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Chapter 13: Problem 50

Two spherically symmetric planets with no atmosphere have the same average density, but planet \(B\) has twice the radius of planet \(A\). A small satellite of mass \(m_{A}\) has period \(T_{A}\) when it orbits planet \(A\) in a circular orbit that is just above the surface of the planet. A small satellite of mass \(m_{B}\) has period \(T_{B}\) when it orbits planet \(B\) in a circular orbit that is just above the surface of the planet. How does \(T_{B}\) compare to \(T_{A} ?\) with a density of \(2500 \mathrm{~kg} / \mathrm{m}^{3} ?\)

Short Answer

Expert verified
The orbital period of the satellite orbiting planet B is \(2\sqrt{2}\) times the orbital period of the satellite orbiting planet A.

Step by step solution

01

Write down the formula for the period of an object in orbit

The formula for the period \( T \) of an object in orbit due to gravitational forces is given by \( T = 2\pi\sqrt{\frac{r^{3}}{GM}} \), where \( r \) is the distance from the center of the orbit (radius of the planet in this case), \( G \) is the gravitational constant, and \( M \) is the mass of the planet.
02

Express the mass of the planets in terms of their density and volume

The mass of a spherically symmetric planet can be expressed as \( M = \rho V \), where \( \rho \) is the density and \( V \) is the volume. For a sphere, \( V = \frac{4}{3}\pi r^{3} \). So, we can rewrite mass as \( M = \rho \frac{4}{3}\pi r^{3} \). Substituting this into the formula for the period we get \( T = 2\pi\sqrt{\frac{r^{3}}{G\rho \frac{4}{3}\pi r^{3}}} \). By canceling out \( r^{3} \) and multiplying by the inverse of \( \rho \), we get \( T \propto \frac{1}{\sqrt{\rho}}.\) This means the period \( T \) is inversely proportional to the square root of the density.
03

Compare \( T_{B} \) and \( T_{A} \) considering the double radius

Since the planets have the same density and the radius of Planet B is twice the radius of Planet A (\( r_{B} = 2r_{A} \)), we have \( T_{B} = 2\pi\sqrt{\frac{(2r_{A})^{3}}{G \rho \frac{4}{3}\pi (2r_{A})^{3}}} \). This simplifies to give us \( T_{2} = \sqrt{8}\) or \( \sqrt{2^{3}} \) times \( T_{A} \). Therefore, \( T_{B} = 2\sqrt{2} T_{A} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Forces
Gravitational forces are the invisible glue that holds the universe together. They are the forces of attraction between two masses. For planets and satellites, this force acts to keep the satellite in orbit around the planet. The strength of gravitational force depends on two major factors: the mass of the objects and the distance between them.

- **Mass:** The more massive an object, the stronger its gravitational pull. This is why larger planets like Jupiter have stronger gravitational fields than smaller ones like Earth.- **Distance:** The further apart two objects are, the weaker the gravitational force.
To calculate the period of a satellite orbiting a planet, we use the formula:\[T = 2\pi \sqrt{\frac{r^{3}}{GM}}\]Here, **r** is the distance from the center of planet to the satellite, **G** is the gravitational constant, and **M** is the mass of the planet. This formula shows how gravitational forces determine the orbital characteristics of satellites.
Spherically Symmetric Planets
A spherically symmetric planet is one that can be described by a perfect sphere, implying that its density, shape, and composition appear uniformly distributed in all directions. This assumption simplifies calculations in celestial mechanics.

When dealing with spherically symmetric planets, it is crucial to understand the role of symmetry in the distribution of mass and how it impacts gravitational forces. Such planets can be characterized by:
  • Uniform Density: The average density remains the same throughout the planet.
  • Center of Mass: The gravitational attraction can be considered as if all mass is concentrated at the center.
This assumption makes it easier to calculate orbital periods because the gravitational force at any point on the surface depends only on the distance from the center.
For satellites orbiting just above a planet鈥檚 surface, the radius of the planet becomes the key distance influencing orbital mechanics, as in our exercise where Planet B鈥檚 radius is twice that of Planet A.
Density and Volume Relationship
Density and volume are interconnected concepts critical in understanding planets and their gravitational fields. Density (\(\rho\)) is defined as mass per unit volume and plays a key role in determining many planetary characteristics including its gravitational force.

