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Chapter 13: Problem 45

The Martian satellite Phobos travels in an approximately circular orbit of radius \(9.4 \times 10^{6} \mathrm{~m}\) with a period of \(7 \mathrm{~h} 39 \mathrm{~min}\). Calculate the mass of Mars from this information.

Short Answer

Expert verified
The mass of Mars is approximately \(6.41 \times 10^{23}\) kg.

Step by step solution

01

Understand the Relationship

The formula we will use to calculate the mass of Mars is derived from Newton's law of gravitation and centripetal force: \[ F = rac{G M m}{r^2} = rac{m v^2}{r} \]where \( G \) is the gravitational constant \( 6.674 imes 10^{-11} \ ext{Nm}^2/ ext{kg}^2 \), \( M \) is the mass of Mars, \( m \) is the mass of Phobos, \( r \) is the orbit radius, and \( v \) is the orbital velocity of Phobos.
02

Use Orbital Velocity Formula

We can relate the orbital speed \( v \) to the orbital period \( T \) and radius \( r \) using the formula: \[ v = \frac{2 \pi r}{T} \]where \( T \) is the period of orbit in seconds. First, convert 7 hours and 39 minutes into seconds: \[ T = 7 \times 3600 + 39 \times 60 = 27540 \text{ seconds} \]
03

Calculate Orbital Velocity

Using the period \( T = 27540 \text{ s} \) and radius \( r = 9.4 \times 10^6 \text{ m} \), we can calculate the velocity \( v \) of Phobos as follows: \[ v = \frac{2 \pi \times 9.4 \times 10^6}{27540} \approx 2144.7 \text{ m/s} \]
04

Relate Forces to Find Mars' Mass

Now that we have \( v \), equate centripetal force to gravitational force to solve for the mass of Mars \( M \):\[ \frac{G M}{r^2} = \frac{v^2}{r} \]Rearranging the equation gives:\[ M = \frac{v^2 r}{G} \]
05

Calculate the Mass of Mars

Substitute the known values \( v = 2144.7 \text{ m/s}, r = 9.4 \times 10^6 \text{ m}, \) and \( G = 6.674 \times 10^{-11} \ ext{Nm}^2/ ext{kg}^2 \) into the equation:\[ M \approx \frac{(2144.7)^2 \times 9.4 \times 10^6}{6.674 \times 10^{-11}} \approx 6.41 \times 10^{23} \text{ kg} \]
06

Verify the Units and Answer

Ensure that all quantities are used in the correct units to get mass in kilograms. The calculated mass appears reasonable compared to known values from literature, confirming our approach was consistent throughout.

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Key Concepts

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

Newton's Law of Gravitation
Understanding gravitational attraction is key to orbital mechanics. Newton's Law of Gravitation tells us how two masses attract each other based on their mass and the distance between them. The formula states that the gravitational force (\( F \)) felt by a mass is:
\[ F = \frac{G M_1 M_2}{r^2} \]
  • \( G \) is the universal gravitational constant, \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \).
  • \( M_1 \) and \( M_2 \) represent the two masses involved鈥攊n this case, Mars and its moon Phobos.
  • \( r \) stands for the distance between the centers of the two masses.
Newton's law allows us to understand the gravitational pull between celestial bodies. It underpins many calculations in orbital mechanics.
Centripetal Force
Centripetal force is what keeps a satellite in orbit, driving it around a larger body in a circular path. For Phobos, this force keeps it moving around Mars. We define centripetal force (\( F_c \)) as:
\[ F_c = \frac{m v^2}{r} \]
  • \( m \) is the mass of the body experiencing the force; in this case, the mass of Phobos (although it cancels out later in the calculations).
  • \( v \) is the velocity of the body as it orbits.
  • \( r \) is the radius of the circular path.
For the stationary orbit, centripetal force arises from gravitational interaction. This means gravitational force equals centripetal force. Thus, centripetal force can be crucial for calculating unknown variables in orbits.
Orbital Velocity
As a satellite moves in orbit, its speed is known as orbital velocity. This speed, \( v \), depends on the orbital period and the radius characterizing the orbit. Therefore, the formula to derive this velocity has importance:
\[ v = \frac{2 \pi r}{T} \]
  • \( r \) is the radius of the orbit.
  • \( T \) is the orbital period, made uniform in seconds for calculations.
Understanding orbital velocity helps us determine how fast a satellite like Phobos moves around its planet, ensuring it stays in a stable orbit. By calculating orbital velocity, we pave the way for other crucial computations, such as the mass of the central body.
Mass Calculation
When we want to find the mass of a planet from satellite data, we utilize the relationship between gravitational and centripetal forces. This is done by rearranging the formula: \( \frac{G M}{r^2} = \frac{v^2}{r} \).
So to find the mass (\( M \)) of the planet, the rearranged equation becomes:
\[ M = \frac{v^2 r}{G} \]
  • \( v \) is the orbital velocity, which we've already calculated.
  • \( r \) is the radius of the orbit.
  • \( G \) is the gravitational constant.
This calculation encapsulates how we derive the mass of Mars using Phobos' orbital mechanics. Validating this process ensures that our result is reasonable compared to recognized mass measurements.

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Most popular questions from this chapter

Several planets (Jupiter, Saturn, Uranus) are encircled by rings, perhaps composed of material that failed to form a satellite. In addition, many galaxies contain ring-like structures. Consider a homogeneous thin ring of mass \(M\) and outer radius \(R\) (Fig. 13-52). (a) What gravitational attraction does it exert on a particle of mass \(m\) located on the ring's central axis a distance \(x\) from the ring center? (b) Suppose the particle falls from rest as a result of the attraction of the ring of matter. What is the speed with which it passes through the center of the ring?

In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constructed. Ignoring Earth's rotation, find the change in your weight if you were to ride an elevator from the street level, where you weigh \(600 \mathrm{~N},\) to the top of the building.

Zero, a hypothetical planet, has a mass of \(5.0 \times 10^{23} \mathrm{~kg},\) a radius of \(3.0 \times 10^{6} \mathrm{~m}\) and no atmosphere. A \(10 \mathrm{~kg}\) space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial energy of \(5.0 \times 10^{7} \mathrm{~J},\) what will be its kinetic energy when it is \(4.0 \times 10^{6} \mathrm{~m}\) from the center of Zero? (b) If the probe is to achieve a maximum distance of \(8.0 \times 10^{6} \mathrm{~m}\) from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?

(a) What is the escape speed on a spherical asteroid whose radius is \(500 \mathrm{~km}\) and whose gravitational acceleration at the surface is \(3.0 \mathrm{~m} / \mathrm{s}^{2} ?\) (b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of \(1000 \mathrm{~m} / \mathrm{s} ?\) (c) With what speed will an object hit the asteroid if it is dropped from \(1000 \mathrm{~km}\) above the surface?B

An asteroid, whose mass is \(2.0 \times 10^{-4}\) times the mass of Earth, revolves in a circular orbit around the Sun at a distance that is twice Earth's distance from the Sun. (a) Calculate the period of revolution of the asteroid in years. (b) What is the ratio of the kinetic energy of the asteroid to the kinetic energy of Earth?
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