Gravitational Energy

We have already become acquainted with gravitational energy through a simple example. A body raised to height above the Earth possesses potential energy .

However, this formula may be used only when height is much smaller than the Earth’s radius.

Gravitational energy is an important quantity, and it would be interesting to obtain a formula for it which would apply to a body raised to an arbitrary height above the Earth and also, more generally, for two masses attracting each other in accordance with the universal law: Let us assume that the bodies approached each other somewhat under the action of their mutual attraction. The distance between them was , but it became . Moreover, the work is performed. The value of the force must be taken at some intermediate point. Thus, If and do not differ much, we may replace by the product . We obtain: This work is performed at the expense of the gravitational energy: where is the initial and the final value of the gravitational potential energy.

Comparing these two formulas, we find the following expression for the potential energy: It resembles the formula for the gravitational force, but is raised to the first power in the denominator.

According to this formula, the potential energy for very large ’s. This is reasonable, since the attraction will no longer be felt at such distances. But when the bodies approach each other, the potential energy should decrease. After all, the work takes place at its expense. But in what direction can it decrease from zero? In the negative direction. Hence there is a minus sign in the formula. After all, -5 is less than zero, and -10 is less than -5.

If we are dealing with motion near the Earth’s surface, we may replace the general expression for the gravitational force by . Then with greater accuracy we have But on the Earth’s surface, a body has potential energy , where is the Earths radius. ’Therefore, at height above the Earth’s surface,

When we first introduced the formula for potential energy, , we agreed to measure height and energy from the Earth’s surface. Using the formula , we discard the constant term , regarding it as conditionally equal to zero. Since we are interested only in energy differences, for it is work which is an energy difference that is ordinarily measured, the presence of the constant term in the potential energy formula does not play any role.

Gravitational energy determines the strength of the “chains” binding a body to the Earth. How can we break these “chains”? How can we ensure that a body thrown from the Earth will not return to the Earth? It is clear that to do this we must impart a large initial velocity to the body. But what is the minimum velocity that is required.

As a body (missile, rocket) thrown from the Earth increases its distance from the Earth, its potential energy will rise (the absolute value of will fall); its kinetic energy will fall. If its kinetic energy becomes equal to zero prematurely, before we break the Earth’s gravitational “chain”, the missile that was thrown will fall back to the Earth.

It is necessary for the body to conserve its kinetic energy until its potential energy practically vanishes. Before its departure, a missile had potential energy ( and are the mass and radius of the Earth). Therefore, the missile must be given the velocity which would make its total energy positive. A body with a negative total energy (the magnitude of its potential energy is greater than that of its kinetic energy) will not get beyond the bounds of gravity.

Hence, we arrive at the simple condition. In order for a body of mass to break away from the Earth, it must, as has been already said, overcome the gravitational potential energy For this, the speed of the missile should be increased to the value of the escape velocity from the Earth, , which is easily computed .By equating its kinetic and potential energies: or, since ,

The value of computed by means of this formula is , of course, without taking air resistance into account. This speed is times as great as the orbital velocity of an artificial satellite whose orbit is near the Earth’s surface, i.e. .

The mass of the Moon is 81 times as small as that of the Earth; the radius of the Moon is four times as small as that of the Earth. Consequently, the gravitational energy on the Moon is twenty times less than that on the Earth, and a speed of is sufficient to break away from the Moon.

Kinetic energy is spent in order to break the gravitational “chains” to the planet—the take-off station. If we want the rocket which has overcome gravity to move with speed , then additional energy is needed for this. In such a case, when launching the rocket, it is necessary to impart it energy Therefore, the three speeds in question are connected by the simple relation: What should be the speed necessary for overcoming the gravitation of the Earth and the Sun—the minimum speed of a missile sent to distant stars? We denoted this speed by because it is called the escape velocity from the solar system.

First of all, let us determine the speed necessary for overcoming only the single attraction of the Sun.

As we have just shown, the speed needed to escape from the Earth’s attraction by a missile sent on a flight is times as great as the speed with which an Earth satellite is sent into orbit. Our reasoning is equally valid or the Sun, i.e. the speed needed to escape from the Sun is times as great as the speed of a satellite of the Sun (i.e. the Earth). Since the speed of the Earth’s motion around the Sun is about , the speed needed to escape from the sphere of the Sun’s attraction is . This is a very great speed, but for sending a missile to distant stars, we must, of course, use the Earth’s motion and launch the body in the direction in which the Earth is moving. We then need to add only .

Now we can finally compute the escape velocity from the solar system. This is the speed with which a rocket must "be launched in order that, escaping from the Earth’s attraction, it have a speed of . Using the formula just adduced, we obtain: from which .

Thus, having a speed of about , a body will leave the Earth, but such a missile will not go “far” away; the Earth let it go, but the Sun will not free it. It will turn into a satellite of the Sun.

It turns out that the speed necessary for interstellar travel is only one and a half times as great as the speed needed for travelling through the solar system within the Earth’s orbit. True, as has been already said, every appreciable increase in the initial speed of a missile is accompanied by many technical difficulties (see here).