The orbital speed of a satellite revolving around a planet

A rocket must accelerate to at least 25,039 mph (40,320 kph) to completely escape Earth's gravity and fly off into space (for more on escape velocity, visit this article at NASA).

Earth's escape velocity is much greater than what's required to place an Earth satellite in orbit. With satellites, the object is not to escape Earth's gravity, but to balance it. Orbital velocity is the velocity needed to achieve balance between gravity's pull on the satellite and the inertia of the satellite's motion -- the satellite's tendency to keep going. This is approximately 17,000 mph (27,359 kph) at an altitude of 150 miles (242 kilometers). Without gravity, the satellite's inertia would carry it off into space. Even with gravity, if the intended satellite goes too fast, it will eventually fly away. On the other hand, if the satellite goes too slowly, gravity will pull it back to Earth. At the correct orbital velocity, gravity exactly balances the satellite's inertia, pulling down toward Earth's center just enough to keep the path of the satellite curving like Earth's curved surface, rather than flying off in a straight line.

The orbital velocity of the satellite depends on its altitude above Earth. The nearer to Earth, the faster the required orbital velocity. At an altitude of 124 miles (200 kilometers), the required orbital velocity is a little more than 17,000 mph (about 27,400 kph). To maintain an orbit that is 22,223 miles (35,786 kilometers) above Earth, the satellite must orbit at a speed of about 7,000 mph (11,300 kph). That orbital speed and distance permit the satellite to make one revolution in 24 hours. Since Earth also rotates once in 24 hours, a satellite at 22,223 miles altitude stays in a fixed position relative to a point on Earth's surface. Because the satellite stays right over the same spot all the time, this kind of orbit is called "geostationary." Geostationary orbits are ideal for weather satellites and communications satellites.

In general, the higher the orbit, the longer the satellite can stay in orbit. At lower altitudes, a satellite runs into traces of Earth's atmosphere, which creates drag. The drag causes the orbit to decay until the satellite falls back into the atmosphere and burns up. At higher altitudes, where the vacuum of space is nearly complete, there is almost no drag and a satellite like the moon can stay in orbit for centuries.

Window of Opportunity

A launch window is a particular period during which it will be easier to place the satellite in the orbit necessary to perform its intended function. With the space shuttle, an extremely important factor in choosing the launch window was the need to bring down the astronauts safely if something went wrong. The astronauts had to be able to reach a safe landing area with rescue personnel standing by. For other types of flights, including interplanetary exploration, the launch window must permit the flight to take the most efficient course to its very distant destination. If weather is bad or a malfunction occurs during a launch window, the flight must be postponed until the next launch window appropriate for the flight. If a satellite were launched at the wrong time of the day in perfect weather, the satellite could end up in an orbit that would not pass over any of its intended users. Timing is everything!

The Moon orbits Earth. In turn, Earth and the other planets orbit the Sun. The space directly above our atmosphere is filled with artificial satellites in orbit. We examine the simplest of these orbits, the circular orbit, to understand the relationship between the speed and period of planets and satellites in relation to their positions and the bodies that they orbit.

As noted at the beginning of this chapter, Nicolaus Copernicus first suggested that Earth and all other planets orbit the Sun in circles. He further noted that orbital periods increased with distance from the Sun. Later analysis by Kepler showed that these orbits are actually ellipses, but the orbits of most planets in the solar system are nearly circular. Earth’s orbital distance from the Sun varies a mere 2%. The exception is the eccentric orbit of Mercury, whose orbital distance varies nearly 40%.

Determining the orbital speed and orbital period of a satellite is much easier for circular orbits, so we make that assumption in the derivation that follows. As we described in the previous section, an object with negative total energy is gravitationally bound and therefore is in orbit. Our computation for the special case of circular orbits will confirm this. We focus on objects orbiting Earth, but our results can be generalized for other cases.

Consider a satellite of mass m in a circular orbit about Earth at distance r from the center of Earth ((Figure)). It has centripetal acceleration directed toward the center of Earth. Earth’s gravity is the only force acting, so Newton’s second law gives

[latex] \frac{Gm{M}_{\text{E}}}{{r}^{2}}=m{a}_{\text{c}}=\frac{m{v}_{\text{orbit}}^{2}}{r}. [/latex]

Figure 13.12 A satellite of mass m orbiting at radius r from the center of Earth. The gravitational force supplies the centripetal acceleration.

