electrons by the Coulomb force is about $$\num{4.2e42}$$ as strong as their attraction by the gravitational force.

## Gravitational acceleration

In mechanics we often calculate with the gravitational acceleration $$g = \SI{9.81}{\frac{m}{s^2}}$$. This acceleration is caused by the gravitational force.

Setzt man die Formel für die Gewichtskraft eines Körpers $$F = m \cdot g$$ gleich der Gravitationskraft und stellt nach der Erdbeschleunigung um, so erhält man: By equating the formula for the weight of a body $$F = m \cdot g$$ with the formula for the gravitational force and solve for the gravitational acceleration, we obtain:

\begin{aligned} \cancel m \cdot g &= G \cdot \dfrac{m_\rm{Erde} \cdot \cancel m}{(r_\rm{Erde})^2} \\ \\ g &= G \cdot \dfrac{m_\rm{Erde}}{(r_\rm{Erde})^2} \\ \\ g &\approx \rm \SI{9.80}{\frac{m}{s^2}} \\ \end{aligned}

## Astronomical mass determination

Knowing the distance and orbital period of a celestial body that revolves around a heavier celestial body, the so-called central body, one can determine the mass $$M$$ of the central body. The gravitational force acts as a radial force:

\begin{aligned} F &= F_\rm{R} \\ \\ G \cdot \dfrac{M \cdot \cancel m}{r^2} &= \cancel m \cdot \omega^2 \cdot r \\ \\ G \cdot \dfrac{M}{r^2} &= \omega^2 \cdot r \\ \\ M &= \dfrac{\omega^2 \cdot r^3}{G} \qquad | \qquad \omega = \dfrac{2 \,\, \pi}{T} \\ \\ M &= \dfrac{4 \,\, \pi^2 \cdot r^3}{G \cdot T^2} \\ \end{aligned}

### Mass of the Earth

The Moon is about $$\SI{384400}{km}$$ away from Earth and orbiting it in $$\num{27.322}$$ days. As such we can calculate the mass of the Earth:

$$M_\rm{Earth} = \dfrac{4 \,\, \pi^2 \cdot ( \SI{3.844e8}{m} )^3}{G \cdot (\num{27.322} \cdot 24 \cdot 60 \cdot 60 \,\, \rm{s})^2} = \SI{6.03e24}{kg}$$

However, this value deviates greatly from the real value for the Earth's mass ($$m_\rm{Earth} = \SI{5.976e24}{kg}$$). This is because the center of the Moon's orbit is not in the center of the Earth, but instead both celestial bodies orbit around their common center of gravity.

### Mass of the Sun

The center of gravity of the system Sun-Earth lies almost in the center of the Sun because their much greater distance and the much larger mass of the Sun. Thus, the calculation of the Sun's mass using the above formula is more accurate. The Earth orbits the Sun in $$149,6e6 \,\, \rm km$$ distance in one year (365.26 days).

$$M_\rm{Sun} = \dfrac{4 \,\, \pi^2 \cdot (\SI{149.6e9}{m})^3}{G \cdot (\num{365.26} \cdot 24 \cdot 60 \cdot 60 \,\, \rm{s})^2} = \SI{1.989e30}{kg}$$

This value is fairly accurate.

### Sources

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