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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, \(D = 2R\). The table gives the values of \(D\) and \(g\) for the eight major planets: (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 g 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 nonuniformity 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
Gravitational acceleration \( g = \frac{2}{3} \pi G \rho D \), linear if \( \rho \) is constant.

Step by step solution

01

Expression for Gravitational Acceleration

The acceleration due to gravity at the surface of a spherical planet is given by Newton's law of gravitation: \[ g = \frac{GM}{R^2} \]where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.
02

Relate Mass and Density

The average density \( \rho \) of the planet is defined as \[ \rho = \frac{M}{V} \].For a sphere, the volume \( V \) is given by \( V = \frac{4}{3} \pi R^3 \).Thus, the mass \( M \) in terms of density and volume is \[ M = \rho \times \frac{4}{3} \pi R^3 \].
03

Substitute and Simplify

Substitute the expression for \( M \) from step 2 into the formula for \( g \) in step 1:\[ g = \frac{G(\rho \times \frac{4}{3} \pi R^3)}{R^2} \]. Simplify this expression to get:\[ g = \frac{4}{3} \pi G \rho R \].
04

Express in Terms of Diameter

Since the diameter \( D = 2R \), substitute to express \( g \) in terms of \( D \):\[ g = \frac{2}{3} \pi G \rho D \].
05

Graphical Interpretation

The equation \( g = \frac{2}{3} \pi G \rho D \) implies that \( g \) is linearly dependent on \( D \) if \( \rho \) is constant. Plotting \( g \) vs. \( D \) for the eight planets will give a straight line if the density \( \rho \) is indeed constant across planets.
06

Calculate Average Density

Calculate the average density \( \rho = \frac{M}{V} \) for each planet using known mass and volume data. List planets in order of decreasing density based on these calculations.
07

Consider Nonuniform Density

Discuss that if a planet has greater density towards its center (non-uniform density), the assumption of constant \( \rho \) might lead to deviations from the linear relationship between \( g \) and \( D \).
08

Hypothetical Calculation for Saturn

If Saturn had the same average density as Earth, the new \( \rho \) would result in a recalculated gravitational acceleration:1. Use Earth's density to find Saturn's \( g \) with \( \rho_S = \rho_E \) using \( g = \frac{4}{3} \pi G \rho_S R_S \).

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Key Concepts

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

Average Density
To understand gravitational acceleration on a spherical planet, a crucial concept is the *average density*. Average density simplifies the complexity of a body's mass distribution into a single value that represents mass per unit volume. For a planet, this is calculated using the formula \( \rho = \frac{M}{V} \), where \( M \) is the planet's mass and \( V \) is its volume. This is especially useful for spherical planets, as their volume can be easily calculated using \( V = \frac{4}{3} \pi R^3 \), where \( R \) is the radius.

Knowing the average density helps in estimating the gravitational force on the planet's surface, which is produced from its mass. Essentially, by determining the average density, we can simplify the calculation of how strong a planet's gravity would be, giving us insights into not just its mass, but how that mass is spread out in space.
Newton's Law of Gravitation
Newton's Law of Gravitation is vital for understanding how planets' gravity works. This law states that every point mass attracts every other point mass by a force acting along the line joining the two points. The gravitational acceleration \( g \) at the surface of a planet of mass \( M \) and radius \( R \) can be expressed as \( g = \frac{GM}{R^2} \), where \( G \) is the gravitational constant.

This formula shows that the gravitational force is directly proportional to the mass of the planet and inversely proportional to the square of its radius. Therefore, larger or denser planets exert more gravitational pull. By substituting the planetary mass with its equivalent expression using average density, \( M = \rho V = \rho \times \frac{4}{3} \pi R^3 \), we derive a formula expressing \( g \) in terms of density and size.
Spherical Planets
The assumption that planets are spherical simplifies many calculations and makes it easier to apply formulas uniformly. A spherical shape allows for the determination of volume using \( V = \frac{4}{3} \pi R^3 \), simplifying calculations regarding gravitational forces detected at the surface.

In these models, the radius \( R \) directly influences calculations for diameter and volume, central to understanding the planet's gravitational acceleration. Because the gravitational force is dependent on mass distribution, a spherical shape ensures that this distribution is uniform relative to the center, aiding in accurate calculations using average density.