We solve for the speed of the orbit, noting that m cancels, to get the orbital speed

[latex] {v}_{\text{orbit}}=\sqrt{\frac{G{M}_{\text{E}}}{r}}. [/latex]

Consistent with what we saw in (Figure) and (Figure), m does not appear in (Figure). The value of g, the escape velocity, and orbital velocity depend only upon the distance from the center of the planet, and not upon the mass of the object being acted upon. Notice the similarity in the equations for [latex] {v}_{\text{orbit}} [/latex] and [latex] {v}_{\text{esc}} [/latex]. The escape velocity is exactly [latex] \sqrt{2} [/latex] times greater, about 40%, than the orbital velocity. This comparison was noted in (Figure), and it is true for a satellite at any radius.

To find the period of a circular orbit, we note that the satellite travels the circumference of the orbit [latex] 2\pi r [/latex] in one period T. Using the definition of speed, we have [latex] {v}_{\text{orbit}}=2\pi r\text{/}T [/latex]. We substitute this into (Figure) and rearrange to get

[latex] T=2\pi \sqrt{\frac{{r}^{3}}{G{M}_{\text{E}}}}. [/latex]

We see in the next section that this represents Kepler’s third law for the case of circular orbits. It also confirms Copernicus’s observation that the period of a planet increases with increasing distance from the Sun. We need only replace [latex] {M}_{\text{E}} [/latex] with [latex] {M}_{\text{Sun}} [/latex] in (Figure).

We conclude this section by returning to our earlier discussion about astronauts in orbit appearing to be weightless, as if they were free-falling towards Earth. In fact, they are in free fall. Consider the trajectories shown in (Figure). (This figure is based on a drawing by Newton in his Principia and also appeared earlier in Motion in Two and Three Dimensions.) All the trajectories shown that hit the surface of Earth have less than orbital velocity. The astronauts would accelerate toward Earth along the noncircular paths shown and feel weightless. (Astronauts actually train for life in orbit by riding in airplanes that free fall for 30 seconds at a time.) But with the correct orbital velocity, Earth’s surface curves away from them at exactly the same rate as they fall toward Earth. Of course, staying the same distance from the surface is the point of a circular orbit.

Figure 13.13 A circular orbit is the result of choosing a tangential velocity such that Earth’s surface curves away at the same rate as the object falls toward Earth.

We can summarize our discussion of orbiting satellites in the following Problem-Solving Strategy.

Problem-Solving Strategy: Orbits and Conservation of Energy

1. Determine whether the equations for speed, energy, or period are valid for the problem at hand. If not, start with the first principles we used to derive those equations.
2. To start from first principles, draw a free-body diagram and apply Newton’s law of gravitation and Newton’s second law.
3. Along with the definitions for speed and energy, apply Newton’s second law of motion to the bodies of interest.

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

Strategy

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

Solution

Using (Figure), the orbital velocity is

[latex] {v}_{\text{orbit}}=\sqrt{\frac{G{M}_{\text{E}}}{r}}=\sqrt{\frac{6.67\,×\,{10}^{-11}\,\text{N}·{\text{m}}^{2}{\text{/kg}}^{2}(5.96\,×\,{10}^{24}\,\text{kg})}{(6.36\,×\,{10}^{6}+4.00\,×\,{10}^{5}\,\text{m})}}=7.67\,×\,{10}^{3}\,\text{m/s} [/latex]

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

[latex] T=2\pi \sqrt{\frac{{r}^{3}}{G{M}_{\text{E}}}}=2\pi \sqrt{\frac{{(6.37\,×\,{10}^{6}+4.00\,×\,{10}^{5}\,\text{m})}^{3}}{(6.67\,×\,{10}^{-11}\,\text{N}·{\text{m}}^{2}{\text{/kg}}^{2})(5.96\,×\,{10}^{24}\,\text{kg})}}=5.55\,×\,{10}^{3}\,\text{s} [/latex]

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.

By what factor must the radius change to reduce the orbital velocity of a satellite by one-half? By what factor would this change the period?

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

Strategy

We use (Figure), solve for [latex] {M}_{\text{E}} [/latex], 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 [latex] =2.36\,×\,{10}^{6}\,\text{s} [/latex], and the average distance between the centers of Earth and the Moon is 384,000 km.

