# Does gravity have an opposite force

## Gravitational force

In this section the Gravitational force explained. In the section Weight force / spring force the special case of a body near the surface of the earth has already been listed as a weight force.

\$ G = m \ cdot g \$ weight force

The gravitational force is based on the fact that two bodies and thus two masses attract each other. Here again the law of interaction applies and the example with the ball, which is attracted to the earth and with the same but opposite force, the ball also attracts the earth. The mass attraction is universal. So this means that it is not a special phenomenon of the earth, but basically applies to two masses. For example, also for the planets (e.g. sun and moon, moon and earth, etc.).

Gravity keeps the moon in orbit around the earth

In the general law of gravitation, Newton formulates this mass attraction as follows:

\$ F_ {grav} = G \ frac {m_1 \ cdot m_2} {x ^ 2} \$ Gravitational force

With

\$ G = 6.67 · 10 ^ {- 11} N \ frac {m ^ 2} {kg ^ 2} \$ constant of gravity

\$ m_1 \$ mass of the body 1

\$ m_2 \$ mass of the body 2

\$ x \$ Distance between the two centers of gravity of the body

From the above equation it can be seen that the attraction between two bodies increases with increasing mass of the bodies and decreases with the square of the distance between the centers of gravity.

#### Gravity on the surface of planets

For a body that is on the surface of a planet, \$ x \$ is equal to the radius \$ r \$ of the planet:

\$ F_ {grav} = G \ frac {M \ cdot m} {r ^ 2} \$

With

\$ G = 6.67 · 10 ^ {- 11} N \ frac {m ^ 2} {kg ^ 2} \$ constant of gravity

\$ M \$ mass of the planet

\$ r \$ radius of the planet

\$ m \$ mass of the body on the surface of the planet

The planet's mass \$ M \$, its radius \$ r \$ and the universal gravitational constants \$ G \$ can now be combined into a single constant:

\$ g = \ frac {M \ cdot G} {r ^ 2} \$

It then follows:

\$ F_ {grav} = m \ cdot g \$

With

\$ g = \ frac {M \ cdot G} {r ^ 2} \$

\$ G = 6.67 · 10 ^ {- 11} N \ frac {m ^ 2} {kg ^ 2} \$ constant of gravity

For the earth, the mass of the earth of \$ M_E = 5.972 \ cdot 10 ^ {24} kg \$ and the radius of the earth \$ r_E = 6,371 km = 6,371,000m \$ results in the already known acceleration due to gravity:

\$ g_E = \ frac {M_E \ cdot G} {r_E ^ 2} = \ frac {5.972 \ times 10 ^ {24} kg \ times 6.67 · 10 ^ {- 11} N \ frac {m ^ 2} { kg ^ 2}} {(6,371,000 m) ^ 2} \$

\$ g_E = 9.81 \ frac {m} {s ^ 2} \$

#### Averages of other planets

• Moon: \$ M_ {moon} = 7.349 \ cdot 10 ^ {22} kg \$, \$ r_ {moon} = 1,738 km \$, \$ g_ {moon} = 1.62 \ frac {m} {s ^ 2} \$

The attraction on the surface of the earth is about 6 times as great as the attraction on the surface of the moon.

Should a lunar station ever be built, the astronauts will have to undergo additional strength training if they stay longer on the moon. Because of the lower attraction on the moon, the muscles are no longer used as much. If no additional strength training were done, the muscles would regress and lead to severe movement problems later on on earth. The astronauts did not have enough muscle mass to counteract the gravitational pull. This means, for example, that your legs can no longer carry you because there are not enough leg muscles to push your body upwards and to lift your legs against the gravity.

What is the weight of a person weighing 70 kg on earth and on the moon?

\$ G = m \ cdot g \$

With

\$ g_E = 9.81 \ frac {m} {s ^ 2} \$

\$ g_ {moon} = 1.62 \ frac {m} {s ^ 2} \$

\$ G_E = 70 kg \ cdot 9.81 \ frac {m} {s ^ 2} = 686.7 N \$

\$ G_ {moon} = 70 kg \ cdot 1.62 \ frac {m} {s ^ 2} = 113.4 N \$

How much would a person on earth have to weigh to obtain a weight of \$ G = 113.4 N \$?

\$ 113.4 N = x \ cdot 9.81 \ frac {m} {s ^ 2} \$

\$ x = \ frac {113.4 N} {9.81 \ frac {m} {s ^ 2}} = 11.56 kg \$

• Sun: \$ M_S = 1.989 \ cdot 10 ^ {30} kg \$, \$ r_S = 695.700 km \$ and thus \$ g_S = 274.1 \ frac {m} {s ^ 2} \$

The attraction on the surface of the sun is approximately 28 times as great as the attraction on the earth's surface.

Apart from the temperatures on the sun, it would not be possible to move on the sun or to exist at all. Our bodies could not withstand the strong attraction of the sun.

What is the weight of a person weighing 70 kg on earth and on the sun?

\$ G = m \ cdot g \$

With

\$ g_E = 9.81 \ frac {m} {s ^ 2} \$

\$ g_S = 274.1 \ frac {m} {s ^ 2} \$

\$ G_E = 70 kg \ cdot 9.81 \ frac {m} {s ^ 2} = 686.7 N \$

\$ G_S = 70 kg \ cdot 274.1 \ frac {m} {s ^ 2} = 19,187 N \$

How much would a person on earth have to weigh to obtain a weight of \$ G = 19,187N \$?

\$ 19.187N = x \ cdot 9.81 \ frac {m} {s ^ 2} \$

\$ x = \ frac {19.187N} {9.81 \ frac {m} {s ^ 2}} = 1.955 kg \$

• Mars: \$ M_ {Mars} = 6.39 · 10 ^ {23} kg \$, \$ r_ {Mars} = 3,390 km \$ and thus \$ g_ {Mars} = 3.71 \ frac {m} {s ^ 2} \$.

The attraction on the surface of the earth is about 2.5 times as great as the attraction on the surface of Mars.