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In fact, the present relative motion of these two galaxies is such that they are expected to collide in about 4 billion years. Although the density of stars in each galaxy makes a direct collision of any two stars unlikely, such a collision will have a dramatic effect on the shape of the galaxies. Examples of such collisions are well known in astronomy.

Check Your Understanding Galaxies are not single objects. How does the gravitational force of one galaxy exerted on the “closer” stars of the other galaxy compare to those farther away? What effect would this have on the shape of the galaxies themselves?

The stars on the “inside” of each galaxy will be closer to the other galaxy and hence will feel a greater gravitational force than those on the outside. Consequently, they will have a greater acceleration. Even without this force difference, the inside stars would be orbiting at a smaller radius, and, hence, there would develop an elongation or stretching of each galaxy. The force difference only increases this effect.

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See the Sloan Digital Sky Survey page for more information on colliding galaxies.

Energy in circular orbits

In Gravitational Potential Energy and Total Energy , we argued that objects are gravitationally bound if their total energy is negative. The argument was based on the simple case where the velocity was directly away or toward the planet. We now examine the total energy for a circular orbit and show that indeed, the total energy is negative. As we did earlier, we start with Newton’s second law applied to a circular orbit,

G m M E r 2 = m a c = m v 2 r G m M E r = m v 2 .

In the last step, we multiplied by r on each side. The right side is just twice the kinetic energy, so we have

K = 1 2 m v 2 = G m M E 2 r .

The total energy is the sum of the kinetic and potential energies, so our final result is

E = K + U = G m M E 2 r G m M E r = G m M E 2 r .

We can see that the total energy is negative, with the same magnitude as the kinetic energy. For circular orbits, the magnitude of the kinetic energy is exactly one-half the magnitude of the potential energy. Remarkably, this result applies to any two masses in circular orbits about their common center of mass, at a distance r from each other. The proof of this is left as an exercise. We will see in the next section that a very similar expression applies in the case of elliptical orbits.

Energy required to orbit

In [link] , we calculated the energy required to simply lift the 9000-kg Soyuz vehicle from Earth’s surface to the height of the ISS, 400 km above the surface. In other words, we found its change in potential energy. We now ask, what total energy change in the Soyuz vehicle is required to take it from Earth’s surface and put it in orbit with the ISS for a rendezvous ( [link] )? How much of that total energy is kinetic energy?

A demonstration of the ISS and Soyuz in parallel orbits about the earth is shown.
The Soyuz in a rendezvous with the ISS. Note that this diagram is not to scale; the Soyuz is very small compared to the ISS and its orbit is much closer to Earth. (credit: modification of works by NASA)

Strategy

The energy required is the difference in the Soyuz ’s total energy in orbit and that at Earth’s surface. We can use [link] to find the total energy of the Soyuz at the ISS orbit. But the total energy at the surface is simply the potential energy, since it starts from rest. [Note that we do not use [link] at the surface, since we are not in orbit at the surface.] The kinetic energy can then be foundfrom the difference in the total energy change and the change in potential energy found in [link] . Alternatively, we can use [link] to find v orbit and calculate the kinetic energy directly from that. The total energy required is then the kinetic energy plus the change in potential energy found in [link] .

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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