- **Density Formula:** \[\rho = \frac{mass}{volume}\]- **Volume of a Sphere:** A sphere's volume, when simplified for planets, is given by the formula: \[V = \frac{4}{3}\pi r^3\]Combining these formulas helps us express a planet's mass in terms of its density and volume: \[M = \rho V = \rho \frac{4}{3}\pi r^3\]In the orbital period formula, substituting for mass (\(M\)) allows density鈥攁n intrinsic property shared by both planets in our exercise鈥攖o be an important factor.
Since the planet's mass is distributed across a volume shaped by its radius, any increase in radius, like Planet B鈥檚 doubling, affects its total volume significantly, and thus influences its gravitational effects.

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Most popular questions from this chapter

\(\cdot \mathrm{A}\) satellite of mass \(m\) is in a circular orbit around a spherical planet of mass \(m_{\mathrm{p}}\). The kinetic energy of the satellite is \(K_{A}\) when its orbit radius is \(r_{A}\). In terms of \(r_{A}\), what must the orbit radius be in order for the kinetic energy of the satellite to be \(2 K_{A} ?\)

The star Rho \({ }^{1}\) Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a circular orbit around Rho \({ }^{1}\) Cancri with an orbital radius equal to 0.11 times the radius of the earth's orbit around the sun. What are (a) the orbital speed and (b) the orbital period of the planet of Rho " Cancri?

The planet Uranus has a radius of \(25,360 \mathrm{~km}\) and a surface acceleration due to gravity of \(9.0 \mathrm{~m} / \mathrm{s}^{2}\) at its poles. Its moon Miranda (discovered by Kuiper in 1948 ) is in a circular orbit about Uranus at an altitude of \(104,000 \mathrm{~km}\) above the planet's surface. Miranda has a mass of \(6.6 \times 10^{19} \mathrm{~kg}\) and a radius of \(236 \mathrm{~km}\). (a) Calculate the mass of Uranus from the given data. (b) Calculate the magnitude of Miranda's acceleration due to its orbital motion about Uranus. (c) Calculate the acceleration due to Miranda's gravity at the surface of Miranda. (d) Do the answers to parts (b) and (c) mean that an object released \(1 \mathrm{~m}\) above Miranda's surface on the side toward Uranus will fall up relative to Miranda? Explain.

A thin, uniform rod has length \(L\) and mass \(M\). A small uniform sphere of mass \(m\) is placed a distance \(x\) from one end of the rod, along the axis of the rod (Fig. E13.38). (a) Calculate the gravitational potential energy of the rod-sphere system. Take the potential energy to be zero when the rod and sphere are infinitely far apart. Show that your answer reduces to the expected result when \(x\) is much larger than \(L\). (Hint: Use the power series expansion for \(\ln (1+x)\) given in Appendix B.) (b) Use \(F_{x}=-d U / d x\) to find the magnitude and direction of the gravitational force exerted on the sphere by the rod (see Section 7.4 ). Show that your answer reduces to the expected result when \(x\) is much larger than \(L\)

At the Galaxy's Core. Astronomers have observed a small, massive object at the center of our Milky Way galaxy (see Section 13.8). A ring of material orbits this massive object; the ring has a diameter of about 15 light-years and an orbital speed of about \(200 \mathrm{~km} / \mathrm{s}\) (a) Determine the mass of the object at the center of the Milky Way galaxy. Give your answer both in kilograms and in solar masses (one solar mass is the mass of the sun). (b) Observations of stars, as well as theories of the structure of stars, suggest that it is impossible for a single star to have a mass of more than about 50 solar masses. Can this massive object be a single, ordinary star? (c) Many astronomers believe that the massive object at the center of the Milky Way galaxy is a black hole. If so, what must the Schwarzschild radius of this black hole be? Would a black hole of this size fit inside the earth's orbit around the sun?
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