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13.4 Satellite orbits and energy  (Page 3/8)

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Check Your Understanding There is another consideration to this last calculation of M E . We derived [link] assuming that the satellite orbits around the center of the astronomical body at the same radius used in the expression for the gravitational force between them. What assumption is made to justify this? Earth is about 81 times more massive than the Moon. Does the Moon orbit about the exact center of Earth?

The assumption is that orbiting object is much less massive than the body it is orbiting. This is not really justified in the case of the Moon and Earth. Both Earth and the Moon orbit about their common center of mass. We tackle this issue in the next example.

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Galactic speed and period

Let’s revisit [link] . Assume that the Milky Way and Andromeda galaxies are in a circular orbit about each other. What would be the velocity of each and how long would their orbital period be? Assume the mass of each is 800 billion solar masses and their centers are separated by 2.5 million light years.

Strategy

We cannot use [link] and [link] directly because they were derived assuming that the object of mass m orbited about the center of a much larger planet of mass M . We determined the gravitational force in [link] using Newton’s law of universal gravitation. We can use Newton’s second law, applied to the centripetal acceleration of either galaxy, to determine their tangential speed. From that result we can determine the period of the orbit.

Solution

In [link] , we found the force between the galaxies to be

F 12 = G m 1 m 2 r 2 = ( 6.67 × 10 −11 N · m 2 /kg 2 ) [ ( 800 × 10 9 ) ( 2.0 × 10 30 kg ) ] 2 [ ( 2.5 × 10 6 ) ( 9.5 × 10 15 m ) ] 2 = 3.0 × 10 29 N

and that the acceleration of each galaxy is

a = F m = 3.0 × 10 29 N ( 800 × 10 9 ) ( 2.0 × 10 30 kg ) = 1.9 × 10 −13 m/s 2 .

Since the galaxies are in a circular orbit, they have centripetal acceleration. If we ignore the effect of other galaxies, then, as we learned in Linear Momentum and Collisions and Fixed-Axis Rotation , the centers of mass of the two galaxies remain fixed. Hence, the galaxies must orbit about this common center of mass. For equal masses, the center of mass is exactly half way between them. So the radius of the orbit, r orbit , is not the same as the distance between the galaxies, but one-half that value, or 1.25 million light-years. These two different values are shown in [link] .

Two galaxies are illustrated, separated by a distance r that is shown on the diagram. The galaxy on the left is larger than the galaxy on the right. A distance from the center of the galaxy on the left to a point between the two galaxies but closer to the left is shown and labeled as r orbit.
The distance between two galaxies, which determines the gravitational force between them, is r , and is different from r orbit , which is the radius of orbit for each. For equal masses, r orbit = 1 / 2 r . (credit: modification of work by Marc Van Norden)

Using the expression for centripetal acceleration, we have

a c = v orbit 2 r orbit 1.9 × 10 −13 m/s 2 = v orbit 2 ( 1.25 × 10 6 ) ( 9.5 × 10 15 m ) .

Solving for the orbit velocity, we have v orbit = 47 km/s . Finally, we can determine the period of the orbit directly from T = 2 π r / v orbit , to find that the period is T = 1.6 × 10 18 s , about 50 billion years.

Significance

The orbital speed of 47 km/s might seem high at first. But this speed is comparable to the escape speed from the Sun, which we calculated in an earlier example. To give even more perspective, this period is nearly four times longer than the time that the Universe has been in existence.
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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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