/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 In its orbit each day, the Inter... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In its orbit each day, the International Space Station makes 15.65 revolutions around the earth. Assuming a circular orbit, how high is this satellite above the surface of the earth?

Short Answer

Expert verified
The International Space Station orbits approximately 400 km above Earth's surface.

Step by step solution

01

Determine the orbital period

First, we calculate the orbital period of the International Space Station (ISS). Given that ISS makes 15.65 revolutions per day, the orbital period \( T \) in seconds can be calculated as: \[ T = \frac{24 \text{ hours} \times 3600 \text{ seconds/hour}}{15.65} \approx 5508.47 \text{ seconds} \]
02

Use Kepler's third law for circular orbits

Next, we apply Kepler's third law for circular orbits. This law relates the orbital period \( T \), the radius of the orbit \( r \), and the mass of Earth \( M \). The formula is \[ T^2 = \frac{4\pi^2}{GM} r^3 \] where \( G = 6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \) is the gravitational constant, and \( M = 5.972 \times 10^{24} \text{ kg} \) is the mass of Earth. Rearrange to solve for \( r \): \[ r = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3} \] Substituting in the known values and \( T = 5508.47 \text{ seconds} \), \[ r \approx \left(\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24} \times 5508.47^2}{4\pi^2}\right)^{1/3} \]
03

Calculate the radius of the orbit

Compute \( r \) from the rearranged orbital formula. On solving, we find \[ r \approx 6771 \text{ km} \] This is the distance from the center of the Earth to the ISS.
04

Calculate the altitude above Earth's surface

Finally, subtract Earth's radius from the orbital radius to find the altitude above the surface. Earth's average radius is approximately 6371 km. Thus, the altitude \( h \) is: \[ h = r - 6371 \approx 6771 - 6371 = 400 \text{ km} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

International Space Station
The International Space Station (ISS) is a marvel of human engineering orbiting Earth approximately every 90 minutes. It serves as a microgravity and space environment research laboratory where scientific research is conducted in astrobiology, astronomy, meteorology, and physics. The ISS is a joint project between five participating space agencies: NASA, Roscosmos, JAXA, ESA, and CSA.
The space station travels in a low Earth orbit, which means it is much closer to Earth compared to other satellites used for GPS or weather monitoring. Its altitude is roughly 400 kilometers above Earth's surface, allowing it to make 15.65 revolutions around the Earth each day.
These frequent orbits provide opportunities for various experiments and observations that contribute significantly to our understanding of space and Earth from a unique vantage point.
Kepler's Third Law
Kepler's third law of planetary motion reveals the relationship between the orbital period of a planet and its distance from the sun. Although initially applied to planets, this law can also forecast the motion of satellites, like the International Space Station, orbiting Earth.
The law is expressed using the formula: \[ T^2 = \frac{4\pi^2}{GM} r^3 \]Where:
  • \( T \) is the orbital period in seconds.
  • \( G \) is the gravitational constant.
  • \( M \) is the mass of the central body (in this case, Earth).
  • \( r \) is the radius of the orbit.
Kepler's third law allows scientists to deduce the height at which a satellite like the ISS orbits by determining its orbital period. The law emphasizes that the square of the orbital period is directly proportional to the cube of the semi-major axis of its orbit.
By using Kepler's third law, we can establish not only the orbital dynamics but also get a deeper understanding of gravitational effects on the ISS.
Gravitational Constant
The gravitational constant, represented by \( G \), is a fundamental constant that plays a pivotal role in our understanding of gravity. It relates gravitational force to the masses of two objects and the distance between them.
The value of \( G \) is approximately \( 6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \). It is used in Newton's law of universal gravitation, which describes the gravitational attraction between two masses. In the context of orbital mechanics, \( G \) helps determine the gravitational force acting on satellites orbiting a planet.
By integrating the gravitational constant into calculations, such as those in Kepler's laws, scientists can predict the behavior of satellites accurately. Specifically for the ISS, \( G \) assists in calculating both the required speed and altitude necessary for maintaining a stable orbit, ensuring that it can conduct its valuable research without drifting away from its intended path.
Orbital Period
The orbital period of an object, like the ISS, is the time it takes to complete one full revolution around another object, in this case, Earth. For the ISS, this period is approximately 90 minutes, allowing for about 15.65 revolutions each day.
Understanding the orbital period is vital for mission planning and ensuring the International Space Station remains functional and correctly positioned for tasks like monitoring weather systems or relaying communications.
During its orbit, the ISS experiences both day and night multiple times a day, which affects factors like solar power generation and onboard research conditions. To maintain such an orbit, precise calculations using orbital period, derived via Kepler's third law, allow for adjustments and ensure that the ISS stays on track. The smooth operation is achieved by accounting for gravitational interactions, Earth's rotation, and other orbital dynamics continually monitored from Earth.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030-kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer. (a) What is the tension between her ears? Would the astronaut find it difficult to keep from being torn apart by the gravitational forces? (Since her whole body orbits with the same angular velocity, one ear is moving too slowly for the radius of its orbit and the other is moving too fast. Hence her head must exert forces on her ears to keep them in their orbits.) (b) Is the center of gravity of her head at the same point as the center of mass? Explain.

A satellite with mass 848 kg is in a circular orbit with an orbital speed of 9640 m/s around the earth. What is the new orbital speed after friction from the earth's upper atmosphere has done \(-\)7.50 \(\times\) 10\(^9\) J of work on the satellite? Does the speed increase or decrease?

Planet X rotates in the same manner as the earth, around an axis through its north and south poles, and is perfectly spherical. An astronaut who weighs 943.0 N on the earth weighs 915.0 N at the north pole of Planet X and only 850.0 N at its equator. The distance from the north pole to the equator is 18,850 km, measured along the surface of Planet X. (a) How long is the day on Planet X? (b) If a 45,000-kg satellite is placed in a circular orbit 2000 km above the surface of Planet X, what will be its orbital period?

At a certain instant, the earth, the moon, and a stationary 1250-kg spacecraft lie at the vertices of an equilateral triangle whose sides are 3.84 \(\times\) 10\(^5\) km in length. (a) Find the magnitude and direction of the net gravitational force exerted on the spacecraft by the earth and moon. State the direction as an angle measured from a line connecting the earth and the spacecraft. In a sketch, show the earth, the moon, the spacecraft, and the force vector. (b) What is the minimum amount of work that you would have to do to move the spacecraft to a point far from the earth and moon? Ignore any gravitational effects due to the other planets or the sun.

The planet Uranus has a radius of 25,360 km and a surface acceleration due to gravity of 9.0 m/s\(^2\) at its poles. Its moon Miranda (discovered by Kuiper in 1948) is in a circular orbit about Uranus at an altitude of 104,000 km above the planet's surface. Miranda has a mass of 6.6 \(\times\) 10\(^{19}\) kg and a radius of 236 km. (a) Calculate the mass of Uranus from the given data. (b) Calculate the magnitude of Miranda's acceleration due to its orbital motion about Uranus. (c) Calculate the acceleration due to Miranda's gravity at the surface of Miranda. (d) Do the answers to parts (b) and (c) mean that an object released 1 m above Miranda's surface on the side toward Uranus will fall \(up\) relative to Miranda? Explain.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.