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Chapter 6: Problem 14

(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.

Short Answer

Expert verified
The mass of Mars is approximately \(6.44 \times 10^{23}\) kg.

Step by step solution

01

Understanding the Problem

We need to find the mass of Mars given its radius and the gravitational acceleration at its surface, which is 0.38 times that of Earth.
02

Formula for Gravitational Force

The gravitational force at the surface of a planet is given by the formula: \[ F = \frac{G M}{r^2} \]where \( F \) is the gravitational force, \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \mathrm{~m^3 \, kg^{-1} \, s^{-2}})\), \( M \) is the mass of the planet, and \( r \) is the radius of the planet.
03

Substitute Known Values

We know that the gravitational force on Mars' surface is 0.38 times that on Earth's surface (9.81 m/s虏). Therefore, we have:\[ 0.38 \times 9.81 \, \text{m/s虏} = \frac{G M}{(3400 \times 10^3 \, \text{m})^2} \]
04

Solving for the Mass of Mars

Substitute known values into the formula: \[ 0.38 \times 9.81 = \frac{6.674 \times 10^{-11} \times M}{(3400 \times 10^3)^2} \]Rearrange to solve for \( M \):\[ M = \frac{0.38 \times 9.81 \times (3400 \times 10^3)^2}{6.674 \times 10^{-11}} \]
05

Calculate the Mass of Mars

Perform the calculations:\[ M = \frac{0.38 \times 9.81 \times 11.56 \times 10^{12}}{6.674 \times 10^{-11}} \]\[ M \approx 6.44 \times 10^{23} \text{ kg} \]

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Key Concepts

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

Gravitational Force
Gravitational force is an invisible force that pulls objects toward each other. It acts between any two objects that have mass. The strength of this force depends on two factors: the masses of the objects and the distance between them.

Key points about gravitational force:
  • The larger the masses, the stronger the gravitational force. This means, the more massive the objects, the greater the pull.
  • The closer the objects, the stronger the force. As distance increases, gravitational force decreases rapidly. This relationship is described by Newton's law of universal gravitation.
  • In planets, gravitational force keeps the atmosphere and holds the planets in their orbits around the star.
In planetary science, we often use the formula: \[ F = \frac{G M}{r^2} \] where:
  • \( F \) is the gravitational force exerted on objects on the planet's surface.
  • \( G \) is the gravitational constant, approximately \(6.674 \times 10^{-11} \, \mathrm{m^3 \, kg^{-1} \, s^{-2}}\).
  • \( M \) is the mass of the planet.
  • \( r \) is the radius of the planet.
Understanding gravitational force helps us determine how much a planet can pull objects towards it, thus affecting the weight and motion of objects on or near its surface.
Acceleration Due to Gravity
Acceleration due to gravity refers to how quickly an object speeds up as it falls freely, under the influence of gravity. This acceleration varies depending on which planet you are on, because it is directly related to the mass of the planet and its radius.

Important notes on acceleration due to gravity:
  • On Earth, this acceleration is approximately \(9.81 \, \text{m/s}^2\). It means every second, a freely falling object speeds up by about 9.81 meters per second.
  • On Mars, it's about 0.38 times that on Earth, which makes it around \(3.73 \, \text{m/s}^2\).
  • Different planets have different gravitational accelerations due to differences in their masses and sizes.
We use the concept of acceleration due to gravity when calculating a planet's gravitational pull. The formula for gravitational force can also be adapted to elaborate this acceleration. It helps us understand why we weigh differently on different planets. For Mars, its lower gravity results in a lower acceleration compared to Earth's, which makes things feel lighter.
Planetary Radius
Planetary radius is the distance from the center of a planet to its surface. It plays a crucial role in calculations related to planetary gravity and their effects on objects on the planet's surface.

Significance of planetary radius:
  • A larger radius means the surface is further from the planet's center, which can lead to weaker surface gravity, assuming mass is constant.
  • It impacts escape velocity, meaning, the further you are from center, the less energy you need to escape the planet's gravity.
  • In the formula \( F = \frac{G M}{r^2} \), radius \(r\) affects the force significantly; doubling the radius results in a force one quarter as strong.
For Mars, with a radius of \(3400 \, \text{km}\), this provides essential data for calculating the mass of Mars when considering gravitational force and allows for drawing comparisons with Earth and other celestial bodies. Understanding radius helps comprehend why planets with similar mass might have different gravitational pulls based on their size.

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Most popular questions from this chapter

A science-fiction tale describes an artificial "planet" in the form of a band completely encircling a sun (Fig. \(6-31\) ). The inhabitants live on the inside surface (where it is always noon). Imagine that this sun is exactly like our own, that the distance to the band is the same as the Earth-Sun distance (to make the climate temperate), and that the ring rotates quickly enough to produce an apparent gravity of \(g\) as on Earth. What will be the period of revolution, this planet's year, in Earth days?

Astronomers using the Hubble Space Telescope deduced the presence of an extremely massive core in the distant galaxy M87, so dense that it could be a black hole (from which no light escapes). They did this by measuring the speed of gas clouds orbiting the core to be \(780 \mathrm{~km} / \mathrm{s}\) at a distance of 60 light-years \(\left(5.7 \times 10^{17} \mathrm{~m}\right)\) from the core. Deduce the mass of the core, and compare it to the mass of our Sun.

(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?

(II) Four \(8.5-\mathrm{kg}\) spheres are located at the corners of a square of side \(0.80 \mathrm{~m}\). Calculate the magnitude and direction of the gravitational force exerted on one sphere by the other three.

The Navstar Global Positioning System (GPS) utilizes a group of 24 satellites orbiting the Earth. Using "triangulation" and signals transmitted by these satellites, the position of a receiver on the Earth can be determined to within an accuracy of a few centimeters. The satellite orbits are distributed evenly around the Earth, with four satellites in each of six orbits, allowing continuous navigational "fixes" The satellites orbit at an altitude of approximately \(11,000\) nautical miles \([1\) nautical mile \(=\) \(1.852 \mathrm{km}=6076 \mathrm{ft} \mathrm{t}\) . \((a)\) Determine the speed of each satellite. (b) Determine the period of each satellite.
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