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Chapter 13: Problem 31

The mean diameters of Mars and Earth are \(6.9 \times 10^{3} \mathrm{~km}\) and \(1.3 \times 10^{4} \mathrm{~km}\), respectively. The mass of Mars is \(0.11\) times Earth's mass. (a) What is the ratio of the mean density (mass per unit volume) of Mars to that of Earth? (b) What is the value of the gravitational acceleration on Mars? (c) What is the escape speed on Mars?

Short Answer

Expert verified
The density ratio is approximately 0.71. Gravitational acceleration on Mars is about 3.7 m/s虏. The escape speed is around 5.0 km/s.

Step by step solution

01

Calculate the Volume of Mars and Earth

The volume of a sphere is given by the formula \(V = \frac{4}{3}\pi r^3\), where \(r\) is the radius. First, calculate the radius of Mars and Earth from their diameters:\[r_{\text{Mars}} = \frac{6.9 \times 10^3}{2} = 3.45 \times 10^3 \, \text{km}\] \[r_{\text{Earth}} = \frac{1.3 \times 10^4}{2} = 6.5 \times 10^3 \, \text{km}\] Now, calculate the volumes:\[V_{\text{Mars}} = \frac{4}{3}\pi (3.45 \times 10^3)^3\] \[V_{\text{Earth}} = \frac{4}{3}\pi (6.5 \times 10^3)^3\]
02

Calculate the Mass and Density of Mars and Earth

Assume Earth's mass is \(M_{\text{Earth}}\). The mass of Mars is given by \(M_{\text{Mars}} = 0.11 \times M_{\text{Earth}}\). The density \(\rho\) is mass per unit volume, \(\rho = \frac{M}{V}\). Now, calculate the density:\[\rho_{\text{Mars}} = \frac{M_{\text{Mars}}}{V_{\text{Mars}}} = \frac{0.11 \times M_{\text{Earth}}}{V_{\text{Mars}}}\] \[\rho_{\text{Earth}} = \frac{M_{\text{Earth}}}{V_{\text{Earth}}}\]
03

Calculate Density Ratio

The ratio of the density of Mars to Earth's density is:\[\frac{\rho_{\text{Mars}}}{\rho_{\text{Earth}}} = \frac{0.11 \times M_{\text{Earth}} / V_{\text{Mars}}}{M_{\text{Earth}} / V_{\text{Earth}}} = 0.11 \times \frac{V_{\text{Earth}}}{V_{\text{Mars}}}\] Substitute the volumes from Step 1 to find this ratio.
04

Calculate Gravitational Acceleration on Mars

Gravitational acceleration \(g\) is given by \(g = \frac{G M}{r^2}\), where \(G\) is the gravitational constant. For Mars:\[g_{\text{Mars}} = \frac{G \times M_{\text{Mars}}}{r_{\text{Mars}}^2}\] Substitute \(M_{\text{Mars}} = 0.11 \times M_{\text{Earth}}\) and calculate \(g_{\text{Mars}}\), knowing that Earth's gravitational acceleration \(g_{\text{Earth}} = 9.8\, \text{m/s}^2\).
05

Calculate Escape Speed on Mars

Escape speed \(v_{\text{esc}}\) is given by \(v_{\text{esc}} = \sqrt{\frac{2 G M}{r}}\). For Mars, substitute for \(M_{\text{Mars}}\) and \(r_{\text{Mars}}\) to find:\[v_{\text{esc, Mars}} = \sqrt{\frac{2 G \times 0.11 M_{\text{Earth}}}{r_{\text{Mars}}}}\] Calculate \(v_{\text{esc, Mars}}\), using the radius calculated in Step 1.

