 Gravitational Energy Wells Printer Friendly Version
Warped Space and Orbital Paths

In his General Theory of Relativity (1916) Einstein suggested that Newton's “gravitational fields” should be replaced with the warping of spacetime around massive celestial objects. His rationale was based on the fact that he did not approve of the concept of “action at a distance” - a mechanism that is fundamental to fields. This concept would require that interactions occurring in one location in a field would require that information be sent "instantaneously" throughout the remainder of the field requiring speeds that exceeded the speed of light - Einstein's universal speed limit. image courtesy of Guide to the Cosmos

Often used to model spacetime around a massive object a stretched or hyperbolically curved surface. Envision a circular rubber sheet stretched tightly about a circular hoop. When a massive ball is placed on its surface, a “gravitational well” is immediately created. image courtesy of Gravity Visualized

If a second, lighter mass, is released with a velocity tangent to the center of the well (along a gravitational equipotential) it will attempt to travel in a straight line path. However, the distortion of spacetime through which it is moving will result in the object orbiting the central celestial body and eventually spiraling down as it loses energy. Under ideal conditions smaller objects should maintain a circular orbit with a constant period. image courtesy of The Stand News

We say that orbiting satellites are trapped in an "energy well" created by the more passive parent body distorting spacetime in its immediate vicinity. To obtain the total energy of an orbiting satellite we must add its potential and kinetic energies.

Gravitational Potential Energy

Let's first examine a satellite's potential energy. The potential energy function used for distances that exceed 1.2 radii from the center of a celestial body is In the formula, the negative preceding the ratio indicates that on astronomical scales the potential energy of a satellite is negative until it reaches infinity where it approaches zero. In the graph shown below, the three curves represents a satellite's kinetic energy (blue), its potential energy (orange), and its total energy (green). Note the linearity of the potential energy function inside the highlighted box. Recall that the gravitational field strength of the Earth is given by The potential energy at the surface of the earth is As mentioned earlier, due to the almost linear nature of our graph near the planet's surface, we can approximate the potential energy for small heights (h<< RE) above the earth's surface as The change in potential energy for a falling mass would be Refer to the following information for the next four questions.

6.4 x 106 meters. The mass of the Earth equals 5.98 x 1024 kg.
 Calculate U(6.403200 x 106) using the formula U(r) = -GmM/r.

 Calculate U(6.400000 x 106) using the formula U(r) = -GmM/r.

 Calculate ΔU = U(6.403200 x 106) - U(6.400000 x 106)

 Calculate the gravitational potential energy at a height of 3200 meters above the surface of the earth using the formula PEg = mgh.

Upon inspection you can see that we should only use the expression U(r) = PEg = mgh to calculate an object’s gravitational potential energy when it is located close to the surface of the central body. Such as an apple falling from a tree branch or a projectile released from a cliff. Not satellites in orbit.

Kinetic Energy and Total Energy

As stated earlier, when asked to calculate the total energy of an orbiting satellite you must include both its gravitational potential energy and its kinetic energy. We use the relationship that the gravitational force between the central body (planet/star) and the satellite provides the satellite's centripetal force to derive an expression for the satellite's kinetic energy, Combining the two expressions of U(r) and KE(r) gives us the satellite's total energy. Notice that the satellite's total energy is negative. This tells us that the satellite is "trapped" in an energy well around the central planet. Like an electron "orbiting" its central nucleus, the satellite would require a minimum "ionization energy" to free it from the influence of the central celestial body.

If a satellite were to lose all of its translational kinetic energy (that is, its tangential velocity) then the satellite would plunge down towards the central body's surface. It would impact the surface with a kinetic energy equal in magnitude to its original gravitational potential energy. Refer to the following information for the next two questions.

The mass of the moon is 7.342 x 1022 kg and its average orbital radius is 3.844 x 108 meters from the center of the earth.
 Calculate the total energy of the moon as it orbits the Earth.

 If the moon were to be struck by a meteorite that resulted in its total loss of linear momentum, at what speed would fragments of the moon impact the Earth's surface? Related Documents