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Chapter 6: Problem 12

(II) Estimate the acceleration due to gravity at the surface of Europa (one of the moons of Jupiter) given that its mass is \(4.9 \times 10^{22} \mathrm{~kg}\) and making the assumption that its density is the same as Earth's.

Short Answer

Expert verified
The acceleration due to gravity at the surface of Europa is approximately 1.314 m/s虏.

Step by step solution

01

Understand the Density Relationship

We are given that Europa's density is the same as Earth's. Earth's density is approximately \(5.52 \, \text{g/cm}^3\). We will use this density to find Europa's radius since density \( \rho \) is mass \( M \) divided by volume \( V \).
02

Compute Volume of Europa

The formula for density is \( \rho = \frac{M}{V} \). Volume \( V \) of a sphere is \( \frac{4}{3} \pi r^3 \). Solving for \( V \) gives \( V = \frac{M}{\rho} \). Substitute \( M = 4.9 \times 10^{22} \, \text{kg} \) and Earth's density \( 5.52 \, \text{g/cm}^3 \) (converted to \( 5520 \, \text{kg/m}^3 \)). Thus, \( V = \frac{4.9 \times 10^{22}}{5520} \approx 8.88 \times 10^{18} \, \text{m}^3 \).
03

Calculate Radius of Europa

Use the volume \( V = \frac{4}{3} \pi r^3 \) to solve for radius \( r \). Thus, \( r = \sqrt[3]{\frac{3V}{4\pi}} \). Plugging in the volume, \( r = \sqrt[3]{\frac{3 \times 8.88 \times 10^{18}}{4\pi}} \approx 1.56 \times 10^6 \, \text{m} \).
04

Apply the Formula for Gravitational Acceleration

Gravitational acceleration \( g \) at the surface is given by \( g = \frac{GM}{r^2} \), where \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2 \). Substituting \( M = 4.9 \times 10^{22} \, \text{kg} \) and \( r = 1.56 \times 10^6 \, \text{m} \), we find \( g = \frac{6.674 \times 10^{-11} \times 4.9 \times 10^{22}}{(1.56 \times 10^6)^2} \approx 1.314 \text{ m/s}^2 \).

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Key Concepts

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

Surface Gravity of Moons
When we discuss the surface gravity of moons, we're talking about the acceleration that a small object would experience due to the moon's gravity when close to its surface.

Surface gravity is an important concept because it determines how strongly an object will be pulled towards the moon. The value depends on a few key factors: the moon's mass, its radius, and fundamentally, the gravitational pull that it exerts.
  • The greater the mass of the moon, the stronger its gravitational pull.
  • The closer you are to the center of the moon, the stronger the gravity鈥攐r in simpler terms, the smaller the radius, the stronger the surface gravity, assuming mass remains constant.
This concept is critical not only for academic purposes but also for practical applications like calculating the force you would feel when standing on a moon like Europa.
For instance, we calculated Europa's surface gravity to be approximately 1.314 m/s虏, which is much less than Earth's gravity, showing how much lighter you would feel standing there.
Europa
Europa is one of Jupiter's many moons, and it's a fascinating celestial body for many reasons. Unlike other moons in the solar system, Europa is known for its smooth ice-covered surface. This makes it particularly intriguing for scientists who speculate there might be liquid water beneath its icy crust.

Given its mass of approximately 4.9 脳 10虏虏 kg, Europa's gravitational pull is significantly less than that of Earth. This lower gravity affects everything from a spacecraft's landing mechanics to potential biological processes.
  • The lower surface gravity, about 1.314 m/s虏, means activities would require recalibration compared to those on Earth.
  • Europa's density, assumed to be similar to Earth's in calculations, is an interesting point because it helps estimate characteristics like its radius and internal composition.
Europa continues to attract science enthusiasm because of the possibility of extraterrestrial life, given its potential sub-crustal ocean.
Gravitational Constant
The gravitational constant, often represented as \( G \), plays a pivotal role in celestial mechanics. Its value is approximately \( 6.674 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2 \).

This constant essentially links the gravitational force to the masses involved and the distance that separates them. The value remains constant across the universe, making it a cornerstone of calculations involving gravitational forces.
  • It's used in Newton's Law of Universal Gravitation, which permits us to calculate the force exerted by one mass upon another.
  • In the context of calculating surface gravity, \( G \) is essential for determining how much force an object would feel due to gravity on any given celestial body, like Europa.
Simply put, \( G \) allows us to predict and understand how astronomical objects interact, making it an indispensable tool for exploring and understanding our universe.

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Most popular questions from this chapter

(II) What will a spring scale read for the weight of a \(53-\mathrm{kg}\) woman in an elevator that moves \((a)\) upward with constant speed \(5.0 \mathrm{~m} / \mathrm{s},(b)\) downward with constant speed \(5.0 \mathrm{~m} / \mathrm{s}\) (c) upward with acceleration \(0.33 g,(d)\) downward with acceleration \(0.33 g\), and \((e)\) in free fall?

What is the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Earth is \(\frac{1}{10}\) of its value at the Earth's surface?

Astronomers have observed an otherwise normal star, called S2, closely orbiting an extremely massive but small object at the center of the Milky Way Galaxy called SgrA. S2 moves in an elliptical orbit around SgrA with a period of \(15.2 \mathrm{yr}\) and an eccentricity \(e=0.87\) (Fig. \(6-16\) ). In \(2002, \mathrm{~S} 2\) reached its closest approach to SgrA, a distance of only \(123 \mathrm{AU} \quad\left(1 \mathrm{AU}=1.50 \times 10^{11} \mathrm{~m}\right.\) is the mean Earth-Sun distance). Determine the mass \(M\) of SgrA, the massive compact object (believed to be a supermassive black hole) at the center of our Galaxy. State \(M\) in \(\mathrm{kg}\) and in terms of the mass of our Sun.

You are an astronaut in the space shuttle pursuing a satellite in need of repair. You are in a circular orbit of the same radius as the satellite \((400 \mathrm{km}\) above the Earth), but 25 \(\mathrm{km}\) behind it. (a) How long will it take to overtake the satellite if you reduce your orbital radius by 1.0 \(\mathrm{km}\) ? (b) By how much must you reduce your orbital radius to catch up in 7.0 \(\mathrm{h}\) ?

(II) Given that the acceleration of gravity at the surface of Mars is 0.38 of what it is on Earth, and that Mars' radius is \(3400 \mathrm{~km},\) determine the mass of Mars.
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