Solution

Solving for [latex] {M}_{\text{E}} [/latex],

[latex] \begin{array}{ccc}\hfill T& =\hfill & 2\pi \sqrt{\frac{{r}^{3}}{G{M}_{\text{E}}}}\hfill \\ \hfill {M}_{\text{E}}& =\hfill & \frac{4{\pi }^{2}{r}^{3}}{G{Τ}^{2}}=\frac{4{\pi }^{2}{(3.84\,×\,{10}^{8}\,\text{m})}^{3}}{(6.67\,×\,{10}^{-11}\,\text{N}·{\text{m}}^{2}{\text{/kg}}^{2}){(2.36\,×\,{10}^{6}\,\text{m})}^{2}}=6.01\,×\,{10}^{24}\,\text{kg}.\hfill \end{array} [/latex]

Compare this to the value of [latex] 5.96\,×\,{10}^{24}\,\text{kg} [/latex] 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.)

There is another consideration to this last calculation of [latex] {M}_{\text{E}} [/latex]. We derived (Figure) 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?

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

[latex] {F}_{12}=G\frac{{m}_{1}{m}_{2}}{{r}^{2}}=\,(6.67\,×\,{10}^{-11}\,\text{N}·{\text{m}}^{2}{\text{/kg}}^{2})\frac{{[(800\,×\,{10}^{9})(2.0\,×\,{10}^{30}\,\text{kg})]}^{2}}{{[(2.5\,×\,{10}^{6})(9.5\,×\,{10}^{15}\,\text{m})]}^{2}}=3.0\,×\,{10}^{29}\,\text{N} [/latex]

and that the acceleration of each galaxy is

[latex] a=\frac{F}{m}=\,\frac{3.0\,×\,{10}^{29}\,\text{N}}{(800\,×\,{10}^{9})(2.0\,×\,{10}^{30}\,\text{kg})}=1.9\,×\,{10}^{-13}{\text{m/s}}^{2}. [/latex]

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, [latex] {r}_{\text{orbit}} [/latex], 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 [latex] {r}_{\text{orbit}} [/latex], which is the radius of orbit for each. For equal masses, [latex] {r}_{\text{orbit}}=1\text{/}2r [/latex]. (credit: modification of work by Marc Van Norden)

Using the expression for centripetal acceleration, we have

[latex] \begin{array}{ccc}\hfill {a}_{\text{c}}& =\hfill & \frac{{v}_{\text{orbit}}^{2}}{{r}_{\text{orbit}}}\hfill \\ \hfill 1.9\,×\,{10}^{-13}{\text{m/s}}^{2}& =\hfill & \frac{{v}_{\text{orbit}}^{2}}{(1.25\,×\,{10}^{6})(9.5\,×\,{10}^{15}\,\text{m})}.\hfill \end{array} [/latex]

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

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?

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,

[latex] \begin{array}{ccc}\frac{Gm{M}_{\text{E}}}{{r}^{2}}\hfill & =\hfill & m{a}_{c}=\frac{m{v}^{2}}{r}\hfill \\ \frac{Gm{M}_{\text{E}}}{r}\hfill & =\hfill & m{v}^{2}.\hfill \end{array} [/latex]

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

[latex] K=\frac{1}{2}m{v}^{2}=\frac{Gm{M}_{\text{E}}}{2r}. [/latex]

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

[latex] E=K+U=\frac{Gm{M}_{\text{E}}}{2{r}^{}}-\frac{Gm{M}_{\text{E}}}{{r}^{}}=-\frac{Gm{M}_{\text{E}}}{2{r}^{}}. [/latex]

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.

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 [latex] {v}_{\text{orbit}} [/latex] 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

[latex] \begin{array}{cc}\hfill {E}_{\text{orbit}}& ={K}_{\text{orbit}}+{U}_{\text{orbit}}=-\frac{Gm{M}_{\text{E}}}{2{r}^{}}\hfill \\ & =\frac{(6.67\,×\,{10}^{-11}\,\text{N}·{\text{m}}^{2}{\text{/kg}}^{2})(9000\,\text{kg})(5.96\,×\,{10}^{24}\,\text{kg})}{2{(6.36\,×\,{10}^{6}+4.00\,×\,{10}^{5}\,\text{m})}^{}}=-2.65\,×\,{10}^{11}\,\text{J}.\hfill \end{array} [/latex]

The total energy at Earth’s surface is

[latex] \begin{array}{cc}\hfill {E}_{\text{surface}}& ={K}_{\text{surface}}+{U}_{\text{surface}}=0-\frac{Gm{M}_{\text{E}}}{{r}^{}}\hfill \\ & =-\frac{(6.67\,×\,{10}^{-11}\,\text{N}·{\text{m}}^{2}{\text{/kg}}^{2})(9000\,\text{kg})(5.96\,×\,{10}^{24}\,\text{kg})}{{(6.36\,×\,{10}^{6}\,\text{m})}^{}}\hfill \\ & =-5.63\,×\,{10}^{11}\,\text{J}.\hfill \end{array} [/latex]

