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For a satellite to be in a circular orbit 890 km above the surface of the earth, (a) what orbital speed must it be given, and (b) what is the period of the orbit (in hours)?

Short Answer

Expert verified
(a) Orbital speed is approximately 7461 m/s; (b) Orbit period is around 1.71 hours.

Step by step solution

01

Identify Known Values

Firstly, we identify the known values: the altitude of the satellite orbit above the Earth's surface is 890 km. The Earth's radius is approximately 6371 km. Thus, the total orbital radius is \( r = 6371 + 890 = 7261 \text{ km} = 7,261,000 \text{ m} \). The gravitational constant \( G = 6.674 \times 10^{-11} \text{ m}^3\text{kg}^{-1}\text{s}^{-2} \), and the mass of the Earth \( M = 5.972 \times 10^{24} \text{ kg} \).
02

Calculate Orbital Speed

The orbital speed \( v \) can be calculated using the formula \( v = \sqrt{\frac{GM}{r}} \). Substituting the values, we get:\[v = \sqrt{\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{7,261,000}} \approx 7,461 \text{ m/s}\]
03

Calculate Orbital Period

The orbital period \( T \) can be found using the formula \( T = \frac{2\pi r}{v} \). Substituting the radius and the orbital speed:\[T = \frac{2\pi \times 7,261,000}{7,461} \approx 6,168 \text{ seconds}\]
04

Convert Period into Hours

To convert the period from seconds to hours, divide by 3600 (the number of seconds in an hour):\[T = \frac{6,168}{3600} \approx 1.71 \text{ hours}\]

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Key Concepts

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

Orbital Speed
Orbital speed is the velocity a satellite needs to maintain in order to stay in orbit around a celestial body like Earth. When a satellite is positioned, for example, 890 km above Earth's surface, its speed ensures that gravity's pull and the satellite's momentum are balanced. This balance lets it move along a curved path, effectively "falling" around Earth rather than towards it.
To find the orbital speed, we use the formula: \( v = \sqrt{\frac{GM}{r}} \). Here, \( G \) represents the gravitational constant, \( M \) is Earth's mass, and \( r \) is the total radius from Earth's center to the satellite.
Applying these values to our formula provides insights into the kind of speed, approximately 7461 m/s, the satellite requires, ensuring it doesn't drift away or crash down.
Orbital Period
The orbital period is the time it takes for a satellite to complete a full journey around Earth. Essentially, it's the satellite's "day". To ascertain this period, we employ the formula \( T = \frac{2\pi r}{v} \), where \( r \) is the circular orbit's radius and \( v \) is the orbital speed.
By calculating using the provided satellite altitude, we estimate the period to be about 6168 seconds. For clarity, we convert it to hours, resulting in around 1.71 hours.
Understanding the orbital period helps in knowing not just how fast the satellite orbits, but also in planning tasks like communication or imaging, where timing is key.
Gravitational Constant
The gravitational constant, symbolized as \( G \), is a fundamental natural force measuring the attraction between two masses. It plays a vital role in calculating gravitational forces, especially in scenarios involving planetary bodies and their satellites.
In calculations like those for satellite orbits, \( G \) is used along with Earth's mass to determine essential properties like speed and orbital period. The defined value is \( 6.674 \times 10^{-11} \, \text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2} \), which is crucial in formulating the balances of gravitational force that keep satellites in orbit.
Earth's Mass
Earth's mass is a fundamental element when computing satellite orbits. It provides a measure of the gravitational force exerted by Earth on nearby objects. Known to be approximately \( 5.972 \times 10^{24} \) kg, this mass is integral to understanding how Earth wields its gravitational strength to keep satellites from soaring into deep space or plummeting back to its surface.
When plugged into orbital equations, Earth's mass combines with the gravitational constant to determine how fast and how long satellites orbit, enabling us to predict satellite behavior and manage orbital mechanics effectively.

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Most popular questions from this chapter

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