# What is movement under gravity

We remember from the fundamentals of mechanics that when moving under the influence of a force, energy is required or released. Examples are:

- An airplane needs energy to move forward against the drag of the air. A satellite, on the other hand, moves in a vacuum and does not need any energy to maintain its speed.
- When braking a car, the force of the brakes acts on the car and it slows down. Most of the difference in kinetic energy is converted into heat.
- A crane lifts a load in the earth's gravitational field. That takes energy.

In this respect, it is not surprising that when a mass moves through the gravitational field, energy is also required or work is done. After all, every movement takes place under the influence of the force of gravity.

Let's briefly repeat the definition of work:

So that a body, under the influence of a force \ (\ vec F \), covers the path vector \ (\ vec s \), the work \ (E = - \ vec F \ cdot \ vec s \) has to be performed inner product).

That means, when moving in the direction of force, energy is released, when moving against the force, energy is required, and when moving perpendicular to the force, no energy is required!

We already know that a mass \ (M \) exerts the force \ (F = \ frac {GMm} {r ^ 2} \) on a mass \ (m \), where \ (r \) is the distance between the two Masses is. Let us now consider a body in the gravitational field of \ (M \) that we want to lift. If the body is only lifted a little, the gravitational force acting on it hardly changes, we can assume it to be constant. If we denote the acceleration due to gravity there with \ (g \), then the gravitational force is constant \ (gm \). If we lift the body by the height \ (h \), the work \ (E = mgh \) is done.

However, if the body is lifted high enough (hundreds or thousands of kilometers), the gravitational force acting on it will change during the lifting process, and the above formula is no longer correct. With the help of integral calculus (which is taught later in mathematics) one can calculate what the work done now looks like:

If there is a mass \ (m \) in the gravitational field of the mass \ (M \) at a distance \ (r_0 \) to \ (M \), and \ (m \) then at a distance \ (r_1 \) to \ (M \) raised is the work done

\ (\ displaystyle \ Delta E = GMm \ Big (\ frac 1 {r_0} - \ frac 1 {r_1} \ Big) \).

The work done is independent of the chosen path! It depends exclusively on the \ (r_0 \) the starting point and the \ (r_1 \) of the end point!

Conversely, let us now consider a freely falling mass \ (m \). At the distance \ (r_0 \) to the center of \ (M \), \ (m \) initially has the velocity \ (v_0 \). Now \ (m \) falls until it reaches the distance \ (r_1 \) to \ (M \). According to the above, the energy \ (\ Delta E = GMm \ big (\ frac 1 {r_1} - \ frac 1 {r_0} \ big) \)

*free*, and is completely converted into kinetic energy. In \ (r_1 \) the body therefore has a changed speed \ (v_1 \). We recognize:*The change in kinetic energy is equal to \ (\ Delta E \):*\ (\ displaystyle \ frac {mv ^ 2_1} {2} - \ frac {mv_0 ^ 2} 2 = GMm \ Big (\ frac 1 {r_1} - \ frac 1 {r_0} \ Big) \).

This can be transformed equivalent to

\ (\ displaystyle \ frac {mv ^ 2_1} {2} - \ frac {GMm} {r_1} = \ frac {mv ^ 2_0} {2} - \ frac {GMm} {r_0} \).

This shows: The size \ (E: = \ frac {mv ^ 2} {2} - \ frac {GMm} {r} \) is constant during the entire free fall!

One calls \ (E = \ frac {mv ^ 2} {2} - \ frac {GMm} {r} \) the total energy of \ (m \) in the gravitational field of \ (M \). It is a conserved quantity, i.e. it remains constant without external influences.

The total energy is made up of the already known kinetic energy and the potential energy. The potential energy of \ (m \) in the gravitational field of \ (M \) is given by

\ (\ displaystyle E_p = - \ frac {GMm} {r} \).

The potential energy is normalized in such a way that it becomes zero at infinity. In the above derivation we could also have added any constant. Physically relevant is mainly the difference in the potential energy.

The potential energy is measured in Joule \ ([\ text {J}] \).

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