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An Earth satellite moves in a circular orbit \(640 \mathrm{~km}\) (uniform circular motion) 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 about 7,491 m/s, and its centripetal acceleration is about 8.00 m/s².

Step by step solution

01

Determine Earth's Radius

The average radius of the Earth is approximately \( R_{E} = 6371 \text{ km} \).
02

Calculate Orbit Radius

The orbit radius is the sum of Earth's radius and the altitude of the satellite above Earth. Thus, \( R = 640 \text{ km} + 6371 \text{ km} = 7011 \text{ km} \).
03

Convert Kilometers to Meters

To be consistent with unit SI systems, convert the orbit radius from kilometers to meters. \( R = 7011 \times 1000 = 7,011,000 \text{ meters} \).
04

Calculate Orbit Circumference

The circumference of the satellite's orbit can be found using the formula \( C = 2\pi R \). Substituting in the radius gives \( C = 2\pi \times 7,011,000 \approx 44,022,304 \text{ meters} \).
05

Convert Period to Seconds

Convert the period from minutes to seconds for use in subsequent calculations. Thus, \( T = 98 \text{ minutes} \times 60 \text{s/min} = 5,880 \text{ seconds} \).
06

Calculate Satellite Speed

Speed is distance over time, so the speed of the satellite can be calculated by the formula \( v = \frac{C}{T} \). Substituting in values gives \( v = \frac{44,022,304}{5,880} \approx 7,491 \text{ m/s} \).
07

Calculate Centripetal Acceleration

Using the formula for centripetal acceleration \( a_c = \frac{v^2}{R} \), we substitute in the values found: \( a_c = \frac{(7,491)^2}{7,011,000} \approx 8.00 \text{ m/s}^2 \).

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Key Concepts

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

Centripetal Acceleration
Centripetal acceleration is an essential concept in circular motion. When an object moves in a circle, it constantly changes direction, even if it maintains a constant speed. This change in direction is due to an inward acceleration called centripetal acceleration.
It is calculated using the formula:
  • \( a_c = \frac{v^2}{R} \)
where \( v \) is the velocity of the object and \( R \) is the radius of the circle.
It points towards the center of the circle, providing the necessary force to keep the object in motion along the circular path.
When dealing with satellites in orbit, understanding centripetal acceleration helps us determine how gravity provides the required force to keep satellites in a stable orbit around the Earth.
Satellite Orbit
Satellites move in paths or trajectories around planets, known as orbits. These paths are commonly circular or elliptical.
A satellite in a circular orbit experiences consistent velocity and distance from the Earth at all points along its path.
The altitude of the orbit directly affects the gravitational force and the required velocity to maintain that orbit.

To calculate the orbit of a satellite, the radius, total circumference, and period are key factors:
  • The radius, \( R \), is the sum of the Earth's radius and the satellite's altitude above Earth's surface.
  • The circumference, \( C \), is calculated using \( C = 2\pi R \), determining how much distance the satellite covers per orbit.
  • The period represents the time taken for one complete orbit, crucial for understanding satellite speed and stability.
Uniform Motion
Uniform circular motion is a type of motion where an object moves around a circle with a constant speed.
Even though the speed remains the same, the object accelerates because its direction changes, resulting in a constantly changing velocity vector.
For uniform circular motion, both the distance traveled and the timing are consistent, allowing us to calculate crucial properties like speed and acceleration.
  • Speed \( (v) \) can be calculated by dividing the circumference by the orbital period: \( v = \frac{C}{T} \).
  • The centripetal force, essential for maintaining uniform motion, equates to the gravitational pull in satellite orbits, maintaining their respective paths.
Earth Radius
The Earth's radius is a fundamental parameter when calculating satellite motion. It defines the baseline from which we determine the orbital radius of any satellite.
Known to be approximately 6,371 km, the Earth's radius is added to the satellite's altitude to yield the distance from the Earth's center to the satellite in orbit.
The importance of this measurement comes into play when calculating gravitational forces and the centripetal acceleration needed for satellite stability.
Considering the Earth’s own gravitational influence, knowing its size helps in understanding how various altitudes affect the speed needed for orbits.
Every elevation change impacts the dynamics and energy requirements of satellites maintaining a stable orbit.

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Most popular questions from this chapter

A moderate wind accelerates a pebble over a horizontal \(x y\) plane with a constant acceleration \(\vec{a}=\left(5.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(7.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\) . At time \(t=0,\) the velocity is \((4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}} .\) What are the (a) magnitude and (b) angle of its velocity when it has been displaced by \(12.0 \mathrm{~m}\) parallel to the \(x\) axis?

The current world-record motorcycle jump is \(77.0 \mathrm{~m}\), set by Jason Renie. Assume that he left the take-off ramp at \(12.0^{\circ}\) to the horizontal and that the take-off and landing heights are the same. Neglecting air drag, determine his take-off speed.

A graphing surprise. At time \(t=0,\) a burrito is launched from level ground, with an initial speed of \(16.0 \mathrm{~m} / \mathrm{s}\) and launch angle \(\theta_{0}\). Imagine a position vector \(\vec{r}\) continuously directed from the launching point to the burrito during the flight. Graph the magnitude \(r\) of the position vector for (a) \(\theta_{0}=40.0^{\circ}\) and (b) \(\theta_{0}=80.0^{\circ} .\) For \(\theta_{0}=40.0^{\circ},\) (c) when does \(r\) reach its maximum value, (d) what is that value, and how far (e) horizontally and (f) vertically is the burrito from the launch point? For \(\theta_{0}=80.0^{\circ},(\mathrm{g})\) when does \(r\) reach its maximum value, (h) what is that value, and how far (i) horizontally and (j) vertically is the burrito from the launch point?

A centripetal-acceleration addict rides in uniform circular motion with radius \(r=3.00 \mathrm{~m}\). At one instant his acceleration is \(\vec{a}=\left(6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(-4.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}} .\) At that instant, what are the values of (a) \(\vec{v} \cdot \vec{a}\) and (b) \(\vec{r} \times \vec{a} ?\)

A car travels around a flat circle on the ground, at a constant speed of \(12.0 \mathrm{~m} / \mathrm{s}\). At a certain instant the car has an acceleration of \(3.00 \mathrm{~m} / \mathrm{s}^{2}\) toward the east. What are its distance and direction from the center of the circle at that instant if it is traveling (a) clockwise around the circle and (b) counterclockwise around the circle?

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