Example

The International Space Station

Determine the orbital speed and period for the International Space Station (ISS).

Strategy

Since the ISS orbits $$ 4.00\,×\,{10}^{2}\text{km} $$ above Earth’s surface, the radius at which it orbits is $$ {R}_{\text{E}}+4.00\,×\,{10}^{2}\text{km}$$. We use (Figure) and (Figure) to find the orbital speed and period, respectively.

Solution

Using (Figure), the orbital velocity is

which is about 17,000 mph. Using (Figure), the period is

which is just over 90 minutes.

Significance

The ISS is considered to be in low Earth orbit (LEO). Nearly all satellites are in LEO, including most weather satellites. GPS satellites, at about 20,000 km, are considered medium Earth orbit. The higher the orbit, the more energy is required to put it there and the more energy is needed to reach it for repairs. Of particular interest are the satellites in geosynchronous orbit. All fixed satellite dishes on the ground pointing toward the sky, such as TV reception dishes, are pointed toward geosynchronous satellites. These satellites are placed at the exact distance, and just above the equator, such that their period of orbit is 1 day. They remain in a fixed position relative to Earth’s surface.

Example

Determining the Mass of Earth

Determine the mass of Earth from the orbit of the Moon.

Strategy

We use (Figure), solve for $$ {M}_{\text{E}}$$, and substitute for the period and radius of the orbit. The radius and period of the Moon’s orbit was measured with reasonable accuracy thousands of years ago. From the astronomical data in Appendix D, the period of the Moon is 27.3 days $$ =2.36\,×\,{10}^{6}\,\text{s}$$, and the average distance between the centers of Earth and the Moon is 384,000 km.

Solution

Solving for $$ {M}_{\text{E}}$$,

Significance

Compare this to the value of $$ 5.96\,×\,{10}^{24}\,\text{kg} $$ that we obtained in (Figure), using the value of g at the surface of Earth. Although these values are very close (~0.8%), both calculations use average values. The value of g varies from the equator to the poles by approximately 0.5%. But the Moon has an elliptical orbit in which the value of r varies just over 10%. (The apparent size of the full Moon actually varies by about this amount, but it is difficult to notice through casual observation as the time from one extreme to the other is many months.)

Example

Galactic Speed and Period

Let’s revisit (Figure). 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 (Figure) and (Figure) 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 (Figure) 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 (Figure), we found the force between the galaxies to be

and that the acceleration of each galaxy is

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, $$ {r}_{\text{orbit}}$$, 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 (Figure).

Figure 13.14 The distance between two galaxies, which determines the gravitational force between them, is r, and is different from $$ {r}_{\text{orbit}}$$, which is the radius of orbit for each. For equal masses, $$ {r}_{\text{orbit}}=1\text{/}2r$$. (credit: modification of work by Marc Van Norden)

Using the expression for centripetal acceleration, we have

Solving for the orbit velocity, we have $$ {v}_{\text{orbit}}=47\,\text{km/s}$$. Finally, we can determine the period of the orbit directly from $$ T=2\pi r\text{/}{v}_{\text{orbit}}$$, to find that the period is $$ T=1.6\,×\,{10}^{18}\,\text{s}$$, 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.

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.

Example

Energy Required to Orbit

In (Figure), 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 ((Figure))? How much of that total energy is kinetic energy?

Figure 13.15 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 (Figure) 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 (Figure) at the surface, since we are not in orbit at the surface.] The kinetic energy can then be found from the difference in the total energy change and the change in potential energy found in (Figure). Alternatively, we can use (Figure) to find $$ {v}_{\text{orbit}} $$ 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 (Figure).

Solution

From (Figure), the total energy of the Soyuz in the same orbit as the ISS is

The total energy at Earth’s surface is

The change in energy is $$ \text{Δ}E={E}_{\text{orbit}}-{E}_{\text{surface}}=2.98\,×\,{10}^{11}\,\text{J}$$. To get the kinetic energy, we subtract the change in potential energy from (Figure), $$ \text{Δ}U=3.32\,×\,{10}^{10}\,\text{J}$$. That gives us $$ {K}_{\text{orbit}}=2.98\,×\,{10}^{11}-3.32\,×\,{10}^{10}=2.65\,×\,{10}^{11}\,\text{J}$$. As stated earlier, the kinetic energy of a circular orbit is always one-half the magnitude of the potential energy, and the same as the magnitude of the total energy. Our result confirms this.

The second approach is to use (Figure) to find the orbital speed of the Soyuz, which we did for the ISS in (Figure).

So the kinetic energy of the Soyuz in orbit is

the same as in the previous method. The total energy is just

Significance

The kinetic energy of the Soyuz is nearly eight times the change in its potential energy, or 90% of the total energy needed for the rendezvous with the ISS. And it is important to remember that this energy represents only the energy that must be given to the Soyuz. With our present rocket technology, the mass of the propulsion system (the rocket fuel, its container and combustion system) far exceeds that of the payload, and a tremendous amount of kinetic energy must be given to that mass. So the actual cost in energy is many times that of the change in energy of the payload itself.