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DATA For a spherical planet with mass \(M,\) volume \(V,\) and radius \(R,\) derive an expression for the acceleration due to gravity at the planet's surface, \(g\), in terms of the average density of the planet, \(\rho=M / V,\) and the planet's diameter, \(D=2 R .\) The table gives the values of \(D\) and \(g\) for the eight major planets: $$ \begin{array}{lrc} \text { Planet } & D(\mathrm{~km}) & g\left(\mathrm{~m} / \mathrm{s}^{2}\right) \\ \hline \text { Mercury } & 4879 & 3.7 \\ \text { Venus } & 12,104 & 8.9 \\ \text { Earth } & 12,756 & 9.8 \\ \text { Mars } & 6792 & 3.7 \\ \text { Jupiter } & 142,984 & 23.1 \\ \text { Saturn } & 120,536 & 9.0 \\ \text { Uranus } & 51,118 & 8.7 \\ \text { Neptune } & 49.528 & 11.0 \end{array} $$ (a) Treat the planets as spheres. Your equation for \(g\) as a function of \(\rho\) and \(D\) shows that if the average density of the planets is constant, a graph of \(g\) versus \(D\) will be well represented by a straight line. Graph 8 as a function of \(D\) for the eight major planets. What does the graph tell you about the variation in average density? (b) Calculate the average density for each major planet. List the planets in order of decreasing density, and give the calculated average density of each. (c) The earth is not a uniform sphere and has greater density near its center. It is reasonable to assume this might be true for the other planets. Discuss the effect this has on your analysis. (d) If Saturn had the same average density as the earth, what would be the value of \(g\) at Saturn's surface?

Short Answer

Expert verified
By plotting a graph of \(g\) against \(D\), it's clear that the variation in the average density causes the graph to be other than a straight line. The average densities of each planet can be calculated using the formula \(\rho = \frac{M}{V}\), which also gives us a list of the planets by decreasing densities. A nonuniform sphere would result in a slightly different true value of \(g\), due to the higher gravitational effect at the surface from the denser core. Using the given formula and Earth's density, \(g\) for Saturn can be calculated.

Step by step solution

01

Deriving Expression for Gravitational Acceleration

To start, we know from physics that the acceleration due to gravity \(g\) on a planet's surface is given by: \(g = \frac{GM}{R^{2}}\). Given that \(M = \rho V\) and \(V = \frac{4}{3} \pi R^{3}\), we substitute these values into the equation to find our expression for \(g\): \(g = \frac{4 \pi G \rho R}{3}\). \nSince Diameter \(D = 2R\) , our final equation is \(g = \frac{2 \pi G \rho D}{3}\). This equation shows gravity \(g\) is directly proportional to the Diameter \(D\) if \(\rho\) is constant.
02

Graphing Variation in Gravity

Using the derived equation, plot a graph of \(g\) against the Diameter \(D\) for the provided data of the eight planets. If the planets all have the same average density, then the graph will be a straight line. The slope of this line indicates the variation in average density.
03

Calculating Average Density

To calculate the average density for each planet, use the equation \(\rho = \frac{M}{V}\). For a sphere, the volume is given by \(V = \frac{4}{3}\pi R^{3}\). You will have to estimate the Mass \(M\) of each planet and then calculate the density. Order the planets according to decreasing densities.
04

Effects of Non-Uniform Sphere

If the planets are non-uniform spheres with higher density near the centre, the derived equation from Step 1 would slightly differ due to the increased gravitational effect at the planet surface from the greater inner density.
05

Gravitational Acceleration on Saturn

Calculate \(g\) for Saturn using the equation from Step 1, but instead of using Saturn's average density, use Earth's average density: \(\rho_{earth}\)

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Key Concepts

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

Planetary Average Density
Understanding the concept of planetary average density is crucial in astrophysics and helps to decipher various attributes of a planet, including its gravitational pull. The average density, denoted by the symbol \( \rho \), is mathematically expressed as the ratio of a planet's mass \( M \) to its volume \( V \). In formula terms, \( \rho = \frac{M}{V} \).

For spherical planets, which most planets are approximated to be, the volume can be calculated using the formula \( V = \frac{4}{3}\pi R^3 \), where \( R \) is the radius of the planet. By substituting the volume in the density equation, we shift our focus from mass to observable quantities like a planet's diameter \( D \) and gravitation. Since \( D = 2R \), the diameter is an easily measurable physical property, making density calculations more feasible.

When we explore the step-by-step solution to calculate average density, we can better appreciate the relevance of this concept. The variation in density across different planets can affect gravitational acceleration, thereby impacting the planet's ability to hold onto its atmosphere, influence tides, and maintain orbital stability.
Gravitational Force in Physics
The concept of gravitational force in physics is pivotal in understanding the cosmic ballet of planets, moons, and celestial bodies. Isaac Newton's law of universal gravitation introduces us to the force that attracts two bodies towards each other. It's governed by the equation \( F = G\frac{m_1m_2}{r^2} \), where \( F \) is the gravitational force, \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the two bodies, and \( r \) is the distance between the centers of the masses.

For a person standing on the surface of a planet, this gravitational force translates into the acceleration due to gravity, denoted as \( g \) - it's the rate at which an object will accelerate when falling towards the surface of the planet. This force is so fundamental that it affects not just the trajectory of a tossed ball, but the motion of planets and the fabric of spacetime. In the exercise's solution, we tie this force to a planet's mass and radius, demonstrating that \( g \) can be derived from these two variables, leading us also to understand the relationship to the planet's density and diameter.
Planetary Diameters
The planetary diameters are a straightforward measure that give us initial insight into the size and scale of planets within our solar system and beyond. Simply put, the diameter of a planet is the distance through the center of the planet from one point on its surface to the opposite point. For spherical objects, this is twice the radius (\( D = 2R \)).

Diameter holds significance as it relates to other fundamental planetary characteristics—like surface gravity. In our exercise, we learn that if the average density is constant, the diameter of a planet can be directly linked to gravitational acceleration at the planet's surface. This relationship is beautifully illustrated when plotting \( g \) against \( D \) for the major planets, which should yield a linear graph if uniform density is assumed. However, planet diameters can vary widely, from the smaller terrestrial planets like Mercury to the gas giants like Jupiter, reflecting the remarkable diversity of our solar system's celestial bodies.

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Most popular questions from this chapter

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 \(L\)

Two uniform spheres, each with mass \(M\) and radius \(R\), touch each other. What is the magnitude of their gravitational force of attraction?

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.

Falling Hammer. A hammer with mass \(m\) is dropped from rest from a height \(h\) above the earth's surface. This height is not necessarily small compared with the radius \(R_{\mathrm{E}}\) of the earth. Ignoring air resistance, derive an expression for the speed \(v\) of the hammer when it reaches the earth's surface. Your expression should involve \(h, R_{\mathrm{E}},\) and \(m_{\mathrm{E}}\) (the earth's mass).

In March \(2006,\) two small satellites were discovered orbiting Pluto, one at a distance of \(48,000 \mathrm{~km}\) and the other at \(64,000 \mathrm{~km}\). Pluto already was known to have a large satellite Charon, orbiting at \(19,600 \mathrm{~km}\) with an orbital period of 6.39 days. Assuming that the satellites do not affect each other, find the orbital periods of the two small satellites without using the mass of Pluto.

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