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# planet x has a mass of m and a radius of r. planet y has a mass of 3m and a radius of 3r. identical satellites orbit both planets at a distance r above their surfaces, as shown above. the planets are separated by such a large distance that the gravitational forces between them are negligible.

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### James

Guys, does anyone know the answer?

get planet x has a mass of m and a radius of r. planet y has a mass of 3m and a radius of 3r. identical satellites orbit both planets at a distance r above their surfaces, as shown above. the planets are separated by such a large distance that the gravitational forces between them are negligible. from EN Bilgi.

## Solved Ral Rol 3R 3M Planet X Planet Y Planet X has a mass

Answer to Solved Ral Rol 3R 3M Planet X Planet Y Planet X has a mass

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## Physics ap classroom part A

Study Sets for Physics ap classroom part A.

3/4 F0

Meterstick and timer

Experiment 1 only

Yes. Other external forces are exerted on the planet, but they are of negligible magnitude.

Place the object on the disk and measure the distance from the center of the disk to the center of mass of the object by using a meterstick. Slowly increase the rate the disk rotates until the object begins to slide off the disk. Record the time in which

FY=3/4FX

gx=3gy

M1 , M2 , and a0

A proton and neutron located 1.0 mm apart

T1+T2/2M

Accelerometer Force sensor

MRocket 1kg FThrusters 12N MRocket 3kg FThrusters 36N

9T--------------- 4t----------

20 M/S^2

A satellite is in orbit around a planet. An object is in free fall just after it is released from rest.

No, because the slope of the curve of the graph indicates that the acceleration is less than gg, which indicates that a force other than gravity is exerted on the object.

No, because the net centripetal force exerted on the ball is the combination of the tension force from the string and the force due to gravity from Earth.

O---------> F gravity

1x10^19 N

There is another celestial body that exerts a gravitational force on the moon.

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## 13.4 Satellite Orbits and Energy

### Circular Orbits

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

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

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 vorbit${v}_{\text{orbit}}$ and vesc${v}_{\text{esc}}$. The escape velocity is exactly 2$\sqrt{2}$ 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 2πr$2\pi r$ in one period T. Using the definition of speed, we have vorbit=2πr/T${v}_{\text{orbit}}=2\pi r\text{/}T$. We substitute this into (Figure) and rearrange to get

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 ME${M}_{\text{E}}$ with MSun${M}_{\text{Sun}}$ 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.

### Example

#### The International Space Station

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

#### Strategy

Since the ISS orbits 4.00×102km$4.00\phantom{\rule{0ex}{0ex}}×\phantom{\rule{0ex}{0ex}}{10}^{2}\text{km}$ above Earth’s surface, the radius at which it orbits is RE+4.00×102km${R}_{\text{E}}+4.00\phantom{\rule{0ex}{0ex}}×\phantom{\rule{0ex}{0ex}}{10}^{2}\text{km}$. 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.

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?

In (Figure), the radius appears in the denominator inside the square root. So the radius must increase by a factor of 4, to decrease the orbital velocity by a factor of 2. The circumference of the orbit has also increased by this factor of 4, and so with half the orbital velocity, the period must be 8 times longer. That can also be seen directly from (Figure).

### Example

#### Determining the Mass of Earth

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

#### Strategy

We use (Figure), solve for ME${M}_{\text{E}}$, 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 =2.36×106s$=2.36\phantom{\rule{0ex}{0ex}}×\phantom{\rule{0ex}{0ex}}{10}^{6}\phantom{\rule{0ex}{0ex}}\text{s}$, and the average distance between the centers of Earth and the Moon is 384,000 km.

#### Solution

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

#### Significance

Compare this to the value of 5.96×1024kg$5.96\phantom{\rule{0ex}{0ex}}×\phantom{\rule{0ex}{0ex}}{10}^{24}\phantom{\rule{0ex}{0ex}}\text{kg}$ 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.)

There is another consideration to this last calculation of ME${M}_{\text{E}}$. 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?

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.

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

Using the expression for centripetal acceleration, we have

Solving for the orbit velocity, we have vorbit=47km/s${v}_{\text{orbit}}=47\phantom{\rule{0ex}{0ex}}\text{km/s}$. Finally, we can determine the period of the orbit directly from T=2πr/vorbit$T=2\pi r\text{/}{v}_{\text{orbit}}$, to find that the period is T=1.6×1018s$T=1.6\phantom{\rule{0ex}{0ex}}×\phantom{\rule{0ex}{0ex}}{10}^{18}\phantom{\rule{0ex}{0ex}}\text{s}$, 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.