The change in energy is [latex] \text{Δ}E={E}_{\text{orbit}}-{E}_{\text{surface}}=2.98\,×\,{10}^{11}\,\text{J} [/latex]. To get the kinetic energy, we subtract the change in potential energy from (Figure), [latex] \text{Δ}U=3.32\,×\,{10}^{10}\,\text{J} [/latex]. That gives us [latex] {K}_{\text{orbit}}=2.98\,×\,{10}^{11}-3.32\,×\,{10}^{10}=2.65\,×\,{10}^{11}\,\text{J} [/latex]. 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).

[latex] {v}_{\text{orbit}}=\sqrt{\frac{G{M}_{\text{E}}}{r}}=\sqrt{\frac{(6.67\,×\,{10}^{-11}\,\text{N}·{\text{m}}^{2}{\text{/kg}}^{2})(5.96\,×\,{10}^{24}\,\text{kg})}{(6.36\,×\,{10}^{6}+4.00\,×\,{10}^{5}\,\text{m})}}=7.67\,×\,{10}^{3}\,\text{m/s.} [/latex]

So the kinetic energy of the Soyuz in orbit is

[latex] {K}_{\text{orbit}}=\frac{1}{2}m{v}_{\text{orbit}}^{2}=\frac{1}{2}(9000\,\text{kg}){(7.67\,×\,{10}^{3}\,\text{m/s})}^{2}=2.65\,×\,{10}^{11}\,\text{J,} [/latex]

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

[latex] {E}_{\text{orbit}}={K}_{\text{orbit}}+\text{Δ}U=2.65\,×\,{10}^{11}+3.32\,×\,{10}^{10}=2.95\,×\,{10}^{11}\,\text{J.} [/latex]

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.

If a planet with 1.5 times the mass of Earth was traveling in Earth’s orbit, what would its period be?

Two planets in circular orbits around a star have speeds of v and 2v. (a) What is the ratio of the orbital radii of the planets? (b) What is the ratio of their periods?

Using the average distance of Earth from the Sun, and the orbital period of Earth, (a) find the centripetal acceleration of Earth in its motion about the Sun. (b) Compare this value to that of the centripetal acceleration at the equator due to Earth’s rotation.

What is the orbital radius of an Earth satellite having a period of 1.00 h? (b) What is unreasonable about this result?

Calculate the mass of the Sun based on data for Earth’s orbit and compare the value obtained with the Sun’s actual mass.

Find the mass of Jupiter based on the fact that Io, its innermost moon, has an average orbital radius of 421,700 km and a period of 1.77 days.

Astronomical observations of our Milky Way galaxy indicate that it has a mass of about [latex] 8.0\,×\,{10}^{11} [/latex] solar masses. A star orbiting on the galaxy’s periphery is about [latex] 6.0\,×\,{10}^{4} [/latex] light-years from its center. (a) What should the orbital period of that star be? (b) If its period is [latex] 6.0\,×\,{10}^{7} [/latex] years instead, what is the mass of the galaxy? Such calculations are used to imply the existence of other matter, such as a very massive black hole at the center of the Milky Way.

(a) In order to keep a small satellite from drifting into a nearby asteroid, it is placed in orbit with a period of 3.02 hours and radius of 2.0 km. What is the mass of the asteroid? (b) Does this mass seem reasonable for the size of the orbit?

The Moon and Earth rotate about their common center of mass, which is located about 4700 km from the center of Earth. (This is 1690 km below the surface.) (a) Calculate the acceleration due to the Moon’s gravity at that point. (b) Calculate the centripetal acceleration of the center of Earth as it rotates about that point once each lunar month (about 27.3 d) and compare it with the acceleration found in part (a). Comment on whether or not they are equal and why they should or should not be.

The Sun orbits the Milky Way galaxy once each [latex] 2.60\,×\,{10}^{8}\,\text{years} [/latex], with a roughly circular orbit averaging a radius of [latex] 3.00\,×\,{10}^{4} [/latex] light-years. (A light-year is the distance traveled by light in 1 year.) Calculate the centripetal acceleration of the Sun in its galactic orbit. Does your result support the contention that a nearly inertial frame of reference can be located at the Sun? (b) Calculate the average speed of the Sun in its galactic orbit. Does the answer surprise you?

A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth’s rotation). Calculate the radius of such an orbit based on the data for Earth in Astronomical Data.