Gravitational Fields II

Introduction

In the first part on gravitational fields, the gravitational force and the gravitational acceleration have been described. The second part is about the work, energy and the movement of bodies in a gravitational field.

Work $$W$$

The mechanical work is already known. It occurs when a force $$F$$ acts along a path $$s$$ on a body. The following applies:

$$W = F \cdot s$$
$$F$$ = Force, $$s$$ = Displacement

When bodies are moved in a gravitational field, either through external influences or through the gravitational force itself, work is being done. It should however be noted that as the path $$s$$ is the path parallel to the field lines in this case. When moving a body along the field lines much work is done, when moving perpendicular to the field lines, however, no work is performed.

Work in homogeneous fields

For homogeneous fields we can use the following formula:

$$W = F \cdot s = m \cdot g \cdot s$$ $$m$$ = Mass of the body, $$g$$ = Gravitational acceleration, $$s$$ = Path parallel to the field lines

The following animation shows 3 bodies ($$m_1 = m_2 = m_3 = 1 \rm kg$$) in a homogeneous field and shows the way parallel to the field lines.

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Although bodies 2 and 3 travel different distances, the path parallel to the field lines is the same, therefore the amount of work performed is also the same. Body 1 is moved perpendicularly to the field lines, that is, the path parallel to the field lines is 0 and as such there is no work done.

Work in inhomogeneous fields

For inhomogeneous fields, the calculation of the work is more difficult, because the gravitational force during the movement is not constant.

In a radial gravitational field, the gravitational force depends on the distance $$r$$. One can determine the work by integrating over the gravitational force:

$$W = \int \limits_{r_1}^{r_2} F(r) \,\,\mathrm{d}r$$ Substituting the gravitational force, we get:

$$W = G \cdot m_1 \cdot m_2 \cdot \int \limits_{r_1}^{r_2} \dfrac{1}{r^2} \,\, \mathrm{d}r = G \cdot m_1 \cdot m_2 \cdot \left( \dfrac{1}{r_1} - \dfrac{1}{r_2} \right)$$ $$m_1, m_2$$ = Masses of the bodies,
$$r_1$$ = Distance at start position, $$r_2$$ = Distance at end position

The following animation shows 2 bodies ($$m_1 = m_2 = 1 \,\, \rm kg$$) in an inhomogeneous field and shows the path parallel to the field lines.

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It can be seen here that the work being done is only determined by the starting and ending point. Whether moving the charges straight or with many curves is not important.

Potential energy $$E_{pot}$$

The potential energy is already known. In the gravitational field of the earth, it is the greater, the higher a body is above the ground.

The following animation shows the potential energy in the homogeneous gravitational field near the Earth's surface.

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In the gravitational field we usually define that the potential energy on the Earth's surface is 0. If one raises a body to a certain height, work is done and stored in the potential energy of the body.

Potential energy in inhomogeneous fields

In an inhomogeneous gravitational field around a mass, we usually define that the potential energy at an infinite distance from the mass is 0.

$$E_{pot} = \Delta W = G \cdot m_1 \cdot m_2 \cdot \left( \dfrac{1}{r_1} - \dfrac{1}{r_2} \right) = G \cdot m_1 \cdot m_2 \cdot \left( \dfrac{1}{r_1} - \dfrac{1}{\infty} \right) = G \cdot m_1 \cdot m_2 \cdot \dfrac{1}{r_1}$$