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Chapter 5: Problem 49

An Earth Satellite An Earth satellite moves in a circular orbit \(640 \mathrm{~km}\) above Earth's surface with a period of \(98.0 \mathrm{~min}\). What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?

Short Answer

Expert verified
The speed of the satellite is approximately 7480 m/s, and the centripetal acceleration is approximately 7.98 m/s虏.

Step by step solution

01

- Determine the orbital radius

Add the Earth's radius (\r{e}=6371\text{k}\text{m}) to the altitude of the satellite (h=640\text{km}). The orbital radius (r) is the distance from the center of the Earth to the satellite.\[ r = r_e + h = 6371 \text{km} + 640 \text{km} = 7011 \text{km} = 7.011 \times 10^6 \text{m} \]
02

- Convert the orbital period to seconds

Convert the orbital period from minutes to seconds. Given the period (T) is 98.0 minutes:\[ T = 98.0 \text{ min} \times 60 \text{ sec/min} = 5880 \text{ sec}\]
03

- Calculate the orbital speed

The speed (v) of the satellite can be found using the formula for the circumference of a circle and the period. \[ v = \frac{2 \pi r}{T} = \frac{2 \pi (7.011 \times 10^6 \text{ m})}{5880 \text{ s}} 鈮� 7480 \text{ m/s} \]
04

- Calculate the centripetal acceleration

The centripetal acceleration (a) can be calculated using the formula: \[ a = \frac{v^2}{r} = \frac{(7480 \text{ m/s})^2}{7.011 \times 10^6 \text{ m}} 鈮� 7.98 \text{ m/s}^2 \]

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Key Concepts

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

Orbital Radius
The **orbital radius** is the distance from the center of the Earth to the satellite in its orbit. It can be found by adding the Earth's radius to the altitude of the satellite.
Given:
- Earth's radius \((r_e)\): 6371 km
- Satellite's altitude above Earth's surface \((h)\): 640 km
The formula to find the orbital radius \((r)\) is:
\[r = r_e + h\]
Using the given values:
\[r = 6371\text{ km} + 640\text{ km} = 7011\text{ km} = 7.011 \times 10^6 \text{ m}\]
This formula helps us determine the complete distance from Earth's center to the satellite in its orbit.
Orbital Period
The **orbital period** is the time it takes for the satellite to complete one full orbit around the Earth.
Given its period is 98.0 minutes, we need to convert this to seconds, as standard units in physics are important for consistent calculations.
- \(1\text{ min} = 60\text{ s}\)
Thus:
\[T = 98.0\text{ min} \times 60\text{ s/min} = 5880\text{ s}\]
This calculation converts the orbital period from minutes into seconds.
Centripetal Acceleration
The satellite is in circular motion around the Earth, which means it is constantly changing direction, requiring a continuous force directed towards the center of its circular path. The **centripetal acceleration** reflects how quickly the satellite changes direction.
It can be calculated using the formula:
\[a = \frac{v^2}{r}\]
Where **\(v\)** is the orbital speed, and **\(r\)** is the orbital radius.
From previous calculations:
- \(v = 7480\text{ m/s}\)
- \(r = 7.011 \times 10^6\text{ m}\)
Substitute these values into the formula:
\[a = \frac{(7480\text{ m/s})^2}{7.011\times 10^6\text{ m}} \approx 7.98\text{ m/s}^2\]
This acceleration keeps the satellite in its circular orbit.
Orbital Speed
The **orbital speed** is the constant speed at which the satellite travels along its path. It can be calculated by considering the distance it covers (the circumference of the orbit) and the time it takes (the orbital period).
Use the formula:
\[v = \frac{2\pi r}{T}\]
Where:
- **\(2\pi r\)** is the circumference of the orbit
- **\(T\)** is the orbital period
Given previously:
- \(r = 7.011 \times 10^6\text{ m}\)
- \(T = 5880\text{ s}\)
Substitute these into the formula:
\[v = \frac{2 \pi (7.011 \times 10^6 \text{ m})}{5880\text{ s}} \approx 7480\text{ m/s}\]
This calculation shows the speed at which the satellite travels to maintain its orbit.

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Most popular questions from this chapter

. Stairway A ball rolls horizontally off the top of a stairway with a speed of \(1.52 \mathrm{~m} / \mathrm{s}\). The steps are \(20.3 \mathrm{~cm}\) high and \(20.3 \mathrm{~cm}\) wide. Which step does the ball hit first?

A particle moves horizontally in uniform circular motion, over a horizontal \(x y\) plane. At one instant, it moves through the point at coordinates \((4.00 \mathrm{~m}, 4.00 \mathrm{~m})\) with a velocity of \((-5.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}\) and an acceleration of \(\left(12.5 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). What are the coordinates of the center of the circular path?

The fast French train known as the TGV (Train 脿 Grande Vitesse) has a scheduled average speed of \(216 \mathrm{~km} / \mathrm{h} .\) (a) If the train goes around a curve at that speed and the magnitude of the acceleration experienced by the passengers is to be limited to \(0.050 \mathrm{~g}\), what is the smallest radius of curvature for the track that can be tolerated? (b) At what speed must the train go around a curve with a \(1.00 \mathrm{~km}\) radius to be at the acceleration limit?

A Small Ball A small ball rolls horizontally off the edge of a tabletop that is \(1.20 \mathrm{~m}\) high. It strikes the floor at a point \(1.52 \mathrm{~m}\) horizontally away from the edge of the table. (a) How long is the ball in the air? (b) What is its speed at the instant it leaves the table?

Dart A dart is thrown horizontally with an initial speed of \(10 \mathrm{~m} / \mathrm{s}\) toward point \(P\), the bull's-eye on a dart board. It hits at point \(Q\) on the rim, vertically below \(P 0.19\) s later. (a) What is the distance \(P Q ?\) (b) How far away from the dart board is the dart released?
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