# Define Newton's universal law of gravity

## Newton's law of gravitation

### introduction

In 1686 Isaac Newton formulated his law of gravitation for the first time in his work Philosophiae Naturalis Principia Mathematica. The law states that every mass point attracts every other mass point with 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 mass

### Gravitational constant

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

### Relationships

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

The force is proportional to the size 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

The force with which two \ (1 \, \, \ rm kg \) heavy bodies are attracted at \ (5 \, \, \ rm m \) distance 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 man weighing 80 kg is attracted to the earth with the following force:

\$\$ F = G \ cdot \ dfrac {m_ \ rm {Earth} \ cdot m_ \ rm {Man}} {(r_ \ rm {Earth}) ^ 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 you can see from the above examples, the gravitational force between two bodies on earth is barely perceptible. The gravitational pull of the earth, on the other hand, is clearly noticeable to us due to its large mass and ensures that we keep our feet on the ground and that the apples fall from the trees in autumn.

### Comparison with the Coulomb force

However, the gravitational force is the weakest of the four basic forces, which is quite clear in comparison with the electrical interaction, the Coulomb force. 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 relationship 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 approximately \ (\ num {4.2e42} \) as strong as their attraction by the gravitational force.

### Acceleration due to gravity

In mechanics one often calculates with the acceleration due to gravity \ (g = \ SI {9.81} {\ frac {m} {s ^ 2}} \). This acceleration is caused by the force of gravity.

If one sets the formula for the weight of a body \ (F = m \ cdot g \) equal to the gravitational force and converts it to the acceleration due to gravity, one gets

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

### Astronomical mass determination

If one knows the distance and the period of revolution of a celestial body, which circles around a heavier celestial body, the so-called central body, then one can determine the mass \ (M \) of the central body. The gravitational force acts as a radial force, whereby the following applies:

\ 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}

### Earth mass

The moon is approximately \ (\ SI {384400} {km} \) away from the earth and orbits it in \ (\ num {27,322} \) days. The mass of the earth is thus:

\$\$ 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 significantly from the table value for the earth's mass (\ (m_ \ rm {Earth} = \ SI {5.976e24} {kg} \)). This is because the center of the lunar orbit is not in the center of the earth, but rather both celestial bodies rotate around their common center of gravity.

### Solar mass

The center of gravity of the sun-earth system is almost in the center of the sun due to the much greater distance and the much greater mass of the sun. As a result, the calculation of the solar mass using the above formula is much more accurate. The earth orbits the sun at a distance of \ (149.6e6 \, \, \ rm km \) 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 corresponds roughly to the table value of the solar mass.

### literature

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