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

Short Answer

Expert verified
The height of the International Space Station above the Earth's surface is approximately 408 km.

Step by step solution

01

Calculate the Orbital Period

First, we need to find the time it takes for one revolution (orbital period) around the Earth. The space station makes 15.65 revolutions per day, so the orbital period can be found using the formula: \( T = \frac{24 \text{ hours}}{15.65} \). Thus, \( T \approx 1.534 \text{ hours} \).
02

Convert Orbital Period to Seconds

Convert the period from hours to seconds since SI units facilitiate subsequent calculations. Since 1 hour is 3600 seconds, we use: \( T_{s} = 1.534 \times 3600 \approx 5522.4 \text{ seconds} \).
03

Using Newton's Law of Gravitation

Using the centripetal force which keeps the satellite in orbit set equal to the gravitational force experienced by the satellite, we use the formula: \( G\frac{M_E m}{r^2} = m\frac{v^2}{r} \). Where \( G \) is the gravitational constant, \( M_E \) is Earth's mass, \( m \) is the mass of the satellite, \( r \) is the orbital radius, and \( v \) is the orbital speed.
04

Orbital Velocity Equation

Simplify and solve for orbital velocity \( v = \frac{2\pi r}{T_s} \) and substitute into our equation from Step 3 to find the orbital radius. This results in: \( r^3 = \frac{GMT_s^2}{4\pi^2} \).
05

Calculate Orbital Radius

Plug in the known constants: \( G = 6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \), \( M_E = 5.972 \times 10^{24} \text{ kg} \), and our previously calculated \( T_s \) into the orbital radius equation. Solve for \( r \).
06

Determine Height Above Earth's Surface

The orbital radius \( r \) we found includes Earth's radius. Calculate the satellite's height \( h \) above Earth's surface: \( h = r - R_E \), where \( R_E \) is Earth's radius, approximately \( 6371 \text{ km} \).

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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 shining example of human ingenuity and international collaboration. Orbiting the Earth, the ISS serves as a massive laboratory in space, allowing scientists from various countries to conduct research that is not possible on Earth. Understanding its orbit is crucial for its operation and the safety of its crew.

The ISS circles the Earth at a high speed, completing approximately 15.65 orbits each day. This frequent orbit allows astronauts and scientists to observe and study Earth's surface from space at regular intervals. Such a rapid orbit also means the ISS is an excellent platform for various scientific experiments and observations.

Students often find it challenging to grasp how such a large structure can remain in orbit consistently. It's fascinating to learn how principles of physics and complex mathematical calculations come together to make this possible.
Orbital Period
The orbital period is a critical concept in understanding how satellites like the ISS stay in orbit. It refers to the time it takes for a satellite to make one complete revolution around the Earth. For example, the International Space Station completes one orbit approximately every 90 minutes.

To calculate the orbital period, we divide the total time by the number of orbits. In the case of the ISS, it makes 15.65 revolutions per day. Thus, by dividing 24 hours by 15.65, we get around 1.534 hours per orbit. This period is not only intriguing but also crucial for scheduling tasks and experiments aboard the ISS.

Converting the period into seconds (SI units) is an important step in making further calculations. It enables precise calculations essential for maintaining the correct orbital parameters and ensuring consistent communication with ground stations.
Newton's Law of Gravitation
Newton's Law of Gravitation is fundamental in understanding orbital mechanics. It helps us determine how the gravitational pull between two masses dictates motion. This is crucial for maintaining the ISS in its orbit around Earth.

The law states that every point mass attracts every other point mass in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers:
  • Formula: \( F = G \frac{M_1 M_2}{r^2} \)
The ISS remains in its stable orbit due to the balance between gravitational forces pulling it towards Earth and its tangential velocity propelling it forward. This delicate balance requires precise calculations and adjustments to maintain.

Using these principles, scientists can predict how satellites orbit, help maintain their paths, and make necessary adjustments to prevent them from deorbiting prematurely.
Centripetal Force
Centripetal force plays a vital role in keeping the ISS in orbit. While gravitational force pulls the ISS towards the Earth, the centripetal force ensures it remains on its circular path.

The formula for centripetal force is given by:
  • \( F_c = m \frac{v^2}{r} \)
where \( m \) is the mass of the ISS, \( v \) is its velocity, and \( r \) is the radius of its orbit from the Earth's center.

This force is necessary for any object traveling in a circular path and prevents the ISS from flying straight off into space. It ensures that the velocity vector is constantly changing direction, thus maintaining a steady orbit.

Understanding centripetal force helps explain why the ISS and other satellites do not need constant propulsion but can stay in orbit with their initial launch velocity and the gravitational pull from Earth.

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Most popular questions from this chapter

An object in the shape of a thin ring has radius \(a\) and mass \(M .\) A uniform sphere with mass \(m\) and radius \(R\) is placed with its center at a distance \(x\) to the right of the center of the ring, along a line through the center of the ring, and perpendicular to its plane (Fig. 12.35 ). What is the gravitational force that the sphere exerts on the ring- shaped object? Show that your result reduces to the expected result when \(x\) is much larger than \(a\) .

What is the escape speed from a \(300-\mathrm{km}\) -diameter asteroid with a density of 2500 \(\mathrm{kg} / \mathrm{m}^{3}\) ?

A uniform, spherical, \(1000.0-\mathrm{kg}\) shell has a radius of 5.00 \(\mathrm{m}\) . (a) Find the gravitational force this shell exerts on a \(2.00-\mathrm{kg}\) point mass placed at the following distances from the center of the shell: (i) \(5.01 \mathrm{m},(\mathrm{ii}) 4.99 \mathrm{m},\) (iii) 2.72 \(\mathrm{m} .\) (b) Sketch a qualitative graph of the magnitude of the gravitational force this sphere exerts on a point mass \(m\) as a function of the distance \(r\) of \(m\) from the center of the sphere. Include the region from \(r=0\) to \(r \rightarrow \infty\) .

In March 2006 , two small satellites were discovered orbiting Pluto, one at a distance of \(48,000 \mathrm{km}\) and the other at \(64,000 \mathrm{km}\) . Pluto already was known to have a large satellite Charon, orbiting at \(19,600 \mathrm{km}\) with an orbital period of 6.39 days. Assuming that the satellites do not affect each other, find the orbital periods of the two small satellites without using the mass of Pluto.

Planet Vulcan. Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to \(\frac{2}{3}\) of the average orbit radius of Mercury. What would be the orbital period of such a planet? (Such a planet was once postulated, in part to explain the precession of Mercury's orbit. It was even given the name Vulcan, although we now have no evidence that it actually exists. Mercury's precession has been explained by general relativity.)

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