While not all planets are perfect spheres due to rotation and other factors causing bulging, treating them as spheres is a useful approximation for many scenarios.
Density Variation Analysis
When calculating gravitational acceleration at the surface, variations in density must be considered. Planets don't always have uniform density; many have higher densities at their centers due to the immense pressure. This is particularly important in calculations because changes in density can cause deviations in expected gravitational force.

A non-uniform density complicates the assumption of a constant \( \rho \), potentially leading to errors if ignored. If a planet's density increases towards the core, its effective gravitational pull might differ from that predicted using simple averages, affecting the straight-line relationship between \( g \) and diameter \( D \).

In practical applications, variations like these must be evaluated to give more accurate portrayals of planetary gravity, influencing how we model not just gravitational forces, but also planetary formation and structure.

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Most popular questions from this chapter

In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. Pluto already was known to have a large satellite Charon, orbiting at 19,600 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.

Your starship, the \(Aimless\) \(Wanderer\), lands on the mysterious planet Mongo. As chief scientist-engineer, you make the following measurements: A 2.50-kg stone thrown upward from the ground at 12.0 m/s returns to the ground in 4.80 s; the circumference of Mongo at the equator is 2.00 \(\times\) 10\(^5\) km; and there is no appreciable atmosphere on Mongo. The starship commander, Captain Confusion, asks for the following information: (a) What is the mass of Mongo? (b) If the \(Aimless\) \(Wanderer\) goes into a circular orbit 30,000 km above the surface of Mongo, how many hours will it take the ship to complete one orbit?

Two stars, with masses \({M_1}\) and \({M_2}\), are in circular orbits around their center of mass. The star with mass \({M_1}\) has an orbit of radius \({R_1}\); the star with mass \({M_2}\) has an orbit of radius \({R_2}\). (a) Show that the ratio of the orbital radii of the two stars equals the reciprocal of the ratio of their masses\(-\)that is, \({R_1}\)/\({R_2}\) \(=\) \({M_2}\)/\({M_1}\). (b) Explain why the two stars have the same orbital period, and show that the period \(T\) is given by \(T = 2\pi\)(R1 + R2)\(^{3/2}\)/\(\sqrt{G(M1 + M2)}\). (c) The two stars in a certain binary star system move in circular orbits. The first star, Alpha, has an orbital speed of 36.0 km/s. The second star, Beta, has an orbital speed of 12.0 km/s. The orbital period is 137 d. What are the masses of each of the two stars? (d) One of the best candidates for a black hole is found in the binary system called A0620-0090. The two objects in the binary system are an orange star, V616 Monocerotis, and a compact object believed to be a black hole (see Fig. 13.28). The orbital period of A0620-0090 is 7.75 hours, the mass of V616 Monocerotis is estimated to be 0.67 times the mass of the sun, and the mass of the black hole is estimated to be 3.8 times the mass of the sun. Assuming that the orbits are circular, find the radius of each object's orbit and the orbital speed of each object. Compare these answers to the orbital radius and orbital speed of the earth in its orbit around the sun.

The acceleration due to gravity at the north pole of Neptune is approximately 11.2 m/s\(^2\). Neptune has mass 1.02 \(\times\) 10\(^{26}\) kg and radius 2.46 \(\times\) 10\(^4\) km and rotates once around its axis in about 16 h. (a) What is the gravitational force on a 3.00-kg object at the north pole of Neptune? (b) What is the apparent weight of this same object at Neptune's equator? (Note that Neptune's "surface" is gaseous, not solid, so it is impossible to stand on it.)

A uniform sphere with mass 50.0 kg is held with its center at the origin, and a second uniform sphere with mass 80.0 kg is held with its center at the point \(x =\) 0, \(y =\) 3.00 m. (a) What are the magnitude and direction of the net gravitational force due to these objects on a third uniform sphere with mass 0.500 kg placed at the point \(x =\) 4.00 m, \(y =\) 0? (b) Where, other than infinitely far away, could the third sphere be placed such that the net gravitational force acting on it from the other two spheres is equal to zero?

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