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Key Concepts

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

Density Calculation
Calculating density is essential in planetary physics to understand the distribution of mass within a celestial body. In simple terms, density is the mass of an object divided by its volume. In the case of Mars and Earth, we can compare their densities to understand their structural differences.
The formula for density \(\rho\) is given by:
  • \(\rho = \frac{M}{V}\)
Where \(M\) represents mass and \(V\) is volume.To find the density of a planet, we first calculate its volume. For a sphere, volume \(V\) is calculated using the formula:
  • \(V = \frac{4}{3}\pi r^3\)
Where \(r\) is the radius. Once the volume is known, density can be determined by substituting the planet's mass into the density formula. For Mars, knowing its relationship to Earth's mass allows us to compute its density by comparing it to Earth's known quantities. The comparison provides insight into differences in core composition and structure between Mars and Earth.
Gravitational Acceleration
Gravitational acceleration is the force that determines how fast objects fall towards a planet's surface. It's crucial for understanding the behavior of planets in their orbits and for planning travel between them. For Mars, calculating its gravitational acceleration helps us determine what a visitor would experience compared to Earth.
The formula for gravitational acceleration \(g\) is:
  • \(g = \frac{G M}{r^2}\)
Here, \(G\) stands for the gravitational constant, \(M\) is the mass of the planet, and \(r\) is its radius.
Using this formula, we see that gravitational acceleration depends on both the mass of the planet and the distance from its center. As Mars has less mass than Earth, the gravitational pull is weaker. By knowing the ratio of Mars's mass to Earth's, we can substitute into the formula to find the gravitational acceleration on Mars. This step helps visualize how much easier it is to jump on Mars compared to Earth.
Escape Speed
Escape speed is the minimum velocity required to break free from a planet's gravitational pull. It鈥檚 vital for understanding space travel and how rockets achieve orbit. Calculating the escape speed for Mars shows us how different factors like mass and radius influence this velocity.
The formula for escape speed \(v_{\text{esc}}\) is:
  • \(v_{\text{esc}} = \sqrt{\frac{2 G M}{r}}\)
Where \(G\) is the gravitational constant, \(M\) is the mass of the planet, and \(r\) is the radius from which escape is attempted. Substituting the given values for Mars, including its mass relative to Earth, lets us calculate the escape speed unique to Mars. This speed is lower than Earth's due to Mars's smaller mass and radius, making it easier for spacecraft to leave Martian gravity. Understanding escape speed is crucial to planning missions and ensuring rockets have enough fuel to return to their journey's origin.

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Most popular questions from this chapter

(a) What will an object weigh on the Moon's surface if it weighs \(100 \mathrm{~N}\) on Earth's surface? (b) How many Earth radii must this same object be from the center of Earth if it is to weigh the same as it does on the Moon?

Three point particles are fixed in place in an \(x y z\) coordinate system. Particle \(A\), at the origin, has mass \(m_{A} .\) Particle \(B\), at \(x y z\) coordinates \((2.00 d, 1.00 d, 2.00 d)\), has mass \(2.00 m_{A}\), and particle \(C\), at coordinates \((-1.00 d, 2.00 d,-3.00 d)\), has mass \(3.00 m_{A}\). A fourth particle \(D\), with mass \(4.00 m_{A}\), is to be placed near the other particles. In terms of distance \(d\), at what (a) \(x,(\mathrm{~b}) y\), and (c) \(z\) coordinate should \(D\) be placed so that the net gravitational force on \(A\) from \(B, C\), and \(D\) is zero?

The mysterious visitor that appears in the enchanting story The Little Prince was said to come from a planet that "was scarcely any larger than a house!" Assume that the mass per unit volume of the planet is about that of Earth and that the planet does not appreciably spin. Approximate (a) the free-fall acceleration on the planet's surface and (b) the escape speed from the planet.

The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just barely provides the centripetal force needed for the rotation. (Why?) (a) Show that the corresponding shortest period of rotation is $$T=\sqrt{\frac{3 \pi}{G \rho}}$$ where \(\rho\) is the uniform density (mass per unit volume) of the spherical planet. (b) Calculate the rotation period assuming a density of \(3.0 \mathrm{~g} / \mathrm{cm}^{3}\), typical of many planets, satellites, and asteroids. No astronomical object has ever been found to be spinning with a period shorter than that determined by this analysis.

In deep space, sphere \(A\) of mass \(20 \mathrm{~kg}\) is located at the origin of an \(x\) axis and sphere \(B\) of mass \(10 \mathrm{~kg}\) is located on the axis at \(x=\) \(0.80 \mathrm{~m}\). Sphere \(B\) is released from rest while sphere \(A\) is held at the origin. (a) What is the gravitational potential energy of the twosphere system just as \(B\) is released? (b) What is the kinetic energy of \(B\) when it has moved \(0.20 \mathrm{~m}\) toward \(A\) ?
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